Question

我们有以下SQL Server 2008示例数据库表，显示每个员工在每个支付期间的工资信息（每周工资为52或53周，在英国纳税年度之后，因此第1周的工资期为4月6日及之后）。

我将样本限制在一个支付期间的一名员工，实际数据可以追溯到很多年。

我需要在运行查询后的最后12周工资中为每位员工生成总计。

+------------+------------+---------+-------+
| EMPLOYEEID | PAYELEMENT | AMOUNT  | HOURS |
+------------+------------+---------+-------+
|     160062 | 1.0 Basic  | 2724.64 |   468 |
+------------+------------+---------+-------+

然而，由于以下原因，我有一些问题可以追溯到12个不同时期......

期间1有两个条目
在度假时没有记录任何时期，例如第30周，缺少38和39。在这种情况下，它需要返回直到使用12个记录周期。
从第52期到第1期结束年终

我尝试使用每位员工的最后12条记录，但上面的第1项只计算了11个期间。

我也尝试在paydate上使用日期差异，但上面的第2项导致时间段丢失。

我是否需要为每位员工添加一个索引以显示12个单独的句点？

+------------+------------+------+--------+--------+-------+------+------------------+
| EMPLOYEEID | PAYELEMENT | YEAR | PERIOD | AMOUNT | HOURS | RATE |     PAYDATE      |
+------------+------------+------+--------+--------+-------+------+------------------+
|     160062 | 1.0 Basic  | 2017 |     29 | 311.22 |    39 | 7.98 | 20/10/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     31 | 311.22 |    39 | 7.98 | 03/11/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     32 | 311.22 |    39 | 7.98 | 10/11/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     33 | 311.22 |    39 | 7.98 | 17/11/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     34 | 311.22 |    39 | 7.98 | 24/11/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     35 | 311.22 |    39 | 7.98 | 01/12/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     36 | 183.54 |    23 | 7.98 | 08/12/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     37 | 311.22 |    39 | 7.98 | 15/12/2016 00:00 |
|     160062 | 1.0 Basic  | 2017 |     40 | 311.22 |    39 | 7.98 | 05/01/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     41 | 311.22 |    39 | 7.98 | 12/01/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     42 | 311.22 |    39 | 7.98 | 19/01/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     43 | 311.22 |    39 | 7.98 | 26/01/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     44 | 311.22 |    39 | 7.98 | 02/02/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     45 | 311.22 |    39 | 7.98 | 09/02/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     46 | 311.22 |    39 | 7.98 | 16/02/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     47 | 311.22 |    39 | 7.98 | 23/02/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     48 | 127.68 |    16 | 7.98 | 02/03/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     49 | 311.22 |    39 | 7.98 | 09/03/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     50 | 247.38 |    31 | 7.98 | 16/03/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     51 | 311.22 |    39 | 7.98 | 23/03/2017 00:00 |
|     160062 | 1.0 Basic  | 2017 |     52 | 311.22 |    39 | 7.98 | 30/03/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      1 | 247.38 |    31 | 7.98 | 06/04/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      1 |      0 |     0 | 7.75 | 06/04/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      2 | 311.22 |    39 | 7.98 | 13/04/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      3 | 255.36 |    32 | 7.98 | 20/04/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      4 | 247.38 |    31 | 7.98 | 27/04/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      5 | 311.22 |    39 | 7.98 | 04/05/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      6 | 127.68 |    16 | 7.98 | 11/05/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      7 | 247.38 |    31 | 7.98 | 18/05/2017 00:00 |
|     160062 | 1.0 Basic  | 2018 |      8 | 277.31 | 34.75 | 7.98 | 25/05/2017 00:00 |
+------------+------------+------+--------+--------+-------+------+------------------+

Answer 1

你应该尝试：

;WITH Top12Periods AS (
    SELECT TOP 12 [YEAR], [PERIOD]
    FROM @employeeTable
    GROUP BY [YEAR], [PERIOD]
    ORDER BY [YEAR] DESC, [PERIOD] DESC
)
SELECT [EMPLOYEEID], [PAYELEMENT], SUM([AMOUNT]) AS TOTAL_AMOUNT, SUM([HOURS]) AS TOTAL_HOURS, AVG([RATE]) AS AVERAGE_RATE, MIN ([PAYDATE]) [MIN_PAYDATE]
FROM @employeeTable et
    JOIN Top12Periods p ON et.[YEAR] = p.[YEAR] AND et.[PERIOD] = p.[PERIOD]
GROUP BY [EMPLOYEEID], [PAYELEMENT]

Answer 2

这是另一种方法：

SELECT
    EmployeeID,
    PayElement,
    sum(Amount) as Amount,
    sum(Hours) as Hours
FROM
    (
    SELECT *
    ,dense_rank() OVER (PARTITION BY EmployeeId ORDER BY PayDate DESC) as rank
    FROM
        Pay
    ) ranked
WHERE
    rank <= 12
GROUP BY
    EmployeeID,
    PayElement

如果零值导致问题，则只需将WHERE Amount > 0（或者<> 0）添加到内部查询中。根据您的数据，您可能还需要将PayElement添加到Partition By子句（或内部WHERE子句）。

您可以在rextester.com/YDBL47689

对此进行测试

SQL Server：返回不同数量的句点

2 个答案: