非常慢的NOT EXISTS查询问题

Question

有人可以告诉我，为什么以下查询需要1.3秒才能完成，是否有任何明显的问题？

有没有办法加快查询速度？

SELECT COUNT(DISTINCT jud.rel_id) AS count_result
    FROM exp_judging AS jud
    LEFT JOIN exp_submissions AS sub ON jud.rel_id = sub.id 
    WHERE (jud.judge_id != 781 OR jud.judge_id IS NULL)
    AND jud.pre = 1
    AND sub.member_group = 5
    AND NOT EXISTS (SELECT sub2.entry_id 
                            FROM exp_judging AS jud2
                    LEFT JOIN exp_submissions AS sub2 ON jud2.rel_id = sub2.id 
                    WHERE (jud2.judge_id = 781)
                    AND jud2.pre = 1
                    AND sub2.member_group = 5
                    AND jud2.rel_id = jud.rel_id)

Answer 1

我已经将整个查询重新编写为1）首先，选择exp_judging.rel_id，其中没有一行使用judge_id = 781，然后2）然后，对于这些rel_id来说，计数是从exp_judging表中获取。

修改：

SELECT 
    COUNT(DISTINCT jud0.rel_id)
FROM
    exp_judging AS jud0
    INNER JOIN (
        SELECT 
            jud.rel_id as rel_id,
            SUM(
                CASE jud.judge_id
                    WHEN 781 THEN 1
                    ELSE 0
                END) sum_judge_id
        FROM 
            exp_judging AS jud
            INNER JOIN exp_submissions AS sub
                ON jud.rel_id = sub.id 
        WHERE 
            jud.pre = 1
            AND sub.member_group = 5
        GROUP BY 
            jud.rel_id) judge_id_sums
        ON jud0.rel_id = judge_id_sums.rel_id 
WHERE 
    judge_id_sums.sum_judge_id = 0;

仅当exp_judging中存在exp_submissions（作为jud.rel_id）时，内部联接jud2.rel_id和exp_submissions才会计入sub2.id。但是，如果您确实要计算所有jud.rel_id（即使exp_submissions中不存在），则可以使用LEFT JOIN。

Answer 2

您似乎想要根据某些条件计算不同的jud.rel_id：

1. jud.pre = 1
2. sub.member_group = 5
3. 该组不包含法官781

回答一些问题，例如“有多少人”，“第5组中的法官”之前从未向法官781提交过申请？如果是这样，可能有一种更简单的方法来编写查询。

在任何情况下，您的查询都被重写，以便我可以阅读：

SELECT COUNT(DISTINCT jud.rel_id) AS count_result
FROM exp_judging jud LEFT JOIN
     exp_submissions sub
     ON jud.rel_id = sub.id 
WHERE (jud.judge_id != 781 OR jud.judge_id IS NULL) AND
      jud.pre = 1 AND
      sub.member_group = 5 AND
      NOT EXISTS (SELECT sub2.entry_id 
                  FROM exp_judging jud2 LEFT JOIN
                       exp_submissions sub2
                       ON jud2.rel_id = sub2.id 
                  WHERE jud2.judge_id = 781 AND
                        jud2.pre = 1 AND
                        sub2.member_group = 5 AND
                        jud2.rel_id = jud.rel_id
                 )

您希望使用索引来加速查询。我对此查询的索引的最佳猜测是：

exp_judging(judge_id, pre, rel_id)
exp_submissions(id, member_group)

Answer 3

尝试使用INNER JOIN。

INNER JOIN exp_submissions AS sub ON jud.rel_id = sub.id