查找幾列重複專屬ID列

Question

我已經發現了很多關於如何找到重複的，包括PK列或不專注於它，因爲這答案：

如果你有一個名爲T1的表，並且列是c1，c2和c3，則此查詢將顯示重複值。

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

但更常見的要求是獲得具有相同c1，c2，c3值的所有重複項的ID。

，所以我需要以下什麼行不通，因爲ID，必須綜合：

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

（將所有重複的ID必須是不同的，但列必須等於）

編輯 ：

謝謝大家。我總是驚訝人們在Stackoverflow上給出了很好的答案！

Answer 1

這裏有很多版本，但我想我想出了一個新的。

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

我不完全理解你的問題，但在這裏是一個不同風格的解決方案一拍：

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 4

要得到所有的是重複的行：

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（即保留一行並過濾其他重複行）使用此：

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 5

假設至少是SQL 2005的CTE：

;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 6

您可以重複的C1，C2，C3組合存儲到一個臨時表中，然後加入它得到的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