Question

我正在尝试按其状态对帖子进行计数。但是下面的查询给我这个错误。

错误消息：[42000] [1140]在没有GROUP BY的汇总查询中， SELECT列表的表达式＃1包含未聚合的列 'jquzntys.posts.id';这与 sql_mode = only_full_group_by

SELECT COUNT(p.id)  total,
       p1.authors   authors,
       p2.published published,
       p3.pending   pending,
       p4.scheduled scheduled,
       p5.draft     draft,
       p6.deleted   deleted
FROM posts p
         LEFT JOIN (
    SELECT id, COUNT(*) as `authors`
    FROM posts
    WHERE user_id = ?
) AS p1 ON p1.id = p.id
         LEFT JOIN (
    SELECT id, COUNT(*) as `published`
    FROM posts
    WHERE status = ?
) AS p2 ON p2.id = p.id
         LEFT JOIN (
    SELECT id, COUNT(*) as `pending`
    FROM posts
    WHERE status = ?
) AS p3 ON p3.id = p.id
         LEFT JOIN (
    SELECT id, COUNT(*) as `scheduled`
    FROM posts
    WHERE status = ?
) AS p4 ON p4.id = p.id
         LEFT JOIN (
    SELECT id, COUNT(*) as `draft`
    FROM posts
    WHERE status = ?
) AS p5 ON p5.id = p.id
         LEFT JOIN (
    SELECT id, COUNT(*) as `deleted`
    FROM posts
    WHERE status = ?
) AS p6 ON p6.id = p.id

没有需要分组的数据。为什么我会收到此错误？

Answer 1

嗯。。。这行得通，对我来说很有意义：

SELECT p.id, COUNT(*) as num_posts,
       p1.authors   authors,
       p2.published published,
       p3.pending   pending,
       p4.scheduled scheduled,
       p5.draft     draft,
       p6.deleted   deleted
FROM posts p . . .
GROUP BY p.id, p1.authors, p2.published, p3.pending, p4.scheduled, p5.draft, p6.deleted

但是，我想我会像其他人一样将聚合纳入其中：

SELECT p.id, p.num_posts,
       p1.authors   authors,
       p2.published published,
       p3.pending   pending,
       p4.scheduled scheduled,
       p5.draft     draft,
       p6.deleted   deleted
FROM (SELECT p.id, COUNT(*) as cnt
      FROM posts p
      GROUP BY p.id
     ) p LEFT JOIN
     . . .

话说回来，这似乎很奇怪。 id在posts中不是唯一的吗？

编辑：

或者，我认为这可能是您真正想要的：

SELECT COUNT(*) as num_posts,
       SUM(p1.authors) as authors,
       SUM(p2.published) as published,
       SUM(p3.pending) as pending,
       SUM(p4.scheduled) as scheduled,
       SUM(p5.draft) as draft,
       SUM(p6.deleted) as deleted
FROM posts p . . .

Answer 2

如果您使用COUNT()，但未指定GROUP BY子句，则该子句仍符合汇总查询的条件，但整个结果将被视为一个“组”。您可能会认为您没有分组，因为您没有GROUP BY子句，但是在使用COUNT()时它是隐式的。

似乎您想统计每种状态的帖子数，并显示（任何状态的）帖子总数，还显示特定用户的帖子数。

我建议通过两个独立的查询来做到这一点：

SELECT COUNT(*) as count
FROM posts
WHERE user_id = ?;

SELECT status, COUNT(*) as count
FROM posts
GROUP BY status WITH ROLLUP;

第二个查询的优点是，如果将来有任何新的状态值出现，则不必重写查询。 WITH ROLLUP为您提供结果集最后一行的总数（在该行上，status为NULL）。

并非每个任务都需要在单个SQL查询中实现。有时将查询分为两个查询可以使两个查询更加简单明了。

在没有GROUP BY的聚合查询中，SELECT列表的表达式＃1包含非聚合列'jquzntys.posts.id'

2 个答案: