选择具有相同列A但不同列B的行

时间:2012-10-26 20:03:04

标签: sql sql-server sql-server-2008

ID          Zip        Room
----------- ---------- ------
317         94087      S105
318         94087      L603
1739        94404-1801 L603
1823        94401-2129 L603
1824        94401-2129 L603
2135        94404-1801 L603
2268        95136-1459 S604
2269        95136-1459 S604
3704        92673-6417 L402
4479        93454-9670 L402
4480        93454-9670 L402
4782        92395-4681 L402
4783        92395-4681 L402
4852        92886-4411 L402
4853        92886-4411 L402
4959        92673-6417 L402
5153        91773-4028 L402
5202        91773-4028 L402
5211        91765-2959 L402
5212        91765-2959 L402
5388        92336-0605 L402
5392        92336-0605 L402
5727        92870      L402
5728        92870      L402
5831        92557      L402
5916        92557      L402

如何选择具有相同拉链但不同房间的ID?

在上表中,我希望结果为:

ID          Zip        Room
----------- ---------- ------
317         94087      S105
318         94087      L603

使用SQL Server 2008

3 个答案:

答案 0 :(得分:10)

您可以使用EXISTS

SELECT ID, Zip, Room 
FROM dbo.Table t1
WHERE EXISTS
(
   SELECT 1 FROM dbo.Table t2
   WHERE t2.Zip = t1.Zip
   AND t2.Room <> t1.Room
)

答案 1 :(得分:5)

select * 
from table t1 
join table t2 
  on t1.Zip = t2.Zip and t1.Room <> t2.Room

答案 2 :(得分:0)

您可以使用窗口聚合函数MIN()MAX()来相应地获取每Room的最小值和最大值Zip,然后仅选择那些行的WITH MinMax AS ( SELECT *, MinRoom = MIN(Room) OVER (PARTITION BY Zip), MaxRoom = MAX(Room) OVER (PARTITION BY Zip) FROM atable ) SELECT ID, Zip, Room FROM MinMax WHERE MinRoom <> MaxRoom ; 行价值观不同。这里:

{{1}}