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”答案

【问题标题】：In aggregated query without GROUP BY, expression #1 of SELECT list contains nonaggregated column 'jquzntys.posts.id'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”
【发布时间】：2020-03-17 15:53:03
【问题描述】：

我正在尝试按状态计算帖子。但是下面的查询给了我这个错误。

错误消息：[42000][1140] 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';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？

【问题讨论】：

因为和sql_mode=only_full_group_by不兼容。
count() 是一个聚合函数。 id 列既不分组也不聚合
样本数据和期望的结果会有所帮助。

标签： mysql sql

【解决方案1】：

如果您使用COUNT() 但未指定GROUP BY 子句，它仍然符合聚合查询的条件，但整个结果被视为一个“组”。您可能认为您没有任何分组，因为您没有 GROUP BY 子句，但它隐含在您使用 COUNT() 中。

您似乎想计算每个状态的帖子数，并显示（任何状态的）帖子总数，还显示特定用户的帖子数。

我建议这样做，使用两个单独的查询：

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

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

第二个查询的好处是，如果以后出现任何新的状态值，你不必重写查询。 WITH ROLLUP 为您提供结果集最后一行的总数（在该行上，status 将为 NULL）。

并非每个任务都需要在单个 SQL 查询中实现。有时将查询拆分为两个查询可以让两个查询更简单、更清晰。

【讨论】：

【解决方案2】：

嗯。 . .这对我有效并且有意义：

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，我想我会像您对其他聚合一样包含聚合：

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
     . . .

话虽如此，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 . . .

【讨论】：