在同一个mysql查询中多次使用一个子查询结果答案

【问题标题】：Using one sub query result multiple times in the same mysql query在同一个mysql查询中多次使用一个子查询结果
【发布时间】：2017-03-30 13:38:05
【问题描述】：

如何在同一个查询中多次使用一个子查询的结果集

SELECT
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) be ON joa.referred_by = be.id
) AS applicationcount,
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) bf ON joa.referred_by = bf.id
    AND joa.admin_review = '3'
    AND joa.rejection_reason = 'Admin rejected your game'
) AS admin_reject,
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) bg ON joa.referred_by = bg.id
    AND joa. STATUS = '5'
    AND joa.admin_review = '2'
) AS employer_reject,
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) bd ON joa.referred_by = bd.id
    AND joa.admin_review = '1'
) AS admin_review,
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) bc ON joa.referred_by = bc.id
    AND joa.admin_review = '5'
) AS accountmanager_review,
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) ba ON joa.referred_by = ba.id
    AND joa.admin_review = '6'
) AS rp_review,
(
    SELECT
        COUNT(joa.id)
    FROM
        game_applied joa
    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
        JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = 1
    ) bh ON joa.referred_by = bh.id
    AND joa.admin_review = '2'
    AND (
        joa. STATUS = '' || joa. STATUS = 1 || joa. STATUS = 2 || joa.    STATUS = 3 || joa. STATUS = 4
    )
) AS other_status 
  FROM
game_applied ja
JOIN user_user u ON u.id = ja.applied_recruiter_id
INNER JOIN (
SELECT
    jrmm.id
FROM
    game_refer_to_member jrmm
    JOIN game_refer jrr ON jrr.id = jrmm.rid
AND jrr.referby_user_id = 2551
AND jrmm. STATUS = '1'
) bn ON ja.referred_by = bn.id
GROUP BY
applicationcount

如何在同一个查询中多次使用一个子查询的结果集

子查询在这个查询中使用了多次到一次使用

(
SELECT
    jrmm.id
FROM
    game_refer_to_member jrmm
JOIN game_refer jrr ON jrr.id = jrmm.rid
AND jrr.referby_user_id = 2551
AND jrmm. STATUS = '1'
  ) bn ON ja.referred_by = bn.id

【问题讨论】：

标签： mysql sql select join subquery

【解决方案1】：

试试这个：

SELECT COUNT(joa.id) AS applicationcount, 
         SUM(joa.admin_review = '3' AND joa.rejection_reason = 'Admin Rejected your resume') AS admin_reject, 
         SUM(joa.STATUS = '5' AND joa.admin_review = '2') AS employer_reject, 
         SUM(joa.admin_review = '1') AS admin_review, 
         SUM(joa.admin_review = '5') AS accountmanager_review, 
         SUM(joa.admin_review = '6') AS rp_review, 
         SUM(joa.admin_review = '2' AND joa.STATUS != '5') AS other_status, 
FROM game_refer_to_member jrmm 
INNER JOIN game_refer jrr ON jrr.id = jrmm.rid AND jrr.referby_user_id = 2551 
INNER JOIN game_applied joa ON jrmm.id  = joa.referred_by  
INNER JOIN user_user u ON u.id = joa.applied_recruiter_id
WHERE jrmm.STATUS = '1'

【讨论】：

【解决方案2】：

好的...我将尝试首先弄清楚查询的作用。我将替换以下所有实例：

SELECT
    COUNT(joa.id)
FROM
    game_applied joa
INNER JOIN (
    SELECT
        jrmm.id
    FROM
        game_refer_to_member jrmm
    JOIN game_refer jrr ON jrr.id = jrmm.rid
    AND jrr.referby_user_id = 2551
    AND jrmm. STATUS = '1'

用这句话：

( SELECT count(*) FROM joa JOIN (subselect))

...为了清楚起见。

我还删除了 GROUP BY，因为它没有用和/或被误导，除非你能解释为什么它在那里。

我假设 ja.applied_recruiter_id 是一个外键，这意味着......

JOIN user_user u ON u.id = ja.applied_recruiter_id

...总是返回一行。由于实际未选择 user_user 中的列，因此可以删除此连接。现在，这部分：

    INNER JOIN (
        SELECT
            jrmm.id
        FROM
            game_refer_to_member jrmm
            JOIN game_refer jrr ON jrr.id = jrmm.rid
        AND jrr.referby_user_id = 2551
        AND jrmm. STATUS = '1'
    ) bn ON ja.referred_by = bn.id

...目前尚不清楚这是做什么的。由于子选择与先前查询中的子选择相同，因此它不太可能过滤整个查询返回的行。我想说它的唯一效果是无用地重复行，这解释了为什么有一个 GROUP BY... 所以，它消失了。

我们得到：

SELECT
( SELECT count(*) FROM joa JOIN (subselect)) be ON joa.referred_by = be.id ) AS applicationcount,
( SELECT count(*) FROM joa JOIN (subselect)) bf ON joa.referred_by = bf.id
    AND joa.admin_review = '3'
    AND joa.rejection_reason = 'Admin rejected your game'
) AS admin_reject,
( SELECT count(*) FROM joa JOIN (subselect)) bg ON joa.referred_by = bg.id
    AND joa. STATUS = '5'
    AND joa.admin_review = '2'
) AS employer_reject,
( SELECT count(*) FROM joa JOIN (subselect)) bd ON joa.referred_by = bd.id
    AND joa.admin_review = '1'
) AS admin_review,
( SELECT count(*) FROM joa JOIN (subselect)) bc ON joa.referred_by = bc.id
    AND joa.admin_review = '5'
) AS accountmanager_review,
( SELECT count(*) FROM joa JOIN (subselect)) ba ON joa.referred_by = ba.id
    AND joa.admin_review = '6'
) AS rp_review,
( SELECT count(*) FROM joa JOIN (subselect)) bh ON joa.referred_by = bh.id
    AND joa.admin_review = '2'
    AND (joa. STATUS = '' || joa. STATUS = 1 || joa. STATUS = 2 || joa.    STATUS = 3 || joa. STATUS = 4)
) AS other_status 
  FROM
game_applied ja

...并且，使用与 Sarhash 相同的逻辑，我们将其简化为：

SELECT COUNT(joa.id) AS applicationcount, 
     SUM(joa.admin_review = '3' AND joa.rejection_reason = 'Admin Rejected your resume') AS admin_reject, 
     SUM(joa.STATUS = '5' AND joa.admin_review = '2') AS employer_reject, 
     SUM(joa.admin_review = '1') AS admin_review, 
     SUM(joa.admin_review = '5') AS accountmanager_review, 
     SUM(joa.admin_review = '6') AS rp_review, 
     SUM(joa.admin_review = '2' AND joa.STATUS != '5') AS other_status, 
FROM game_refer_to_member jrmm 
INNER JOIN game_refer jrr ON jrr.id = jrmm.rid 
INNER JOIN game_applied joa ON jrmm.id  = joa.referred_by  
WHERE jrmm.STATUS = '1' AND jrr.referby_user_id = 2551

（这是相同的，减去对 USER 的无用连接和在 WHERE 中的清理，谢谢 Sarhash，你得到了所有的功劳）。

【讨论】：

谢谢@peufeu

【解决方案3】：

我使用CREATE TEMPORARY TABLE 找到了一个可行的解决方案。我建议在同一个会话中背靠背运行两个查询。首先，从要多次引用的子查询中创建一些临时表：

CREATE
TEMPORARY TABLE temp_table1
  (SELECT COUNT(joa.id)
   FROM game_applied);

CREATE
TEMPORARY TABLE new_table
  (SELECT jrmm.id
   FROM game_refer_to_member jrmm
   JOIN game_refer jrr ON jrr.id = jrmm.rid
   AND jrr.referby_user_id = 2551
   AND jrmm. STATUS = '1');

使用这些临时表名对原始查询运行简单的查找和替换后，新查询将如下所示：

SELECT
  (SELECT *
   FROM temp_table1 joa
   INNER JOIN temp_table2 be ON joa.referred_by = be.id) AS applicationcount,

  (SELECT *
   FROM temp_table1 joa
   INNER JOIN temp_table2 bf ON joa.referred_by = bf.id
   AND joa.admin_review = '3'
   AND joa.rejection_reason = 'Admin rejected your game') AS admin_reject,

  (SELECT *
   FROM temp_table1 joa
   INNER JOIN temp_table2 bg ON joa.referred_by = bg.id
   AND joa. STATUS = '5'
   AND joa.admin_review = '2') AS employer_reject,

  (SELECT *
   FROM temp_table1 joa
   INNER JOIN temp_table2 bd ON joa.referred_by = bd.id
   AND joa.admin_review = '1') AS admin_review,

  (SELECT *
   FROM temp_table1 joa
   INNER JOIN temp_table2 bc ON joa.referred_by = bc.id
   AND joa.admin_review = '5') AS accountmanager_review,

  (SELECT *
   FROM temp_table1 joa
   INNER JOIN temp_table2 ba ON joa.referred_by = ba.id
   AND joa.admin_review = '6') AS rp_review,

  (SELECT *
   FROM temp_table1 joa
   INNER JOIN
     (SELECT jrmm.id
      FROM game_refer_to_member jrmm
      JOIN game_refer jrr ON jrr.id = jrmm.rid
      AND jrr.referby_user_id = 2551
      AND jrmm. STATUS = 1) bh ON joa.referred_by = bh.id
   AND joa.admin_review = '2'
   AND (joa. STATUS = '' || joa. STATUS = 1 || joa. STATUS = 2 || joa. STATUS = 3 || joa. STATUS = 4)) AS other_status
FROM game_applied ja
JOIN user_user u ON u.id = ja.applied_recruiter_id
INNER JOIN temp_table2 bn ON ja.referred_by = bn.id
GROUP BY applicationcount

我不保证此查询没有错误，因为我没有您的数据库，因此无法轻松测试它，但这至少概述了我认为应该改进的体面策略性能，并且至少使查询大小减少了一半。

【讨论】：