sql - 跨多个列删除具有重复值的行

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

我需要结果集不包含跨两列的用户不唯一的第二行。有没有一种方法可以在不使用 where 子句检查每个单独的列组合的情况下进行检查？因为当结果集跨越 6 列之类的东西时，要检查的东西太多了。

结果集也可能返回为:

id | user | user | user
1    rob    steve  rob

所以与 <> 进行比较或 !=连接时的运算符不会捕获这样的行，但可以捕获上述结果集中的第 2 行。

谢谢

Answer 1

你可以使用 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

这样，对于 distinct on 子句中提到的列的每个不同组合，您只会获得 1 条记录