Question

我找到了很多关于如何找到重复项的答案，包括PK-column或没有关注它：

如果您有一个名为T1的表，并且列是c1，c2和c3，那么此查询将显示重复的值。

SELECT C1, C2, C3, count(*)as DupCount
 from T1
 GROUP BY C1, C2, C3
 HAVING COUNT(*) > 1

但更常见的要求是获取具有相等c1，c2，c3值的所有重复项的ID。

所以我需要关注什么不起作用，因为必须聚合id：

SELECT ID
 from T1
 GROUP BY C1, C2, C3
 HAVING COUNT(*) <> 1

（所有重复项的ID必须不同，但列必须相等）

修改：

谢谢大家。我总是很惊讶人们在Stackoverflow上给出了极好的答案！

Answer 1

这里有很多版本，但我想我想出了一个新版本。

select *
from @T as T1
where exists (select *
              from @T as T2
              where
                T1.ID <> T2.ID and
                T1.C1 = T2.C1 and
                T1.C2 = T2.C2 and
                T1.C3 = T2.C3)

Answer 2

;WITH CTE
     AS (SELECT ID,
                C1,
                C2,
                C3,
                COUNT(*) OVER (PARTITION BY C1, C2, C3) AS Cnt
         FROM   T1)
SELECT ID,
       C1,
       C2,
       C3
FROM   CTE
WHERE  Cnt > 1

Answer 3

获取所有重复的行：

使用此：

WITH Dups AS
(
    SELECT *, 
           COUNT(1) OVER(PARTITION BY C1, C2, C3) AS CNT
      FROM T1  
)
SELECT * 
  FROM Dups
 WHERE CNT > 1

和唯一行（即保留一行并过滤其他重复行）使用：

WITH NoDups AS
(
    SELECT *, 
         ROW_NUMBER() OVER(PARTITION BY C1, C2, C3 ORDER BY ID) AS RN
      FROM T1  
)
SELECT * 
  FROM NoDups
WHERE RN = 1

Answer 4

假设CTE至少有SQL 2005：

;with cteDuplicates as (
    select c1, c2, c3
        from t1
        group by c1, c2, c3
        having count(*) > 1
)
select id
    from t1
        inner join cteDuplicates d
            on t1.c1 = d.c1
                and t1.c2 = d.c2
                and t1.c3 = d.c3

Answer 5

我并不完全理解你的问题，但这里有一个不同风格的解决方案：

select id
from t1 a
join t1 b on a.c1 = b.c2
join t1 c on b.c2 = c.c3
where a.id <> b.id and b.id <> c.id and a.id <> c.id

Answer 6

您可以将重复的C1，C2，C3组合存储到临时表中，然后将其加入以获取ID。

select C1, C2, C3
into #duplicates
from T1
group by C1, C2, C3
having count(*) > 1

select ID
from T1 t
inner join #duplicates d
    on  t.C1 = d.C1
    and t.C2 = d.C2
    and t.C3 = d.C3

查找多个列独有ID列的重复项

6 个答案: