Question

我的数据与此类似。

    ID        Age
    ------------
    101       60  
    102       40  
    102       40  
    103       25  
    103       35  
    104       28  
    104       28  
    104       28

需要sql只输出具有相同id但不同年龄值的行。

    ID        Age
    ------------
    103       25  
    103       35

对此有任何帮助表示赞赏！

Answer 1

如果你只想要id：

 select id, count(distinct age) nb
 from yourtable
 group by id
 having count(distinct age)>1

如果您也想要年龄，请再次加入您的表格

 with doubleage as (
 select id, count(distinct age) nb
 from yourtable
 group by id
 having count(distinct age)>1
 )
 select f1.* from yourtable f1 inner join doubleage f2 on f1.id=f2.id

Answer 2

其他解决方案

select * from yourtable f1
where exists
(
select * from yourtable f2
where f1.id=f2.id and f1.age<>f2.age
)

Answer 3

其他解决方案

select * from yourtable f1
inner join lateral
(
     select count(*) nb from yourtable f2
     where f1.id=f2.id and f1.age<>f2.age
) f3 on 1=1
where f3.nb>0

Answer 4

SELECT * FROM F1 ID在哪里（ SELECT F1.id 来自F1 GROUP BY F1.ID 有计数（不同的F1.ID）＆gt; 1）

这将为所有Id（s）和Age提供重复记录
内部查询将帮助您只获取ID

Answer 5

如果我正确理解了这个问题并且你只有一张桌子，那么这就是你要找的东西：

select * from tablename group by ID where count(ID)>1

输出相同：

ID        Age
------------
103       25  
103       35

Answer 6

SELECT
     [a].[id]
    ,[a].[age]     
FROM
    @tbl    AS  [a]
INNER JOIN
    @tbl    AS  [b] 
ON
        [a].[id]    =   [b].[id]
    AND [a].[age]   <>  [b].[age]
GROUP BY
     [a].[id]
    ,[a].[age];

P.S. if you have data like:
     (101, 60)  
    ,(102, 40)  
    ,(102, 40) 
    ,(103, 25)  
    ,(103, 35)
    ,(103, 35)
    ,(104, 28)  
    ,(104, 28)  
    ,(104, 28)
Result is:
    ,(103, 25)  
    ,(103, 35)
Not:
    ,(103, 25)  
    ,(103, 35)
    ,(103, 35)
Keep this in mind...

需要与分组列相关的sql

6 个答案: