Question

我在下面有以下查询。问题是我在执行时抛出了这个错误：

column "v4.id" must appear in the GROUP BY clause or be used in an aggregate function

这没有多大意义，因为我没有在聚合函数中使用该字段。关于这是为什么的任何想法？

SELECT          c0."id", 
                c0."assigned_user_id", 
                c0."assigned_team_id", 
                c0."last_activity", 
                c0."last_message", 
                c3."unread_count", 
                c0."closed", 
                v4."name", 
                v4."avatar", 
                v4."id", 
                Count(c0."id") OVER() 
FROM            "conversations" AS c0 
LEFT OUTER JOIN "apps"          AS a1 
ON              a1."id" = c0."app_id" 
LEFT OUTER JOIN "conversation_users" AS c5 
ON              c5."conversation_id" = c0."id" 
LEFT OUTER JOIN "users" AS u2 
ON              c5."user_id" = u2."id" 
LEFT OUTER JOIN "conversation_users" AS c3 
ON              c3."conversation_id" = c0."id" 
LEFT OUTER JOIN "visitors" AS v4 
ON              v4."id" = c0."visitor_id" 
WHERE           ( a1."id" = $1) 
AND             ( u2."id" = $2) 
AND             ( c3."user_id" = $3) 
GROUP BY        c0."id", 
ORDER BY        c0."last_activity" DESC ["ASnYW1-RgCl0I", "clt8ojxvk0000dp2fyvwq126", "clt8ojxvk0000dp2fyvwq126"]

Answer 1

完全为什么你想要group by中的列。从形式上讲，您应该在group by中确实拥有所有非聚合列：

group by c0."id", c0."assigned_user_id", c0."assigned_team_id",
         c0."last_activity", c0."last_message", c0."closed",
         c3."unread_count", 
         v4."name", v4."avatar", v4."id"

Postgres有一个扩展（由ANSI标准支持），让您可以替换group by中的主键或唯一键，并仍可访问其他列。我没有将它用于多个表，但这可能也有效：

group by c0."id",
         c3."unread_count", 
         v4."id"

此外，您的选择是使用窗口功能。你为什么在那里有over条款？这是一个聚合查询，因此在此上下文中没有意义。

字段未聚合错误所需的分组

1 个答案: