在结果中添加一列，显示按年份分组的序号

如何解决在结果中添加一列，显示按年份分组的序号

这些 sql 能够添加具有递增数字（1、2、3、4 等）的列

select year,quarter,uploaddate,wagecount,SUM(1) 
OVER (ORDER BY year,uploaddate) as SequentialNumbers from SAMPLE_DATA
order by year,uploaddate

或

select your_query.*,rownum
  from (select year,wagecount 
          from SAMPLE_DATA) your_query

结果如下

年	季度	上传日期	工资计数	序列号
2021	1	01-JAN-21	2	1
2021	1	02-JAN-21	2	2
2021	1	03-JAN-21	1	3
2021	1	04-JAN-21	5	4

无论如何要采取下一个大胆的步骤，让每年都有自己的序列号。换句话说，结果最终会像这样

类似的东西

年	季度	上传日期	工资计数	序列号
2019	1	01-JAN-21	2	1
2019	1	02-JAN-21	2	2
2019	1	03-JAN-21	1	3
2019	1	04-JAN-21	5	4
2020	1	01-JAN-20	2	1
2020	1	02-JAN-20	2	2
2020	1	03-JAN-20	1	3
2020	1	04-JAN-20	5	4
2021	1	01-JAN-21	2	1
2021	1	02-JAN-21	2	2
2021	1	03-JAN-21	1	3
2021	1	04-JAN-21	5	4

样本数据如下：

CREATE TABLE SAMPLE_DATA (
    YEAR VARCHAR2(4) NULL,QUARTER NUMBER(1,0) NULL,UPLOADDATE DATE NULL,WAGECOUNT NUMBER(10,0) NULL
);

Insert into SAMPLE_DATA (YEAR,QUARTER,UPLOADDATE,WAGECOUNT) values ('2019','1',to_date('27-MAR-19','DD-MON-RR'),5);
Insert into SAMPLE_DATA (YEAR,to_date('28-MAR-19',8493);
Insert into SAMPLE_DATA (YEAR,to_date('29-MAR-19',15070);
Insert into SAMPLE_DATA (YEAR,to_date('30-MAR-19',1244);
Insert into SAMPLE_DATA (YEAR,WAGECOUNT) values ('2020',to_date('03-JAN-20',0);
Insert into SAMPLE_DATA (YEAR,to_date('05-JAN-20',2);
Insert into SAMPLE_DATA (YEAR,to_date('06-JAN-20',3);
Insert into SAMPLE_DATA (YEAR,to_date('07-JAN-20',6);
Insert into SAMPLE_DATA (YEAR,WAGECOUNT) values ('2021','2',to_date('21-APR-21',59);
Insert into SAMPLE_DATA (YEAR,to_date('22-APR-21',10);
Insert into SAMPLE_DATA (YEAR,to_date('23-APR-21',16);
Insert into SAMPLE_DATA (YEAR,to_date('24-APR-21',1);

解决方法

您需要for (var i = 0; i < example.length - 1; i++) {解析函数：

row_number

结果：

select year,quarter,uploaddate,wagecount,row_number()
         OVER (partition by year
               ORDER BY quarter,uploaddate) as SequentialNumbers 
from SAMPLE_DATA
order by year,uploaddate;