mysql - MySQL如何优化一个NOT IN查询？

Question

我在 mysql 上有以下查询，查询逻辑是正确的，但是问题是因为有超过 10,000 多封求职者电子邮件和 24,000 多封访客电子邮件需要很长时间才能执行，有没有更好的方法来做到这一点？

SELECT g.email, g.name
FROM   guest g
WHERE  g.type='guest' 
AND g.email NOT IN (SELECT email FROM seeker GROUP BY email)
GROUP  BY g.email

Answer 1

尝试这个：

SELECT
    g.email, g.name
FROM
    guest g
LEFT JOIN
    seeker s
ON
    s.email = g.email
WHERE
    g.type = 'guest'
AND
    s.email IS NULL
GROUP BY
    g.email;

http://sqlfiddle.com/#!2/d94bf/5

Answer 2

SELECT DISTINCT g.email, g.name
FROM   guest g
WHERE  g.type='guest' 
AND NOT EXISTS (SELECT 1 FROM seeker s WHERE g.email = s.email)

并确保您在 seeker.email、guest.type、guest.email 上有一个索引，如果列NOT NULL在上面，那就太棒了。

Answer 3

您不需要按内部查询分组。您可以改为添加 DISTINCT。

SELECT g.email, g.name
FROM   guest g
WHERE  g.type='guest' 
       AND g.email NOT IN (SELECT DISTINCT email FROM seeker)
GROUP  BY g.email

即使这样也行

SELECT g.email, g.name
FROM   guest g left outer join seeker s on g.email = s.email
WHERE  g.type='guest' 
       AND s.email is null
GROUP  BY g.email

您的查询中会有很多字符串比较，如果您在表中索引电子邮件会有所帮助，尤其是。寻求者。

---

此外，避免使用未聚合且不存在于 GROUP BY 中的 SELECT 列。结果是不确定的。

服务器可以从每个组中自由选择任何值，因此除非它们相同，否则选择的值是不确定的。此外，从每个组中选择值不会受到添加 ORDER BY 子句的影响。

更多在手册中。

Answer 4

首先，对于您的查询，您不需要group by：

SELECT g.email, g.name
FROM   guest g
WHERE  g.type = 'guest' AND g.email NOT IN (SELECT email FROM seeker)
GROUP  BY g.email

这可能就足够了。使用上的索引seeker(email)，以下内容应该可以优化：

SELECT g.email, g.name
FROM   guest g
WHERE  g.type = 'guest' AND
       not exists (SELECT 1 FROM seeker where seeker.email = g.email)
GROUP  BY g.email;

如果您在大多数表中都有很多重复项email，那么我不会推荐这种left join方法。

Answer 5

SELECT DISTINCT g.email, g.name
FROM   guest g
LEFT OUTER seeker s ON s.email = g.email 
WHERE  g.type='guest' AND s.email IS NULL