SQL提高及优化

一基础单表查询
1.1查询表结构
desc 表名
sql> desc emp
Name Null? Type

---

EMPNO NOT NULL NUMBER(4)
ENAME VARCHAR2(10)
JOB VARCHAR2(9)
MGR NUMBER(4)
HIREDATE DATE
SAL NUMBER(7,2)
COMM NUMBER(7,2)
DEPTNO NUMBER(2)
1.2查找空值
使用 is null
sql> select empno from emp where comm is null;

 EMPNO

---

  7369
  7566
  7698
  7782
  7839
  7900
  7902
  7934

8 rows selected.
1.3 将空值转换成实际值，推荐使用coalesce
sql> select empno,nvl(comm,0) from emp where comm is null;

 EMPNO NVL(COMM,0)

---

  7369           0
  7566           0
  7698           0
  7782           0
  7839           0
  7900           0
  7902           0
  7934           0

8 rows selected.
sql> select empno,nvl2(comm,comm,0) from emp where comm is null;

 EMPNO NVL2(COMM,COMM,0)

---

  7369                 0
  7566                 0
  7698                 0
  7782                 0
  7839                 0
  7900                 0
  7902                 0
  7934                 0

8 rows selected.
sql> select empno,nullif(0,comm) from emp where comm is null;

 EMPNO NULLIF(0,COMM)

---

  7369              0
  7566              0
  7698              0
  7782              0
  7839              0
  7900              0
  7902              0
  7934              0

8 rows selected.

sql> select empno,coalesce(comm,0) from emp where comm is null;

 EMPNO COALESCE(COMM,0)

---

  7369                0
  7566                0
  7698                0
  7782                0
  7839                0
  7900                0
  7902                0
  7934                0

8 rows selected.
NVL(expr1,expr2)
如果expr1和expr2的数据类型一致，则：
如果expr1为空(null),那么显示expr2，
如果expr1的值不为空，则显示expr1。
NVL2(expr1,expr2, expr3)
如果expr1不为NULL，返回expr2； expr1为NULL，返回expr3。
expr2和expr3类型不同的话，expr3会转换为expr2的类型，转换不了，则报错。
NULLIF(expr1,expr2)
如果expr1和expr2相等则返回空(NULL)，否则返回expr1。
coalesce(expr1, expr2, expr3….. exprn)
返回表达式中第一个非空表达式，如果都为空则返回空值。
所有表达式必须是相同类型，或者可以隐式转换为相同的类型，否则报错。
Coalese函数和NVL函数功能类似，只不过选项更多。
1.4 在SELECT语句中使用条件逻辑
sql> select empno,
2 ename,
3 sal,
4 case
5 when sal<=2000 then '过低'
6 when sal>=4000 then '过高'
7 else 'OK'
8 end as status
9 from emp
10 where deptno=10;

 EMPNO ENAME             SAL STATUS

---

  7782 CLARK            2450 OK
  7839 KING             5000 过高
  7934 MILLER           1300 过低

1.5限制返回行数
sql> select empno from emp where rownum<=2;

 EMPNO

---

  7369
  7499

1.6从表中随机返回n条记录
sql> select empno,ename from (select empno,ename from emp order by dbms_random.value()) where rownum<=3;

 EMPNO ENAME

---

  7839 KING
  7521 WARD
  7566 JOnes

sql> select empno,ename from (select empno,ename from emp order by dbms_random.value()) where rownum<=3;

 EMPNO ENAME

---

  7499 ALLEN
  7698 BLAKE
  7654 MARTIN

1.7 TRANSLATE替换
2.sql> select TRANSLATE('ab 你好 abcdef','abcdef','123456') as newstring from dual;

NEWSTRING

12 你好 123456

sql> select TRANSLATE('ab 你好 abcdef','abcdef','1234') as newstring from dual;

NEWSTRING

12 你好 1234

sql> select TRANSLATE('ab 你好 abcdef','acdef','1234') as newstring from dual;

NEWSTRING

1b 你好 1b234
sql> select TRANSLATE('ab 你好 abcdef','acdef','') as newstring from dual;

N

替换值为空，返回空
sql> select TRANSLATE('ab 你好 abcdef','1abcdef','1') as newstring from dual;

NEWSTRING

你好
替换wei位置没有字符则删除
1.8 混合字符串按字母排序
sql> set line 100
sql> col TRANSLATE(EMPNO||''||ENAME,'-1234567890','-') format A40
sql> select empno||' '||ename as data,translate(empno||' '||ename,'- 1234567890','-') from emp e order by 2 ;

DATA TRANSLATE(EMPNO		''		ENAME,'-1234567890'

7499 ALLEN ALLEN
7698 BLAKE BLAKE
7782 CLARK CLARK
7902 FORD FORD
7900 JAMES JAMES
7566 JOnes JOnes
7839 KING KING
7654 MARTIN MARTIN
7934 MILLER MILLER
7369 SMITH SMITH
7844 TURNER TURNER

DATA TRANSLATE(EMPNO		''		ENAME,'-1234567890'

7521 WARD WARD

12 rows selected.
sql> select empno||' '||ename as data from emp e order by translate(empno||' '||ename,'- 1234567890','-') ;

DATA

7499 ALLEN
7698 BLAKE
7782 CLARK
7902 FORD
7900 JAMES
7566 JOnes
7839 KING
7654 MARTIN
7934 MILLER
7369 SMITH
7844 TURNER

DATA

7521 WARD

12 rows selected.
1.9 NULL排序使用NULLS FirsT/LAST
1.10按条件区不同列中值来排序
sql> select empno,
2 ename,
3 sal
4 from emp
5 where deptno=30
6 order by Case
7 when sal>=1000 and sal <2000 then
8 empno
9 else ename
10 end,
11 sal;
ename,
*
ERROR at line 2:
ORA-00932: inconsistent datatypes: expected NUMBER got CHAR

sql> select empno,
2 ename,
3 sal
4 from emp
5 where deptno=30
6 order by Case
7 when sal>=1000 and sal <2000 then 1
8 else 2
9 end, 3;

 EMPNO ENAME             SAL

---

  7654 MARTIN           1250
  7521 WARD             1250
  7844 TURNER           1500
  7499 ALLEN            1600
  7900 JAMES             950
  7698 BLAKE            2850

6 rows selected.
二 多表操作
2.1 union all与空字符串
sql> select 'a' as c1 from dual
2 union all
3 select '' as c1 from dual;

C

a

2.2 union与or
sql> select empno,ename from emp where empno=7782 or ename='WARD';

 EMPNO ENAME

---

  7521 WARD
  7782 CLARK

sql> select empno,ename from emp where empno=7782
2 union
3 select empno,ename from emp where ename='WARD';

 EMPNO ENAME

---

  7521 WARD
  7782 CLARK

sql> alter session set"_b_tree_bitmap_plans"=false;

Session altered.

sql> explain plan for select empno,ename from emp where empno=7782 or ename='WARD';

Explained.

sql> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT

Plan hash value: 3956160932

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Time |

| 0 | SELECT STATEMENT | | 2 | 20 | 3 (0)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| EMP | 2 | 20 | 3 (0)| 00:00:01 |

Predicate information (identified by operation id):

PLAN_TABLE_OUTPUT

1 - filter("EMPNO"=7782 OR "ENAME"='WARD')

13 rows selected.
sql> explain plan for select empno,ename from emp where empno=7782
2 union
3 select empno,ename from emp where ename='WARD';

Explained.

sql> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT

Plan hash value: 1027572458

---

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Ti
me |

---

---

PLAN_TABLE_OUTPUT

| 0 | SELECT STATEMENT | | 2 | 20 | 6 (34)| 00
:00:01 |

| 1 | SORT UNIQUE | | 2 | 20 | 6 (34)| 00
:00:01 |

2	UNION-ALL

| 3 | TABLE ACCESS BY INDEX ROWID| EMP | 1 | 10 | 1 (0)| 00
:00:01 |

PLAN_TABLE_OUTPUT

|* 4 | INDEX UNIQUE SCAN | PK_EMP | 1 | | 0 (0)| 00
:00:01 |

|* 5 | TABLE ACCESS FULL | EMP | 1 | 10 | 3 (0)| 00
:00:01 |

---

---

PLAN_TABLE_OUTPUT

Predicate information (identified by operation id):

4 - access("EMPNO"=7782)
5 - filter("ENAME"='WARD')

18 rows selected.
实际上ENAME也可以建索引那样更快
需要注意的
sql> select deptno from emp where EMPNO=7698 or job='SALESMAN' ORDER BY 1;

DEPTNO

---

    30
    30
    30
    30
    30

sql> select deptno,empno from emp where EMPNO=7698 or job='SALESMAN' ORDER BY 1;

DEPTNO      EMPNO

---

    30       7499
    30       7521
    30       7844
    30       7698
    30       7654
            sql> select deptno from emp where EMPNO=7698

2 union
3 select deptno from emp where job='SALESMAN';

DEPTNO

---

    30

避免这样问题出现可以用唯一列，主键列或rowid
sql> select deptno,empno from emp where EMPNO=7698
2 union
3 select deptno,empno from emp where job='SALESMAN';

DEPTNO      EMPNO

---

    30       7499
    30       7521
    30       7654
    30       7698
    30       7844

sql> with
2 e as (select rownum as sn,deptno,empno,job from emp)
3 select deptno
4 from
5 (
6 select sn,deptno from e where EMPNO=7698
7 union
8 select sn,deptno from e where job='SALESMAN'
9 )
10 order by 1;

DEPTNO

---

    30
    30
    30
    30
    30

2.3 组合相关的行
sql> select e.empno,e.ename,d.dname,d.loc
2 from emp e
3 inner join dept d on (e.deptno=d.deptno)
4 where e.deptno =10;

 EMPNO ENAME      DNAME          LOC

---

  7782 CLARK      ACCOUNTING     NEW YORK
  7839 KING       ACCOUNTING     NEW YORK
  7934 MILLER     ACCOUNTING     NEW YORK
sql> select e.empno,e.ename,d.dname,d.loc

2 from emp e
3 inner join dept d using(deptno)
4 where deptno =10;

 EMPNO ENAME      DNAME          LOC

---

  7782 CLARK      ACCOUNTING     NEW YORK
  7839 KING       ACCOUNTING     NEW YORK
  7934 MILLER     ACCOUNTING     NEW YORK   

2.4 IN,EXISTS和INNER JOIN
sql> alter session set"_b_tree_bitmap_plans"=false;
alter session set"_b_tree_bitmap_plans"=false
*
ERROR at line 1:
ORA-12571: TNS:packet writer failure

sql> conn scott/tiger@clonepdb_plug
Connected.
sql> alter session set"_b_tree_bitmap_plans"=false;

Session altered.
sql> explain plan for select empno,ename,job,deptno,sal
2 from emp
3 where (empno,ename,sal) in (select empno,ename,sal from emp )
4 ;

Explained.

sql> select * from table(dbms_xplan.display());

PLAN_TABLE_OUTPUT

Plan hash value: 3956160932

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Time |

| 0 | SELECT STATEMENT | | 12 | 300 | 3 (0)| 00:00:01 |
|* 1 | TABLE ACCESS FULL| EMP | 12 | 300 | 3 (0)| 00:00:01 |

Predicate information (identified by operation id):

PLAN_TABLE_OUTPUT

1 - filter("ENAME" IS NOT NULL AND "SAL" IS NOT NULL)

13 rows selected.
sql> explain plan for select empno,ename,job,deptno,sal
2 from emp a
3 where exists (select null
4 from emp b
5 where b.ename=a.ename
6 and b.job=a.job
7 and b.sal=a.sal);

Explained.

sql> select * from table(dbms_xplan.display());

PLAN_TABLE_OUTPUT

Plan hash value: 977554918

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Time |

| 0 | SELECT STATEMENT | | 12 | 516 | 6 (0)| 00:00:01 |
|* 1 | HASH JOIN SEMI | | 12 | 516 | 6 (0)| 00:00:01 |
| 2 | TABLE ACCESS FULL| EMP | 12 | 300 | 3 (0)| 00:00:01 |
| 3 | TABLE ACCESS FULL| EMP | 12 | 216 | 3 (0)| 00:00:01 |

PLAN_TABLE_OUTPUT

Predicate information (identified by operation id):

1 - access("B"."ENAME"="A"."ENAME" AND "B"."JOB"="A"."JOB" AND
"B"."SAL"="A"."SAL")

16 rows selected.
sql> select * from table(dbms_xplan.display());

PLAN_TABLE_OUTPUT

Plan hash value: 3638257876

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Time |

| 0 | SELECT STATEMENT | | 12 | 516 | 6 (0)| 00:00:01 |
|* 1 | HASH JOIN | | 12 | 516 | 6 (0)| 00:00:01 |
| 2 | TABLE ACCESS FULL| EMP | 12 | 300 | 3 (0)| 00:00:01 |
| 3 | TABLE ACCESS FULL| EMP | 12 | 216 | 3 (0)| 00:00:01 |

PLAN_TABLE_OUTPUT

Predicate information (identified by operation id):

1 - access("B"."JOB"="A"."JOB" AND "B"."ENAME"="A"."ENAME" AND
"B"."SAL"="A"."SAL")

16 rows selected.
sql> explain plan for select a.empno,ename,job,sal,a.deptno
2 from emp a inner join emp b using(job,ename,sal)
3 ;

Explained.

sql> select * from table(dbms_xplan.display());

PLAN_TABLE_OUTPUT

Plan hash value: 3638257876

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Time |

| 0 | SELECT STATEMENT | | 12 | 516 | 6 (0)| 00:00:01 |
|* 1 | HASH JOIN | | 12 | 516 | 6 (0)| 00:00:01 |
| 2 | TABLE ACCESS FULL| EMP | 12 | 300 | 3 (0)| 00:00:01 |
| 3 | TABLE ACCESS FULL| EMP | 12 | 216 | 3 (0)| 00:00:01 |

PLAN_TABLE_OUTPUT

Predicate information (identified by operation id):

1 - access("A"."SAL"="B"."SAL" AND "A"."ENAME"="B"."ENAME" AND
"A"."JOB"="B"."JOB")

16 rows selected.

2.5 INNER JOIN、LEFT JOIN、RIGHT JOIN、FULL JOIN区别
INNER JOIN 返回必配数据
LEFT JOIN 左表为主，右表只返回左表匹配数据，右表没有显示的为空 等同于右(+)
RIGHT JOIN与上面相反等同于左(+)
FULL JOIN 左右表均返回索引数据，匹配的显示一行
2.6 自关联
sql> run/
1 select a.empno as "员工编号",
2 a.ename as "员工姓名",
3 a.job as "职位",
4 b.empno as "主管编号",
5 b.ename as "主管姓名"
6 from emp a
7 left join emp b on(a.mgr=b.empno)
8* order by 1

员工编号 员工姓名 职位 主管编号 主管姓名

---

  7369 SMITH      CLERK           7902 FORD
  7499 ALLEN      SALESMAN        7698 BLAKE
  7521 WARD       SALESMAN        7698 BLAKE
  7566 JOnes      MANAGER         7839 KING
  7654 MARTIN     SALESMAN        7698 BLAKE
  7698 BLAKE      MANAGER         7839 KING
  7782 CLARK      MANAGER         7839 KING
  7839 KING       PRESIDENT
  7844 TURNER     SALESMAN        7698 BLAKE
  7900 JAMES      CLERK           7698 BLAKE
  7902 FORD       ANALYST         7566 JOnes

员工编号 员工姓名 职位 主管编号 主管姓名

---

  7934 MILLER     CLERK           7782 CLARK

12 rows selected.

2.7 NOT IN、NOT EXISTS和 LEFT JOIN
sql> select count(*) from emp where deptno =40;

COUNT(*)

     0

sql> select * from dept where deptno not in (select deptno from emp where deptno is null);

DEPTNO DNAME          LOC

---

    10 ACCOUNTING     NEW YORK
    20 RESEARCH       DALLAS
    30 SALES          CHICAGO
sql> select * from dept where not exists (select null from emp where emp.deptno=dept.deptno );

no rows selected
sql> select dept.* from dept left join emp on dept.deptno=emp.deptno where emp.deptno is null;

no rows selected
2.8 外连接的条件不能乱放
sql> select dept.* from dept left join emp on(dept.deptno=emp.deptno and emp.deptno is null);

DEPTNO DNAME          LOC

---

    10 ACCOUNTING     NEW YORK
    20 RESEARCH       DALLAS
    30 SALES          CHICAGO

sql> alter session set"_b_tree_bitmap_plans"=false;

Session altered.

sql> explain plan for select dept.* from dept left join emp on(dept.deptno=emp.deptno and emp.deptno is null);

Explained.
sql> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT

Plan hash value: 2251696546

---

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Ti
me |

---

---

PLAN_TABLE_OUTPUT

| 0 | SELECT STATEMENT | | 3 | 69 | 6 (17)| 00
:00:01 |

| 1 | MERGE JOIN OUTER | | 3 | 69 | 6 (17)| 00
:00:01 |

| 2 | TABLE ACCESS BY INDEX ROWID| DEPT | 3 | 60 | 2 (0)| 00
:00:01 |

| 3 | INDEX FULL SCAN | PK_DEPT | 3 | | 1 (0)| 00
:00:01 |

PLAN_TABLE_OUTPUT

|* 4 | SORT JOIN | | 1 | 3 | 4 (25)| 00
:00:01 |

|* 5 | TABLE ACCESS FULL | EMP | 1 | 3 | 3 (0)| 00
:00:01 |

---

---

PLAN_TABLE_OUTPUT

Predicate information (identified by operation id):

4 - access("DEPT"."DEPTNO"="EMP"."DEPTNO"(+))
filter("DEPT"."DEPTNO"="EMP"."DEPTNO"(+))
5 - filter("EMP"."DEPTNO"(+) IS NULL)

19 rows selected.
sql> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT

Plan hash value: 1353548327

---

---

| Id | Operation | Name | Rows | Bytes | Cost (%cpu)| Ti
me |

---

---

PLAN_TABLE_OUTPUT

| 0 | SELECT STATEMENT | | 1 | 23 | 6 (17)| 00
:00:01 |

| 1 | MERGE JOIN ANTI | | 1 | 23 | 6 (17)| 00
:00:01 |

| 2 | TABLE ACCESS BY INDEX ROWID| DEPT | 3 | 60 | 2 (0)| 00
:00:01 |

| 3 | INDEX FULL SCAN | PK_DEPT | 3 | | 1 (0)| 00
:00:01 |

PLAN_TABLE_OUTPUT

|* 4 | SORT UNIQUE | | 12 | 36 | 4 (25)| 00
:00:01 |

| 5 | TABLE ACCESS FULL | EMP | 12 | 36 | 3 (0)| 00
:00:01 |

---

---

PLAN_TABLE_OUTPUT

Predicate information (identified by operation id):

4 - access("DEPT"."DEPTNO"="EMP"."DEPTNO")
filter("DEPT"."DEPTNO"="EMP"."DEPTNO")

18 rows selected.
2.9 检查两个表中数据及对应数据条数是否相等
sql> run
1 select a.empno,a.ename,b.empno,b.ename
2 from emp a
3 full join emp b on(b.empno=a.empno)
4* where b.empno is null or b.empno is null

no rows selected

sql> 4
4 where b.empno is null or b.empno is null
sql> del
sql> run
1 select a.empno,a.ename,b.empno,b.ename
2 from emp a
3 full join emp b on(b.empno=a.empno)

 EMPNO ENAME           EMPNO ENAME

---

  7369 SMITH            7369 SMITH
  7499 ALLEN            7499 ALLEN
  7521 WARD             7521 WARD
  7566 JOnes            7566 JOnes
  7654 MARTIN           7654 MARTIN
  7698 BLAKE            7698 BLAKE
  7782 CLARK            7782 CLARK
  7839 KING             7839 KING
  7844 TURNER           7844 TURNER
  7900 JAMES            7900 JAMES
  7902 FORD             7902 FORD

 EMPNO ENAME           EMPNO ENAME

---

  7934 MILLER           7934 MILLER

12 rows selected.
2.10多表查询的空值处理
比ALLEN提成低的
sql> select a.ename,a.comm
2 from emp a
3 where coalesce(a.comm,0)<(select b.comm from emp b where b.ename='ALLEN');

ENAME COMM

---

SMITH
JOnes
BLAKE
CLARK
KING
TURNER 0
JAMES
FORD
MILLER

9 rows selected.
第三插入、更新与删除
sql> create table test(
2 c1 varchar2(10) default '默认1',
3 c2 varchar2(10) default '默认2',
4 c3 varchar2(10) default '默认3',
5 c4 date default sysdate
6 );

Table created.
sql> insert into test(c1,c2,c3) values(default,null,'test');

1 row created.

sql> select * from test
2 ;

C1 C2 C3 C4

---

默认1 test 2017-12-26 09:46:20
3.1阻止对某几列插入
sql> create or replace view v_test as select c1,c2,c3 from test;

View created.
sql> insert into V_TEST values ('手输1',null,'不改4');

1 row created.
sql> select * from test;

C1 C2 C3 C4

---

默认1 test 2017-12-26 09:46:20
手输1 不改4 2017-12-26 09:57:36
sql> insert into V_TEST values (default,null,'不改4');
insert into V_TEST values (default,null,'不改4')
*
ERROR at line 1:
ORA-32575: Explicit column default is not supported for modifying views

3.2复制表定义与结构
sql> create table test1 as select * from test where 1=2;

Table created.

sql> select * from test1;

no rows selected

sql> create table test2 as select * from test;

Table created.

sql> select * from test2;

C1 C2 C3 C4

---

默认1 test 2017-12-26 09:46:20
手输1 不改4 2017-12-26 09:57:36

3.3利用with check option限制数据输入
sql> alter table test modify c3 not null;

Table altered.

sql> create or replace view v_test1 as select c1,c2,c3 from test with check option;

View created.
sql> insert into V_TEST1 values ('test',null,null);
insert into V_TEST1 values ('test',null,null)
*
ERROR at line 1:
ORA-01400: cannot insert NULL into ("SCott"."TEST"."C3")

3.4多表插入语句
无条件insert
sql> insert all
2 into test1(c1,c2,c3) values ('1','2','3')
3 into test2(c1,c2,c3) values ('1','2','3')
4 into test(c1,c2,c3) values ('1','2','3')
5 select from test1 ;
插入次数取决于select 行数
要一行的话建议用select from dual;
有条件insert
sql> run
1 insert all
2 when job in ('CLERK','SALESMAN') then
3 into test (c1,c2,c3) values (ENAME,JOB,mgr)
4 when job='MANAGER' then
5 into test1 (c1,c2,c3) values (ENAME,JOB,mgr)
6 else
7 into test2 (c1,c2,c3) values (ENAME,JOB,mgr)
8 select from emp

12 rows created.

sql> select * from test
2 ;

C1 C2 C3 C4

---

默认1 test 2017-12-26 09:46:20
手输1 不改4 2017-12-26 09:57:36
1 2 3 2017-12-26 10:29:31
SMITH CLERK 7902 2017-12-26 10:39:54
ALLEN SALESMAN 7698 2017-12-26 10:39:54
WARD SALESMAN 7698 2017-12-26 10:39:54
MARTIN SALESMAN 7698 2017-12-26 10:39:54
TURNER SALESMAN 7698 2017-12-26 10:39:54
JAMES CLERK 7698 2017-12-26 10:39:54
MILLER CLERK 7782 2017-12-26 10:39:54

10 rows selected.

sql> select * from test1;

C1 C2 C3 C4

---

11 12 13
JOnes MANAGER 7839
BLAKE MANAGER 7839
CLARK MANAGER 7839

sql> select * from test2;

C1 C2 C3 C4

---

默认1 test 2017-12-26 09:46:20
手输1 不改4 2017-12-26 09:57:36
21 22 23
KING PRESIDENT
FORD ANALYST 7566

sql> insert first
2 when job in ('CLERK','SALESMAN') then
3 into test (c1,c2,c3) values (ENAME,JOB,mgr)
4 when empno in (7900,7934,7566) then
5 into test1 (c1,c2,c3) values (ENAME,JOB,mgr)
6 else
7 into test2 (c1,c2,c3) values (ENAME,JOB,mgr)
8 select job,ename,mgr,empno from emp;

12 rows created.

sql> select * from test;

C1 C2 C3 C4

---

SMITH CLERK 7902 2017-12-26 10:53:18
ALLEN SALESMAN 7698 2017-12-26 10:53:18
WARD SALESMAN 7698 2017-12-26 10:53:18
MARTIN SALESMAN 7698 2017-12-26 10:53:18
TURNER SALESMAN 7698 2017-12-26 10:53:18
JAMES CLERK 7698 2017-12-26 10:53:18
MILLER CLERK 7782 2017-12-26 10:53:18

7 rows selected.

sql> select * from test1;

C1 C2 C3 C4

---

JOnes MANAGER 7839

sql> select * from test2;

C1 C2 C3 C4

---

BLAKE MANAGER 7839
CLARK MANAGER 7839
KING PRESIDENT
FORD ANALYST 7566

3.5Merge into用法总结
MERGE INTO table_name alias1
USING (table|view|sub_query) alias2
ON (join condition)
WHEN MATCHED THEN
UPDATE table_name
SET col1 = col_val1,
col2 = col_val2
WHEN NOT MATCHED THEN
INSERT (column_list) VALUES (column_values);
严格意义上讲，”在一个同时存在Insert和Update语法的Merge语句中，总共Insert/Update的记录数，就是Using语句中alias2的记录数”。
3.6删除重复记录
sql> insert into test values (1,2,3,default)
2 ;

1 row created.

sql> insert into test values (1,2,3,default);

1 row created.

sql> select * from test;

C1 C2 C3 C4

---

1 2 3 2017-12-26 11:08:14
1 2 3 2017-12-26 11:08:18
sql> select rowid as rid,
2 c1,
3 row_number() over(partition by c1 order by c4) as seq
4 from test
5 order by 2,3;

RID C1 SEQ

---

AAASXpAALAAAAcuaAA 1 1
AAASXpAALAAAAcuaAB 1 2
sql> delete
2 from test
3 where rowid in (select rid
4 from (select rowid as rid,
5 row_number() over(partition by c1 order by c4) as seq
6 from test)
7 where seq>1);

1 row deleted.

sql> select * from test;

C1 C2 C3 C4

---

1 2 3 2017-12-26 11:08:14
sql> delete
2 from test a
3 where exists(select /+hash_sj/ null from test b where b.c1=a.c1 and b.rowid>a.rowid);
保留最新的<保留老的
1 row deleted.

sql> select * from test;

C1 C2 C3 C4

---

1 2 3 2017-12-26 13:32:18
第四字符串
4.1 遍历字符串
sql> select level from dual connect by level<=4;

 LEVEL

---

     1
     2
     3
     4

sql> select "拼音",level,substr("拼音",level,1) from (select 'TTXS' as "拼音" FROM DUAL) connect by level <=4;

拼音 LEVEL SUB

---

TTXS 1 T
TTXS 2 T
TTXS 3 X
TTXS 4 S
4.2 字符串'
sql> select 'g''day mate' qmarks from dual;

QMARKS

g'day mate
下面是10g
sql> select q'[g'day mate]' qmarks from dual;

QMARKS

g'day mate

sql> select q'{g'day mate}' qmarks from dual;

QMARKS

g'day mate

sql> select q'<g'day mate>' qmarks from dual;

QMARKS

g'day mate

sql> select q'(g'day mate)' qmarks from dual;

QMARKS

g'day mate
4.3 统计字符串出现次数
11g
sql> select regexp_count('wo shi wo','o') from dual;

REGEXP_COUNT('WOSHIWO','O')

                      2

sql> select length(translate('wo shi wo','wo shi wo','o')) from dual;

LENGTH(TRANSLATE('WOSHIWO','WOSHIWO','O'))

                                     2                                                  

4.4 从字符里面删除不需要的
sql> select ename,translate(ename,'1AEIoU','1') from emp;

ENAME TRANSLATE(ENAME,'1AEIoU','1')

---

SMITH SMTH
ALLEN LLN
WARD WRD
JOnes JNS
MARTIN MRTN
BLAKE BLK
CLARK CLRK
KING KNG
TURNER TRNR
JAMES JMS
FORD FRD

ENAME TRANSLATE(ENAME,'1AEIoU','1')

---

MILLER MLLR

12 rows selected.
sql> select ename,regexp_replace(ename,'[AEIoU]') from emp;

ENAME REGEXP_REPLACE(ENAME,'[AEIoU]'

---

SMITH SMTH
ALLEN LLN
WARD WRD
JOnes JNS
MARTIN MRTN
BLAKE BLK
CLARK CLRK
KING KNG
TURNER TRNR
JAMES JMS
FORD FRD

ENAME REGEXP_REPLACE(ENAME,'[AEIoU]'

---

MILLER MLLR

12 rows selected.

4.5 将字母与数字分开
sql> select dname||deptno,translate(dname||deptno,'a0123456789','a') as data,translate(DNAME||DEPTNO,'0123456789'||dname||deptno,'0123456789') as data1 from dept;

DNAME		DEPTNO DATA DATA1

ACCOUNTING10 ACCOUNTING 10
RESEARCH20 RESEARCH 20
SALES30 SALES 30
sql> select dname||deptno,regexp_replace(dname||deptno,'[0-9]','') as data,regexp_replace(DNAME||DEPTNO,'[^0-9]','') as data1 from dept;

DNAME		DEPTNO DATA DATA1

ACCOUNTING10 ACCOUNTING 10
RESEARCH20 RESEARCH 20
SALES30 SALES 30
4.6 ^,$.,+意义

表示自少匹配6零次

4.7 姓名字母首字大写
sql> select regexp_replace('Michael Hartstein','([[:upper:]])(.)([[:upper:]])(.)','\1.\3') from dual;

REG

M.H
4.8 按字符串中数字排序
sql> select dname||deptno||loc from dept order by translate(dname||deptno||loc,'0123456789'||dname||deptno||loc,'0123456789') desc;

DNAME||DEPTNO||LOC

SALES30CHICAGO
RESEARCH20DALLAS
ACCOUNTING10NEW YORK

sql> select dname||deptno||loc from dept order by regexp_replace(dname||deptno||loc,'[^0-9]') desc;

DNAME||DEPTNO||LOC

SALES30CHICAGO
RESEARCH20DALLAS
ACCOUNTING10NEW YORK

4.9 创建分割列表
sql> select deptno,sal,ename from emp;

DEPTNO        SAL ENAME

---

    20        800 SMITH
    30       1600 ALLEN
    30       1250 WARD
    20       2975 JOnes
    30       1250 MARTIN
    30       2850 BLAKE
    10       2450 CLARK
    10       5000 KING
    30       1500 TURNER
    30        950 JAMES
    20       3000 FORD

DEPTNO        SAL ENAME

---

    10       1300 MILLER

12 rows selected.
sql> col TOTAL_SAL format 999999
sql> col TOTAL_name format A100
sql> select deptno,
2 sum(sal) as total_sal,
3 listagg(ename,',') within group(order by ename) as total_name
4 from emp
5 group by deptno;

DEPTNO TOTAL_SAL TOTAL_NAME

---

    10      8750 CLARK,KING,MILLER
    20      6775 FORD,JOnes,SMITH
    30      9400 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

4.10 提取第n个分割子串
sql> run
1 with
2 a as
3 (
4 select listagg(ename,',') within group(order by ename) as name from emp where deptno in(10,20) group by deptno
5 )
6* select regexp_substr(a.name,'[^,]+',1,2) as "子串" from a

子串

KING
JOnes
4.11 分解ip地址
sql> run
1 select regexp_substr(v.ip,'[^.]+',1,1 ) a
2 ,regexp_substr(v.ip,'[^.]+',1,2 ) b
3 ,regexp_substr(v.ip,'[^.]+',1,3) c
4 ,regexp_substr(v.ip,'[^.]+',1,4 ) d
5* from (select '192.168.0.1' as ip from dual) v

A B C D

---

192 168 0 1
4.12 将分个数据转换成多值IN
sql> var v_emps varchar2(30);
sql> exec :v_emps :='CLARK,KING,MILLER';

PL/sql procedure successfully completed.

sql> SET LINE 1000
sql> run
1 SELECT FROM EMP WHERE ENAME IN
2 (
3 SELECT REGEXP_SUBSTR(:v_emps,'[^,]+',1,level) as ename from dual
4 connect by level <=(length(translate(:v_emps,','||:v_emps,','))+1)
5 )

 EMPNO ENAME      JOB              MGR HIREDATE                   SAL       COMM     DEPTNO

---

  7782 CLARK      MANAGER         7839 1981-06-09 00:00:00       2450                    10
  7839 KING       PRESIDENT            1981-11-17 00:00:00       5000                    10
  7934 MILLER     CLERK           7782 1982-01-23 00:00:00       1300                    10

第五 使用数字
5.1 累计和
sql> select empno,
2 ename,
3 sal,
4 sum(sal) over (order by empno)
5 from emp
6 where deptno=30
7 order by empno;

 EMPNO ENAME             SAL SUM(SAL)OVER(ORDERBYEMPNO)

---

  7499 ALLEN            1600                       1600
  7521 WARD             1250                       2850
  7654 MARTIN           1250                       4100
  7698 BLAKE            2850                       6950
  7844 TURNER           1500                       8450
  7900 JAMES             950                       9400

6 rows selected.

5.2 返回各部门排行前三的员工
sql> run
1 select deptno,
2 empno,
3 sal,
4 row_number() over (partition by deptno order by sal desc) as row_num,
5 rank() over (partition by deptno order by sal desc) as rank,
6 dense_rank() over (partition by deptno order by sal desc) as dense_rank
7 from emp
8 where deptno in (20,30)
9* order by 1,3 desc

DEPTNO      EMPNO        SAL    ROW_NUM       RANK DENSE_RANK

---

    20       7902       3000          1          1          1
    20       7566       2975          2          2          2
    20       7369        800          3          3          3
    30       7698       2850          1          1          1
    30       7499       1600          2          2          2
    30       7844       1500          3          3          3
    30       7521       1250          4          4          4
    30       7654       1250          5          4          4
    30       7900        950          6          6          5

9 rows selected.

5.3 返回最大值所在行数据
sql> run
1 select deptno,
2 empno,
3 max(ename) keep(dense_rank first order by sal) over (partition by deptno),
4 max(ename) keep(dense_rank last order by sal) over (partition by deptno),
5 ename,
6 sal
7 from emp
8 where deptno=10
9* order by 1,6 desc

DEPTNO      EMPNO MAX(ENAME) MAX(ENAME) ENAME             SAL

---

    10       7839 MILLER     KING       KING             5000
    10       7782 MILLER     KING       CLARK            2450
    10       7934 MILLER     KING       MILLER           1300
            sql> select deptno,

2 empno,
3 first_value(ename) over (partition by deptno),
4 ename,
5 sal
6 from emp
7 where deptno=10
8 order by 1,5 desc;

DEPTNO      EMPNO FirsT_VALU ENAME             SAL

---

    10       7839 KING       KING             5000
    10       7782 KING       CLARK            2450
    10       7934 KING       MILLER           1300

5.4 求和百分比
sql> run
1 select deptno,
2 empno,
3 ename,
4 sal,
5 round(ratio_to_report(sal) over(partition by deptno)100,2)
6 from emp
7 order by 1,2

DEPTNO      EMPNO ENAME             SAL ROUND(RATIO_TO_REPORT(SAL)OVER(PARTITIONBYDEPTNO)*100,2)

---

    10       7782 CLARK            2450                                                       28
    10       7839 KING             5000                                                    57.14
    10       7934 MILLER           1300                                                    14.86
    20       7369 SMITH             800                                                    11.81
    20       7566 JOnes            2975                                                    43.91
    20       7902 FORD             3000                                                    44.28
    30       7499 ALLEN            1600                                                    17.02
    30       7521 WARD             1250                                                     13.3
    30       7654 MARTIN           1250                                                     13.3
    30       7698 BLAKE            2850                                                    30.32
    30       7844 TURNER           1500                                                    15.96

DEPTNO      EMPNO ENAME             SAL ROUND(RATIO_TO_REPORT(SAL)OVER(PARTITIONBYDEPTNO)*100,2)

---

    30       7900 JAMES             950                                                    10.11

12 rows selected.
第六 日期
6.1 年月日加减
sql> select hiredate,
2 hiredate -5,
3 hiredate +5,
4 add_months(hiredate,-5),
5 add_months(hiredate,5),
6 add_months(hiredate,-512),
7 add_months(hiredate,512)
8 from emp
9 where rownum<=1;

HIREDATE HIREDATE-5 HIREDATE+5 ADD_MONTHS(HIREDATE ADD_MONTHS(HIREDATE ADD_MONTHS(HIREDATE ADD_MONTHS(HIREDATE

---

1980-12-17 00:00:00 1980-12-12 00:00:00 1980-12-22 00:00:00 1980-07-17 00:00:00 1981-05-17 00:00:00 1975-12-17 00:00:00 1985-12-17 00:00:00
6.2 时分秒加减
sql> run
1 select hiredate,
2 hiredate -5/24/60/60,
3 hiredate +5/24/60/60,
4 hiredate -5/24/60,
5 hiredate +5/24/60,
6 hiredate -5/24,
7 hiredate +5/24
8 from emp
9* where rownum<=1

HIREDATE HIREDATE-5/24/60/60 HIREDATE+5/24/60/60 HIREDATE-5/24/60 HIREDATE+5/24/60 HIREDATE-5/24
HIREDATE+5/24

---

1980-12-17 00:00:00 1980-12-16 23:59:55 1980-12-17 00:00:05 1980-12-16 23:55:00 1980-12-17 00:05:00 1980-12-16 19:00:00 1980-12-17 05:00:00
6.3 时间间隔
sql> select max(hiredate)-min(hiredate),
2 (max(hiredate)-min(hiredate))24,
3 (max(hiredate)-min(hiredate))2460,
4 (max(hiredate)-min(hiredate))246060
5 from emp
6 where ename in('WARD','ALLEN')
7 ;

MAX(HIREDATE)-MIN(HIREDATE) (MAX(HIREDATE)-MIN(HIREDATE))24 (MAX(HIREDATE)-MIN(HIREDATE))2460 (MAX(HIREDATE)-MIN(HIREDATE))246060

---

                      2                               48                                2880

172800
6.4 日期间隔
sql> run
1 select max(hiredate)-min(hiredate),
2 months_between(max(hiredate),min(hiredate)),
3 months_between(max(hiredate),min(hiredate))/12
4* from emp

MAX(HIREDATE)-MIN(HIREDATE) MONTHS_BETWEEN(MAX(HIREDATE),MIN(HIREDATE)) MONTHS_BETWEEN(MAX(HIREDATE),MIN(HIREDATE))/12

---

                    402                                  13.1935484                                     1.09946237

6.5 当前记录和下一条记录差
sql> run
1 select deptno,
2 ename,
3 hiredate,
4 lead(hiredate) over(order by hiredate)
5 from emp
6* where deptno=10

DEPTNO ENAME      HIREDATE            LEAD(HIREDATE)OVER(

---

    10 CLARK      1981-06-09 00:00:00 1981-11-17 00:00:00
    10 KING       1981-11-17 00:00:00 1982-01-23 00:00:00
    10 MILLER     1982-01-23 00:00:00
            sql> run

1 select deptno,
2 ename,
3 hiredate,
4 lag(hiredate) over(order by hiredate)
5 from emp
6* where deptno=10

DEPTNO ENAME      HIREDATE            LAG(HIREDATE)OVER(O

---

    10 CLARK      1981-06-09 00:00:00
    10 KING       1981-11-17 00:00:00 1981-06-09 00:00:00
    10 MILLER     1982-01-23 00:00:00 1981-11-17 00:00:00

6.6 sysdate

6.7 interval
sql> select interval '50' month as month from dual;

MONTH

+04-02

sql> select interval '99' day as day from dual;

DAY

+99 00:00:00

sql> select interval '80' hour as hour from dual;

HOUR

+03 08:00:00

sql> select interval '5' year as year from dual;

YEAR

+05-00

6.8 extract
sql> select extract(year from systimestamp) as year from dual;

  YEAR

---

  2017

sql> select extract(month from systimestamp) as month from dual;

 MONTH

---

    12

sql> select extract(day from systimestamp) as day from dual;

   DAY

---

    27

sql> select extract(hour from systimestamp) as hour from dual;

  HOUR

---

     3  

第七 报表和数据仓库
7.1 行转列
sql> select job,
2 case deptno when 10 then sal end as deptno10,
3 case deptno when 20 then sal end as deptno20,
4 case deptno when 30 then sal end as deptno30,
5 sal
6 from emp
7 order by 1;

JOB DEPTNO10 DEPTNO20 DEPTNO30 SAL

---

ANALYST 3000 3000
CLERK 1300 1300
CLERK 950 950
CLERK 800 800
MANAGER 2975 2975
MANAGER 2850 2850
MANAGER 2450 2450
PRESIDENT 5000 5000
SALESMAN 1500 1500
SALESMAN 1250 1250
SALESMAN 1600 1600

JOB DEPTNO10 DEPTNO20 DEPTNO30 SAL

---

SALESMAN 1250 1250

12 rows selected.
sql> select job,
2 sum(case deptno when 10 then sal end) as deptno10,
3 sum(case deptno when 20 then sal end) as deptno20,
4 sum(case deptno when 30 then sal end) as deptno30,
5 sum(sal) as sal
6 from emp
7 group by job
8 order by 1;

JOB DEPTNO10 DEPTNO20 DEPTNO30 SAL

---

ANALYST 3000 3000
CLERK 1300 800 950 3050
MANAGER 2450 2975 2850 8275
PRESIDENT 5000 5000
SALESMAN 5600 5600
sql> select *
2 from (select job,
3 sal
4 ,deptno
5 from emp)
6 pivot(sum(sal) as s
7 for deptno in (10 as d10,
8 20 ,
9 30 as d30)
10 )
11 order by 1;

JOB D10_S 20_S D30_S

---

ANALYST 3000
CLERK 1300 800 950
MANAGER 2450 2975 2850
PRESIDENT 5000
SALESMAN 5600
7.2 控制结果集重复值
sql> select job ,ename from emp where deptno=30 order by emp.job,ename;

JOB ENAME

---

CLERK JAMES
MANAGER BLAKE
SALESMAN ALLEN
SALESMAN MARTIN
SALESMAN TURNER
SALESMAN WARD

6 rows selected.

sql> select case
2 when lag(job) over(order by job,ename)=job then
3 null
4 else
5 job
6 end as job,
7 ename
8 from emp
9 where deptno=30
10 order by emp.job,ename;

JOB ENAME

---

CLERK JAMES
MANAGER BLAKE
SALESMAN ALLEN
MARTIN
TURNER
WARD

6 rows selected.
7.3 简单小计
sql> select deptno,sum(sal) as s_sal from emp group by rollup(deptno)
2 ;

DEPTNO      S_SAL

---

    10       8750
    20       6775
    30       9400
            24925

7.4 分组函数详解
sql> select DEPTNO,sum(sal) from emp group by deptno;

DEPTNO   SUM(SAL)

---

    30       9400
    20       6775
    10       8750

            sql> select DEPTNO,sum(sal) from emp group by rollup(deptno);

DEPTNO   SUM(SAL)

---

    10       8750
    20       6775
    30       9400
            24925
     sql> select DEPTNO,job,sum(sal) from emp group by rollup(deptno,job);

DEPTNO JOB         SUM(SAL)

---

    10 CLERK           1300
    10 MANAGER         2450
    10 PRESIDENT       5000
    10                 8750
    20 CLERK            800
    20 ANALYST         3000
    20 MANAGER         2975
    20                 6775
    30 CLERK            950
    30 MANAGER         2850
    30 SALESMAN        5600

DEPTNO JOB         SUM(SAL)

---

    30                 9400
                      24925

13 rows selected.

grouping值为0时说明这个值是数据库中本来的值，为1说明是统计的结果
sql> select DEPTNO,sum(sal) from emp group by cube(deptno) order by 1;

DEPTNO   SUM(SAL)

---

    10       8750
    20       6775
    30       9400
            24925
    sql> select DEPTNO,job,sum(sal) from emp group by cube(deptno,job) order by 1;

DEPTNO JOB         SUM(SAL)

---

    10 CLERK           1300
    10 MANAGER         2450
    10 PRESIDENT       5000
    10                 8750
    20 ANALYST         3000
    20 CLERK            800
    20 MANAGER         2975
    20                 6775
    30 CLERK            950
    30 MANAGER         2850
    30 SALESMAN        5600

DEPTNO JOB         SUM(SAL)

---

    30                 9400
       ANALYST         3000
       CLERK           3050
       MANAGER         8275
       PRESIDENT       5000
       SALESMAN        5600
                      24925

18 rows selected.

仔细观察一下，CUBE与ROLLUP之间的细微差别
rollup(a,b) 统计列包含：(a,b)、(a)、()
rollup(a,b,c) 统计列包含：(a,b,c)、(a,b)、(a)、()
……以此类推ing……

cube(a,b) 统计列包含：(a,b)、(a)、(b)、()
cube(a,b,c) 统计列包含：(a,b,c)、(a,b)、(a,c)、(b,c)、(a)、(b)、(c)、()
CUBE在ROLLUP的基础上进一步从各种维度上给出细化的统计汇总结果。
sql> select DEPTNO,job,sum(sal) from emp group by grouping sets(deptno,job) order by 1;

DEPTNO JOB         SUM(SAL)

---

    10                 8750
    20                 6775
    30                 9400
       ANALYST         3000
       MANAGER         8275
       SALESMAN        5600
       CLERK           3050
       PRESIDENT       5000
grouping sets就是对參数中的每一个參数做grouping。假设使用group by grouping sets(a,b)。则对(a),(b)进行group by

Grouping_id()的返回值事实上就是參数中的每列的grouping()值的二进制向量。假设grouping(a)=1，grouping(b)=1，则grouping_id(A,B)的返回值就是二进制的11。转成10进制就是3。

參数能够是多个，但必须为group by中出现的列。

7.5 不同组进行统计
sql> select ename,deptno,count() over(partition by deptno),job,count()over(partition by job),count(*) over()
2 from emp;

ENAME DEPTNO COUNT()OVER(PARTITIONBYDEPTNO) JOB COUNT()OVER(PARTITIONBYJOB) COUNT(*)OVER()

---

KING 10 3 PRESIDENT 1 12
CLARK 10 3 MANAGER 3 12
MILLER 10 3 CLERK 3 12
JOnes 20 3 MANAGER 3 12
SMITH 20 3 CLERK 3 12
FORD 20 3 ANALYST 1 12
ALLEN 30 6 SALESMAN 4 12
WARD 30 6 SALESMAN 4 12
TURNER 30 6 SALESMAN 4 12
MARTIN 30 6 SALESMAN 4 12
JAMES 30 6 CLERK 3 12

ENAME DEPTNO COUNT()OVER(PARTITIONBYDEPTNO) JOB COUNT()OVER(PARTITIONBYJOB) COUNT(*)OVER()

---

BLAKE 30 6 MANAGER 3 12

12 rows selected.
7.6 移动范围内值计算
sql> select hiredate,
2 sal,
3 sum(sal) over(order by hiredate range between interval '3' month preceding and current row)
4 from emp
5 where deptno=30
6 order by 1;

HIREDATE SAL SUM(SAL)OVER(ORDERBYHIREDAteraNGEBETWEENINTERVAL'3'MONTHPRECEDINGANDCURRENTROW)

---

1981-02-20 00:00:00 1600 1600
1981-02-22 00:00:00 1250 2850
1981-05-01 00:00:00 2850 5700
1981-09-08 00:00:00 1500 1500
1981-09-28 00:00:00 1250 2750
1981-12-03 00:00:00 950 3700

6 rows selected.
第八 分层查询
8.1 简单树形结构
sql> run
1 select empno,
2 ename,
3 mgr,
4 prior ename
5 from emp
6 start with empno=7566
7 connect by(prior empno)=mgr
8*

 EMPNO ENAME             MGR PRIORENAME

---

  7566 JOnes            7839
  7902 FORD             7566 JOnes
  7369 SMITH            7902 FORD

8.2 根节点，分支节点，叶子节点
sql> run
1 select lpad('-',(level-1)2,'-')||empno as empno,
2 ename,
3 mgr,
4 level,
5 decode(level,1,1) as root,
6 decode(connect_by_isleaf,1,1) as leaf,
7 case
8 when(connect_by_isleaf=0 and level>1) then
9 1
10 end as fenzi
11 from emp
12 start with empno=7566
13 connect by (prior empno)=mgr

EMPNO ENAME MGR LEVEL ROOT LEAF FENZI

---

7566 JOnes 7839 1 1
--7902 FORD 7566 2 1
----7369 SMITH 7902 3 1
8.3 sys_connect_by_path ==listagg
sql> run
1 select empno,
2 ename,
3 mgr,
4 sys_connect_by_path(ename,',') as enames
5 from emp
6 start with empno=7566
7* connect by (prior empno)=mgr

EMPNO ENAME MGR ENAMES

---

7566 JOnes 7839 ,JOnes
7902 FORD 7566 ,JOnes,FORD
7369 SMITH 7902 ,JOnes,FORD,SMITH
8.4 树形查询排序
sql> select lpad('-',(level-1)*2,'-')||empno as empno,
2 ename,
3 mgr
4 from emp
5 start with empno=7839
6 connect by (prior empno)=mgr
7 order siblings by emp.empno desc;

EMPNO ENAME MGR

---

7839 KING
--7782 CLARK 7839
----7934 MILLER 7782
--7698 BLAKE 7839
----7900 JAMES 7698
----7844 TURNER 7698
----7654 MARTIN 7698
----7521 WARD 7698
----7499 ALLEN 7698
--7566 JOnes 7839
----7902 FORD 7566

EMPNO ENAME MGR

---

------7369 SMITH 7902

12 rows selected.
8.5 树型查询使用where
sql> select empno,
2 mgr,
3 ename,
4 deptno
5 from(select * from emp where deptno=20) emp
6 start with mgr is null
7 connect by(prior empno)=mgr;

no rows selected
8.6 查询树型的一个分支
sql> run
1 select empno,
2 mgr,
3 ename,
4 level
5 from emp
6 start with empno=7698
7* connect by (prior empno)=mgr

EMPNO MGR ENAME LEVEL

---

7698 7839 BLAKE 1
7499 7698 ALLEN 2
7521 7698 WARD 2
7654 7698 MARTIN 2
7844 7698 TURNER 2
7900 7698 JAMES 2
8.7 减去一个分支
sql> run
1 select empno,
2 mgr,
3 ename,
4 level
5 from emp
6 start with mgr is NULL
7* connect by (prior empno)=mgr

EMPNO MGR ENAME LEVEL

---

7839 KING 1
7566 7839 JOnes 2
7902 7566 FORD 3
7369 7902 SMITH 4
7698 7839 BLAKE 2
7499 7698 ALLEN 3
7521 7698 WARD 3
7654 7698 MARTIN 3
7844 7698 TURNER 3
7900 7698 JAMES 3
7782 7839 CLARK 2

EMPNO MGR ENAME LEVEL

---

7934 7782 MILLER 3

12 rows selected.
sql> RUN
1 select empno,
2 mgr,
3 ename,
4 level
5 from emp
6 start with mgr is NULL
7 connect by (prior empno)=mgr
8* and empno !=7698

EMPNO MGR ENAME LEVEL

---

7839 KING 1
7566 7839 JOnes 2
7902 7566 FORD 3
7369 7902 SMITH 4
7782 7839 CLARK 2
7934 7782 MILLER 3

6 rows selected.
第九 调优案例分享
9.1 不建议使用标量子查询，使用left join优化标量子查询
sql> select empno,
2 ename,
3 sal,
4 deptno,
5 (select dname from dept where dept.deptno=emp.deptno)
6 from emp;

EMPNO ENAME SAL DEPTNO (SELECTDNAMEFR

---

7369 SMITH 800 20 RESEARCH
7499 ALLEN 1600 30 SALES
7521 WARD 1250 30 SALES
7566 JOnes 2975 20 RESEARCH
7654 MARTIN 1250 30 SALES
7698 BLAKE 2850 30 SALES
7782 CLARK 2450 10 ACCOUNTING
7839 KING 5000 10 ACCOUNTING
7844 TURNER 1500 30 SALES
7900 JAMES 950 30 SALES
7902 FORD 3000 20 RESEARCH

EMPNO ENAME SAL DEPTNO (SELECTDNAMEFR

---

7934 MILLER 1300 10 ACCOUNTING

12 rows selected.

sql> select e.empno,
2 e.ename,
3 e.sal,
4 e.deptno,
5 d.dname
6 from emp e
7 left join dept d on(e.deptno=d.deptno);

EMPNO ENAME SAL DEPTNO DNAME

---

7782 CLARK 2450 10 ACCOUNTING
7839 KING 5000 10 ACCOUNTING
7934 MILLER 1300 10 ACCOUNTING
7369 SMITH 800 20 RESEARCH
7566 JOnes 2975 20 RESEARCH
7902 FORD 3000 20 RESEARCH
7499 ALLEN 1600 30 SALES
7521 WARD 1250 30 SALES
7654 MARTIN 1250 30 SALES
7698 BLAKE 2850 30 SALES
7844 TURNER 1500 30 SALES

EMPNO ENAME SAL DEPTNO DNAME

---

7900 JAMES 950 30 SALES

12 rows selected.
sql> run
1 select /+use_nl(e,d)/
2 e.ename,
3 e.sal,
4 e.deptno,
5 d.dname
6 from emp e
7* left join dept d on(e.deptno=d.deptno)

ENAME SAL DEPTNO DNAME

---

SMITH 800 20 RESEARCH
ALLEN 1600 30 SALES
WARD 1250 30 SALES
JOnes 2975 20 RESEARCH
MARTIN 1250 30 SALES
BLAKE 2850 30 SALES
CLARK 2450 10 ACCOUNTING
KING 5000 10 ACCOUNTING
TURNER 1500 30 SALES
JAMES 950 30 SALES
FORD 3000 20 RESEARCH

ENAME SAL DEPTNO DNAME

---

MILLER 1300 10 ACCOUNTING

12 rows selected.
9.2 使用left jion 优化标量子查聚合
sql> select d.department_id,
2 d.department_name,
3 d.location_id,
4 nvl((select sum(e.salary)
5 from employees e
6 where e.department_id=d.department_id),
7 0) as sum_sal
8 from departments d;

DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL

---

       10 Administration                        1700       4400
       20 Marketing                             1800      19000
       30 Purchasing                            1700      24900
       40 Human Resources                       2400       6500
       50 Shipping                              1500     156400
       60 IT                                    1400      28800
       70 Public Relations                      2700      10000
       80 Sales                                 2500     304500
       90 Executive                             1700      58000
      100 Finance                               1700      51608
      110 Accounting                            1700      20308

DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL

---

      120 Treasury                              1700          0
      130 Corporate Tax                         1700          0
      140 Control And Credit                    1700          0
      150 Shareholder Services                  1700          0
      160 Benefits                              1700          0
      170 Manufacturing                         1700          0
      180 Construction                          1700          0
      190 Contracting                           1700          0
      200 Operations                            1700          0
      210 IT Support                            1700          0
      220 NOC                                   1700          0

DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL

---

      230 IT Helpdesk                           1700          0
      240 Government Sales                      1700          0
      250 Retail Sales                          1700          0
      260 Recruiting                            1700          0
      270 Payroll                               1700          0

27 rows selected.
sql> select d.department_id,
2 d.department_name,
3 d.location_id,
4 COALESCE(e.sum_sal,0) as sum_sal
5 from departments d
6 left join (select e.department_id,sum(e.salary) as sum_sal
7 from employees e
8 group by e.department_id) e on ( e.department_id=
9 d.department_id);

DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL

---

       10 Administration                        1700       4400
       20 Marketing                             1800      19000
       30 Purchasing                            1700      24900
       40 Human Resources                       2400       6500
       50 Shipping                              1500     156400
       60 IT                                    1400      28800
       70 Public Relations                      2700      10000
       80 Sales                                 2500     304500
       90 Executive                             1700      58000
      100 Finance                               1700      51608
      110 Accounting                            1700      20308

DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL

---

      120 Treasury                              1700          0
      130 Corporate Tax                         1700          0
      140 Control And Credit                    1700          0
      150 Shareholder Services                  1700          0
      160 Benefits                              1700          0
      170 Manufacturing                         1700          0
      180 Construction                          1700          0
      190 Contracting                           1700          0
      200 Operations                            1700          0
      210 IT Support                            1700          0
      220 NOC                                   1700          0

DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL

---

      230 IT Helpdesk                           1700          0
      240 Government Sales                      1700          0
      250 Retail Sales                          1700          0
      260 Recruiting                            1700          0
      270 Payroll                               1700          0

27 rows selected.