Question

当我尝试此查询时：

SELECT * 
FROM sds_posts 
WHERE topic_id = '2439' 
  AND author = ANY (SELECT mid 
                    FROM sds_actions 
                    WHERE whoami = '710' AND type = 'block')
  AND status = '1' 
  AND deleted = '0' 
ORDER BY 
  id ASC 
LIMIT 50

它正常工作。

但我需要这个：

SELECT * 
FROM sds_posts 
WHERE topic_id = '2439' 
  AND author <> ANY (SELECT mid 
                    FROM sds_actions 
                    WHERE whoami = '710' AND type = 'block')
  AND status = '1' 
  AND deleted = '0' 
ORDER BY 
  id ASC 
LIMIT 50

这次查询必须选择第一个查询的相反，但它只是选择所有作者。我尝试了!=和NOT IN，但结果是一样的。

为什么？为什么<>无法按预期工作？

Answer 1

我认为会改变

and author = any...
to
and NOT author = any...

会起作用......但如果没有，那么我会尝试做一个左连接并寻找null。由于作者是＆＃34; mid＆＃34;从sds_actions中，我会把它写成......

SELECT 
      sp.* 
   FROM 
      sds_posts sp
         LEFT JOIN sds_actions sa
            on sp.author = sa.mid
           AND sa.whoami = '710'
           AND sa.type = 'block'
   WHERE 
          sp.topic_id = '2439' 
      AND sp.status = '1' 
      AND sp.deleted = '0' 
      AND sa.mid IS NULL
   ORDER by 
      sp.id ASC 
   LIMIT 50

Answer 2

您可以尝试将author = any(...)更改为author IN (...)

并将author <> any(...)更改为author NOT IN (...)

mysql不等于运算符不工作

2 个答案: