SQL 服务器 : return distinct number of periods

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周期内的支付信息(每周支付 52 或 53 周，在英国纳税年度之后，因此支付周期的第 1 周是 4 月 6 日及以后)。

我将样本限制为在一系列支付周期内的一名员工，实际数据可以追溯到很多年前。

我需要为每个员工生成从运行查询时起过去 12 周的工资总额。

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

但是，由于以下原因，我在回溯 12 个不同时期时遇到了一些问题......

1. 第 1 期有两个条目
2. 休假时没有月经记录，因此例如缺少第 30、38 和 39 周。在这种情况下，它需要返回直到使用 12 个记录周期。
3. 年末从第 52 期到第 1 期

我尝试使用每个员工的最后 12 条记录，但上面的项目 1 导致只计算 11 个周期。

我也尝试过在发薪日上使用日期差异，但上面的第 2 项会导致句点缺失。

我是否需要为每位员工添加一个索引以显示 12 个单独的时期？

+------------+------------+------+--------+--------+-------+------+------------------+
| 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

你应该试试:

;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]