返回多列不相同的所有行/列

时间:2013-11-21 19:52:27

标签: sql sql-server

我有一个查询,我想查找其中ActivityDate和TaskId同时有多个条目的行:

SELECT
    ActivityDate, taskId
FROM
    [DailyTaskHours]
GROUP BY
    ActivityDate, taskId
HAVING 
    COUNT(*) > 1

以上查询似乎有效。但是我希望所有列现在只返回两个(ActivityDate,taskId)。这不起作用:

SELECT *
FROM
    [DailyTaskHours]
GROUP BY
    ActivityDate, taskId
HAVING 
    COUNT(*) > 1

因为许多列不在group by子句中。我不希望任何列受到HAVING COUNT(*)>的影响。除ActivityDate之外的1,taskId。

我如何实现这一目标?

4 个答案:

答案 0 :(得分:2)

WITH sel as(
SELECT
    ActivityDate, taskId
FROM
    [DailyTaskHours]
GROUP BY
    ActivityDate, taskId
HAVING 
    COUNT(*) > 1
)
SELECT * 
      FROM [DailyTaskHours] d
           INNER JOIN sel ON d.ActivityDate = sel.ActivityDate AND d.taskId = sel.taskId

答案 1 :(得分:1)

SELECT t1.*
FROM
    [DailyTaskHours] t1
INNER JOIN (
  SELECT
      ActivityDate, taskId
  FROM
      [DailyTaskHours]
  GROUP BY
      ActivityDate, taskId
  HAVING 
      COUNT(*) > 1
) t2 ON (
  t1.ActivityDate = t2.ActivityDate AND
  t1.taskId = t2.taskId
)

答案 2 :(得分:0)

以下是显示以下查询的SQL Fiddle

SELECT DISTINCT m.* 
FROM 
(
  SELECT s.ActivityDate, s.taskId
  FROM DailyTaskHours s
  GROUP BY s.ActivityDate, s.taskId
  HAVING COUNT(*) > 1
) sub
JOIN DailyTaskHours m 
ON m.taskId = sub.taskId
AND m.ActivityDate = sub.ActivityDate

答案 3 :(得分:0)

-- fully functional example.
DECLARE @table TABLE ( ActivityDate DATE, TaskID INT);

SET NOCOUNT ON;
INSERT @table VALUES ('01/01/2013',1);
INSERT @table VALUES ('01/02/2013',1);
INSERT @table VALUES ('01/02/2013',2);
INSERT @table VALUES ('01/03/2013',1);
INSERT @table VALUES ('01/03/2013',2);
INSERT @table VALUES ('01/03/2013',5); -- duplicate date,taskid
INSERT @table VALUES ('01/03/2013',5); -- duplicate date,taskid
SET NOCOUNT OFF;

SELECT A.*
FROM @table A
    INNER JOIN (
        SELECT [ActivityDate], TaskId
        FROM @table
        GROUP BY [ActivityDate], TaskId
        HAVING Count(*) > 1
    ) AS B ON B.[ActivityDate]=A.ActivityDate AND B.TaskId=A.TaskId;
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