编写一个SQL查询，检索列中具有相同值的所有不同行，其中这些行的计数大于1

基本上，我需要找到超过一个名称的所有ID 所以输出应该是

 (2,a)(2,b)(4,a)(4,b)(4,c) 

身份证名称 1A 2A 2 b 3 c 3 c 4A 4 b 4 c

我有个问题要问

select * from (select distinct * from test) t group by t.id having count(*) > 1

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但我认为可能有一个更简单的解决方案，所以是的，有一个简单的解决方案吗，如果有，是什么？提前感谢。

以下是一些您可以尝试的东西：

Select
  Id,
  Name
FROM test 
Group By Id, Name
Having Count(Distinct(Name)) > 1

select id, name
from T
where exists (
  select * from T as Tcopy
  where Tcopy.id = T.id
  and Tcopy.name <> T.name
);

with IDmultipleNames(id) as (
  select id
  from T
  group by id
  having min(name) <> max(name)
)
  select id, name
  from T
  where T.id in (
    select id from IDmultipleNames
  )

---

以下是您可以尝试的几件事情：

select id, name
from T
where exists (
  select * from T as Tcopy
  where Tcopy.id = T.id
  and Tcopy.name <> T.name
);

with IDmultipleNames(id) as (
  select id
  from T
  group by id
  having min(name) <> max(name)
)
  select id, name
  from T
  where T.id in (
    select id from IDmultipleNames
  )

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Thanx对于输入，我在Sybase中尝试过，删除了Groupby works中的名称，Thanx.Thanx对于输入，我在Sybase中尝试过，删除了Groupby works中的名称，Thanx。