MySQL PHP：优化排名查询和计数子查询

这是原始数据,并希望根据得分(count(tbl_1.id))对它们进行排名.

[tbl_1]
===========
id | name  
===========
1  | peter
2  | jane
1  | peter
2  | jane
3  | harry
3  | harry
3  | harry
3  | harry
4  | ron

因此,制作临时表(tbl_2)来计算每个id的分数.

SELECT id, name, COUNT( id ) AS score
FROM tbl_1
GROUP BY id
ORDER BY score DESC;
LIMIT 0, 30;

然后结果是;

[tbl_2]
===================
id | name  | score
===================
3  | harry | 4
1  | peter | 2
2  | jane  | 2
4  | ron   | 1

然后查询这个;

SELECT v1.id, v1.name, v1.score, COUNT( v2.score ) AS rank
FROM Votes v1
JOIN Votes v2 ON v1.score < v2.score
OR (
v1.score = v2.score
AND v1.id = v2.id
)
GROUP BY v1.id, v1.score
ORDER BY v1.rank ASC, v1.id ASC
LIMIT 0, 30;

然后结果是;

==========================
id | name  | score | rank
==========================
3  | harry | 4     |  1
1  | peter | 2     |  2
2  | jane  | 2     |  2
4  | ron   | 1     |  4

是否可以很好地在一个事务(查询)中执行此操作？

解决方法:

是的,可以在单个查询中执行此操作.但它在MysqL中是一个完整的毛球,因为MysqL没有简单的ROWNUM操作,你需要一个用于排名计算.

这是您显示排名的投票查询. @ranka变量用于对行进行编号.

SELECT @ranka:=@ranka+1 AS rank, id, name, score
  FROM
  (
     SELECT id, 
            name, 
            COUNT( id ) AS score
       FROM tbl_1
       GROUP BY id
       ORDER BY score DESC, id
   ) Votes,
  (SELECT @ranka:=0) r

正如您已经发现的那样,您需要自行加入此项以获得正确的排名(正确处理关系).因此,如果您接受查询并将两个引用替换为您的投票表,每个引用都有自己的子查询版本,那么您将获得所需的内容.

SELECT v1.id,
       v1.name,
       v1.score,
       COUNT( v2.score ) AS rank
  FROM (
         SELECT @ranka:=@ranka+1 AS rank,
                id,
                name,
                score
           FROM
              (
                SELECT id,
                       name,
                       COUNT( id ) AS score
                  FROM tbl_1
                 GROUP BY id
                 ORDER BY score DESC, name
               ) Votes,
         (SELECT @ranka:=0) r) v1
  JOIN (
         SELECT @rankb:=@rankb+1 AS rank, 
                id, 
                name, 
                score
           FROM
              (
                SELECT id,
                       name,
                       COUNT( id ) AS score
                  FROM tbl_1
                  GROUP BY id
                  ORDER BY score DESC, name
              ) Votes,
        (SELECT @rankb:=0) r) v2 
    ON (v1.score < v2.score) OR 
       (v1.score = v2.score  AND v1.id = v2.id)
 GROUP BY v1.id, v1.score
 ORDER BY v1.rank ASC, v1.id ASC
 LIMIT 0, 30;

告诉你这是一个毛球.请注意,在您自行加入的子查询的两个版本中需要不同的@ranka和@rankb变量,以使行编号正常工作：这些变量在MysqL中具有连接范围,而不是子查询范围.

http://sqlfiddle.com/#!2/c5350/1/0显示了这个工作.

编辑：使用Postgresql的RANK()函数更容易做到这一点.

SELECT name, Votes, rank() over (ORDER BY Votes)
  FROM (
        SELECT name, count(id) Votes
          FROM tab
         GROUP BY name
       )x

http://sqlfiddle.com/#!1/94cca/18/0