Question

我有如下查询：

SELECT
    c.id, 
c.user,
c1.user,
c2.user
FROM (
    SELECT 
    id,
    user
    FROM 
        table_x
) c
INNER JOIN table_x c1 ON c.id = c1_id AND c.user = 'steve'
INNER JOIN table_x c2 ON c.id = c2.id AND c1.user = 'rob'
INNER JOIN table_x c3 ON c.id = c3.id AND c2.user LIKE 'r%'
GROUP BY c.id, c.user, c1.user, c2.user

它可以产生如下结果集：

id | user | user | user
1    steve  rob    rob52
1    steve  rob    rob

我需要结果集不包括跨越两列的用户不唯一的第二行。有没有办法检查这个，而不使用where子句来检查列的每个单独组合？，因为当结果集跨越到6列之类的东西时，检查它将是太多了。

结果集也可以返回：

id | user | user | user
1    rob    steve  rob

因此，在加入时与<>或!=运算符的比较不会捕获这样的行，但可能会在上面的结果集中排第2行。

由于

Answer 1

您可以使用distinct on

SELECT DISTINCT ON (c.id, c.user, c1.user)
    c.id, 
c.user,
c1.user,
c2.user
FROM (
    SELECT 
    id,
    user
    FROM 
        table_x
) c
INNER JOIN table_x c1 ON c.id = c1_id AND c.user = 'steve'
INNER JOIN table_x c2 ON c.id = c2.id AND c1.user = 'rob'
INNER JOIN table_x c3 ON c.id = c3.id AND c2.user LIKE 'r%'
GROUP BY c.id, c.user, c1.user, c2.user

通过这种方式，您只能获得distinct on子句

中提到的每个不同组合列的1条记录

Answer 2

只是出了什么问题：

WHERE c.user <> c1.user AND  c1.user <> c2.user AND c2.user <> c.user

这似乎完全符合您的要求。我想你是在思考你的问题......

删除多列中具有重复值的行

2 个答案: