Question

我有这个示例表，带有示例数据：

+-------------------+-----------------+------------+------------+
| OriginalBeginDate | OriginalEndDate | Start Date |  End Date  |
+-------------------+-----------------+------------+------------+
| 2015-06-01        | 2015-06-30      | 2015-08-01 | 2015-08-31 |
| 2015-06-01        | 2015-06-30      | 2015-09-01 | 2015-09-30 |
| 2015-06-01        | 2015-06-30      | 2015-10-01 | 2015-10-31 |
| 2015-06-01        | 2015-06-30      | 2015-11-01 | 2015-11-30 |
| 2015-06-01        | 2015-06-30      | 2015-12-01 | 2015-12-31 |
| 2015-07-01        | 2015-12-31      | 2015-08-01 | 2015-08-31 |
| 2015-07-01        | 2015-12-31      | 2015-09-01 | 2015-09-30 |
| 2015-07-01        | 2015-12-31      | 2015-10-01 | 2015-10-31 |
| 2015-07-01        | 2015-12-31      | 2015-11-01 | 2015-11-30 |
| 2015-07-01        | 2015-12-31      | 2015-12-01 | 2015-12-31 |
+-------------------+-----------------+------------+------------+

如果您注意到，则会有一组5＆＃39;开始日期和结束日期重复，因为＆＃39; OriginalBeginDate＆＃39;和＆＃39; OriginalEndDate＆＃39;有两套＆＃39;。

我需要只有一组最后两列（它们的不同值）。

这是输出需要的样子：

| OriginalBeginDate | OriginalEndDate | Start Date |  End Date  |
+-------------------+-----------------+------------+------------+
| 2015-06-01        | 2015-06-30      | 2015-08-01 | 2015-08-31 |
| 2015-06-01        | 2015-06-30      | 2015-09-01 | 2015-09-30 |
| 2015-06-01        | 2015-06-30      | 2015-10-01 | 2015-10-31 |
| 2015-06-01        | 2015-06-30      | 2015-11-01 | 2015-11-30 |
| 2015-06-01        | 2015-06-30      | 2015-12-01 | 2015-12-31 |
+-------------------+-----------------+------------+------------+

如果我跑

SELECT DISTINCT * FROM myTable

我得到了整张桌子。如果我跑

SELECT DISTINCT [START DATE], [END DATE] FROM myTable

我不能获得所有4列，我需要。

任何帮助将不胜感激！

Answer 1

只需添加一些MIN分组即可获取其他列的最低值，这看起来就像您所追求的那样：

SELECT MIN(OriginalBeginDate) as OriginalBeginDate, 
       MIN(OriginalEndDate) as OriginalEndDate, 
       [START DATE], 
       [END DATE] 
FROM myTable
GROUP BY [START DATE], [END DATE]

Answer 2

您可以尝试根据开始日期和结束日期对数据进行分组

 DECLARE @mytable TABLE
        (
          OriginalBeginDate DATE ,
          OriginalEndDate DATE NOT NULL ,
          Start_Date DATE NOT NULL ,
          End_Date DATE NOT NULL
        );

INSERT INTO @mytable(OriginalBeginDate,OriginalEndDate,Start_Date,End_Date) 
VALUES ('2015-06-01','2015-06-30','2015-08-01','2015-08-31')
     , ('2015-06-01','2015-06-30','2015-09-01','2015-09-30')
     , ('2015-06-01','2015-06-30','2015-10-01','2015-10-31')
     , ('2015-06-01','2015-06-30','2015-11-01','2015-11-30')
     , ('2015-06-01','2015-06-30','2015-12-01','2015-12-31')
     , ('2015-07-01','2015-12-31','2015-08-01','2015-08-31')
     , ('2015-07-01','2015-12-31','2015-09-01','2015-09-30')
     , ('2015-07-01','2015-12-31','2015-10-01','2015-10-31')
     , ('2015-07-01','2015-12-31','2015-11-01','2015-11-30')
     , ('2015-07-01','2015-12-31','2015-12-01','2015-12-31')

SELECT MIN(OriginalBeginDate), MIN(OriginalEndDate), Start_Date, End_Date
FROM @mytable
GROUP BY Start_Date, End_Date

Answer 3

您可以通过向查询添加条件来仅返回OriginalBeginDate值最低的行来完成此操作。

SELECT * 
FROM myTable
WHERE NOT EXISTS (SELECT TOP 1 OriginalBeginDate 
                  FROM myTable t2 
                  WHERE t2.[Start Date] = myTable.[Start Date]
                  AND t2.[End Date] = myTable.[End Date]
                  AND t2.OriginalBeginDate = myTable.OriginalBeginDate)

Answer 4

您可以尝试使用：

编辑：在订购之前使用分区......

SELECT tbl.*
FROM
(
SELECT ROW_NUMBER() OVER(PARTITION BY [Start Date] ORDER BY OriginalBeginDate) AS Inx1
      ,ROW_NUMBER() OVER(PARTITION BY [End Date] ORDER BY OriginalEndDate) AS Inx2
      ,*
FROM @myTable
) AS tbl
WHERE tbl.Inx1=1 OR tbl.Inx2=1

Answer 5

group by min（或max）的问题是可以从不同的行得到min 有了这些数据，它没有任何区别，但也许这会更安全

SELECT tbl.*
FROM
(  SELECT *, ROW_NUMBER() OVER(PARTITION BY [Start Date], [End Date] 
                               ORDER     BY [OriginalBeginDate], [OriginalEndDate]) AS rn
     FROM myTable
) AS tbl
WHERE tbl.rn = 1

选择具有特定不同列的整行

5 个答案: