Sql 返回多个列不不同的所有行/列

我有一个查询，希望在其中查找ActivityDate和TaskId同时具有多个条目的行：

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

上述查询似乎有效。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GROUPBY子句中。我不希望任何列受到计数（*）大于1（ActivityDate、taskId除外）的影响

如何实现这一点？

以下是显示以下查询的示例：

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

---

您在加入条款中缺少ActivityDate我提供了答案。这恰好与您在大约2000万年前删除的问题的答案完全相同。@Linger我不理解原始查询要求的不同结果。并且假设

dailytaskhurs

有一个主键，这是最佳实践，那么

DISTINCT

将没有效果。但是，当然，如果需要，可以很容易地添加

DISTINCT

。

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

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
)

-- 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;