Question

我想显示第2阶段最新记录的减法和第3阶段的最新记录这是表结构：

   ------------------------------------
      name |stage| date
   ------------------------------------
    | ABC  | 2   | 2016-5-07
    | ABC  | 2   | 2016-6-08  
    | ABC  | 3   | 2016-6-07
    | ABC  | 3   | 2016-12-08 
    | PQR  | 2   | 2016-6-07
    | PQR  | 2   | 2016-8-12  
    | PQR  | 3   | 2016-9-07
    | PQR  | 3   | 2016-10-08

我想从2016-12-08日期减去日期2016-6-08，其中名称是ABC，它们将是183天从2016-10-08减去2016-8-12，名称为PQR

输出应该像

一样

| name | Date |
 ---------------
| ABC  | 183  |
| PQR  | 61   |

Answer 1

WITH cte AS (
    SELECT t.[name], t.[stage], t.[date]
    FROM
    (
        SELECT [name], [stage], [date],
               ROW_NUMBER() OVER (PARTITION BY [name], [stage] ORDER BY [date] DESC) rn
    ) t
    WHERE t.rn = 1
)

SELECT [name], DATEDIFF(day, MIN(t.[date]), MAX(t.[date])) AS [Date]
FROM cte t
GROUP BY [name]

Answer 2

SELECT [NAME], DATEDIFF(DAY, MIN(x.[DATE]), MAX(x.[DATE])) AS [DATE]
FROM
    (SELECT T.[NAME], T.[STAGE], T.[DATE]
    FROM
    (
        SELECT [NAME], [STAGE], [DATE],
               ROW_NUMBER() OVER (PARTITION BY [NAME], [STAGE] ORDER BY [DATE] DESC) RN FROM #TABLE3
    ) T
    WHERE T.RN = 1
)X
GROUP BY [NAME]

或

SELECT NAME, DATEDIFF(DD,MIN([DATE]),MAX([DATE])) AS [DATE] FROM 
(SELECT NAME, STAGE, MAX([DATE]) AS [DATE] FROM #TABLE3 GROUP BY NAME, STAGE) C
 GROUP BY NAME;

Answer 3

with cte as (
select name, stage, max([date]) as [date] from #YourTable group by name, stage
)
select name, datediff(dd,min([date]),max([date])) as [Date] from cte group by name;

Answer 4

其他答案是减去忽略阶段的最小和最大日期，而不是每个阶段的最大日期。试试这个：

SELECT s2.name, DATEDIFF(day, MAX(s2.date), MAX(s3.date)) days
  FROM MyTable s2 JOIN MyTable s3 ON s2.name = s3.name
 WHERE s2.stage = 2 AND s3.stage = 3
 GROUP BY s2.name;

JOIN将表的两个副本组合成一个大表，该表对原始表中具有相同名称的两行的每个组合都有一行。

 name | s2.stage |  s2.date   | s3.stage |  s3.date
------+----------+------------+----------+------------
 ABC  |        2 | 2016-06-08 |        2 | 2016-06-08
 ABC  |        2 | 2016-06-08 |        3 | 2016-06-07
 ABC  |        2 | 2016-06-08 |        3 | 2016-12-08
 ABC  |        2 | 2016-06-08 |        2 | 2016-05-07
 ABC  |        3 | 2016-06-07 |        2 | 2016-06-08
 ABC  |        3 | 2016-06-07 |        3 | 2016-06-07
 ABC  |        3 | 2016-06-07 |        3 | 2016-12-08
 ABC  |        3 | 2016-06-07 |        2 | 2016-05-07
 ABC  |        3 | 2016-12-08 |        2 | 2016-06-08
 ABC  |        3 | 2016-12-08 |        3 | 2016-06-07
 ABC  |        3 | 2016-12-08 |        3 | 2016-12-08
 ABC  |        3 | 2016-12-08 |        2 | 2016-05-07
 ABC  |        2 | 2016-05-07 |        2 | 2016-06-08
 ABC  |        2 | 2016-05-07 |        3 | 2016-06-07
 ABC  |        2 | 2016-05-07 |        3 | 2016-12-08
 ABC  |        2 | 2016-05-07 |        2 | 2016-05-07
 PQR  |        2 | 2016-08-12 |        2 | 2016-08-12
 PQR  |        2 | 2016-08-12 |        3 | 2016-09-07
 PQR  |        2 | 2016-08-12 |        3 | 2016-10-08
 PQR  |        2 | 2016-08-12 |        2 | 2016-06-07
 PQR  |        3 | 2016-09-07 |        2 | 2016-08-12
 PQR  |        3 | 2016-09-07 |        3 | 2016-09-07
 PQR  |        3 | 2016-09-07 |        3 | 2016-10-08
 PQR  |        3 | 2016-09-07 |        2 | 2016-06-07
 PQR  |        3 | 2016-10-08 |        2 | 2016-08-12
 PQR  |        3 | 2016-10-08 |        3 | 2016-09-07
 PQR  |        3 | 2016-10-08 |        3 | 2016-10-08
 PQR  |        3 | 2016-10-08 |        2 | 2016-06-07
 PQR  |        2 | 2016-06-07 |        2 | 2016-08-12
 PQR  |        2 | 2016-06-07 |        3 | 2016-09-07
 PQR  |        2 | 2016-06-07 |        3 | 2016-10-08
 PQR  |        2 | 2016-06-07 |        2 | 2016-06-07

WHERE s2.stage=2 AND s3.stage=3将其限制为s2列对应于阶段2且s3列对应于阶段3的行：

 name | s2.stage | s2.date    | s3.stage |    s3.date
------+----------+------------+----------+------------
 ABC  |        2 | 2016-05-07 |        3 | 2016-06-07
 ABC  |        2 | 2016-05-07 |        3 | 2016-12-08
 ABC  |        2 | 2016-06-08 |        3 | 2016-06-07
 ABC  |        2 | 2016-06-08 |        3 | 2016-12-08
 PQR  |        2 | 2016-06-07 |        3 | 2016-09-07
 PQR  |        2 | 2016-06-07 |        3 | 2016-10-08
 PQR  |        2 | 2016-08-12 |        3 | 2016-09-07
 PQR  |        2 | 2016-08-12 |        3 | 2016-10-08

然后，我们使用MAX(...)函数和GROUP BY name子句为每个名称取每个日期字段的最大值，并使用DATEDIFF函数减去它们。

请注意，您应该使用您正在使用的特定数据库标记问题;上述SQL在Microsoft SQL Server中有效，但其他数据库的日期差异不同，并且支持JOIN ... USING语法，可以略微简化查询。例如，这就是Postgres的样子：

SELECT name, MAX(s3.date) - MAX(s2.date) days
  FROM MyTable s2 JOIN MyTable s3 USING (name)
 WHERE s2.stage = 2 AND s3.stage = 3
 GROUP BY name;

如何使用sql查询获取最新的日期记录，从而减去同一列中的两行值？

4 个答案: