在 SQL 中按 90 天的时间段对用户进行分组

Question

我需要获取过去 4 周内每一周被解雇的最后 90 名员工。这是一个滚动报告。

员工表是这样的：

id      hired_date       term_date
3        07/1/2011        09/01/2011
4        07/18/2011        NULL    (NULL means still active) 
5        01/20/2009       08/23/2011
6        05/30/2011       8/22/2011
7        7/20/2011        7/23/2011

报告将采用这种格式。 ID #4,5 将被忽略，因为 #4 仍处于活动状态且 #5 在 90 天期限之前被雇用。

Week ending     Terminated employees hired within past 90 days
09/03/2011        2   --(id 3,7)
8/27/2011         2   --(id 6,7)
..
..

• 所以 9/03 的第一行是雇用的终止员工人数 2011 年 9 月 3 日过去 90 天（可追溯到 2011 年 6 月 5 日）。这不包括 id #6，因为员工是在 6 月 5 日之前被雇用的。
• 8/27 周的第二行相同，但范围是 8/27 到 5/29。

我有一个日期表，但它只包含 week_start_date、week_end_date 和 week_number。我是否需要创建一个包含 90 天的时间段？

我不知道如何仅在过去 90 天内为员工执行此操作，然后计算过去 4 周的每一周。

SQL Server 2008

编辑： 我想我很接近了。我现在正在测试它。 All_termed_employees 是任何日期范围内被解雇的员工的列表。 week 表现在包含与其关联的每个星期的 Ninety_begin_date 和 Ninety_end_date。

select wk.ninety_end, count(h.id)
FROM @weeks wk
LEFT JOIN all_termed_employees h 
    ON h.hire_date <= wk.ninety_end and h.hire_date >= wk.ninety_begin
          and .termination_date <= wk.ninety_end AND h.termination_date >= wk.ninety_begin
ORDER BY d.id
GROUP BY wk.ninety_end

Answer 1

;WITH n(n) AS 
      -- just 4 rows - makes it easy to extend to 5 weeks, 6 weeks, etc.
      ( SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 ),

d(dt) AS
      -- single row with the end of the current week
      -- this could be a variable but I get a lot of flack for not inlining
      ( SELECT dt = CONVERT(DATE, DATEADD(DAY, 
          7-DATEPART(WEEKDAY, CURRENT_TIMESTAMP), CURRENT_TIMESTAMP))),

w(dt) AS 
      -- get the end of each week based on the rows in n
      ( SELECT DATEADD(WEEK, -n.n, d.dt) FROM n CROSS JOIN d )

SELECT 
    w.dt, SUM(CASE 
    WHEN e.term_date >= DATEADD(DAY, -90, w.dt) 
    AND e.term_date  <  DATEADD(DAY, 1,   w.dt) 
    AND e.hired_date >= DATEADD(DAY, -90, w.dt) 
    AND e.hired_date <  DATEADD(DAY, 1,   w.dt)  
    THEN 1 ELSE 0 END)
FROM dbo.Employees AS e
CROSS JOIN w
GROUP BY w.dt
ORDER BY w.dt DESC; 

Answer 2

这样的事情怎么样？

EMP 模拟您的员工表。 PAST_FOUR_WEEKS 是一个简单的 4 行内存表，用于标识报告期间。然后我计算雇用日期和任期日期之间的差异，如果它小于或等于 90，我们将其计为 1（达到阈值）或 0）。然后，我将在给定时间段内达到的所有阈值相加

; WITH EMP (id, hired_date, term_date) AS
(
    select 3, CAST('2011-07-01' AS datetime), CAST('2011-09-01' AS datetime)
    union all select 4, '2011-06-18', null
    union all select 5, '01/20/2009','08/23/2011'
    union all select 6, '05/30/2011','8/22/2011'
    union all select 7, '7/20/2011','7/23/2011'
)
, PAST_FOUR_WEEKS (period, period_rank) AS
(
    -- magic goes here to determine end of week
    SELECT CURRENT_TIMESTAMP, 1
    UNION ALL SELECT dateadd(week, -1, CURRENT_TIMESTAMP) , 2
    UNION ALL SELECT dateadd(week, -2, CURRENT_TIMESTAMP) , 3
    UNION ALL SELECT dateadd(week, -3, CURRENT_TIMESTAMP) , 4
)
, HIRED_TERMED_DIFF AS
(
    SELECT
        *
    ,   DATEDIFF(d, E.hired_date, coalesce(E.term_date, '9999-12-31T23:59:59.997')) AS duration
    ,   CASE WHEN DATEDIFF(d, E.hired_date, coalesce(E.term_date, '9999-12-31T23:59:59.997')) < 91 THEN 1 ELSE 0 END AS threshold_met
    FROM
        EMP E 
)
SELECT
    PFW.period
,   SUM(HTD.threshold_met) AS [Terminated employees hired within past 90 days]
FROM 
    PAST_FOUR_WEEKS PFW
    LEFT OUTER JOIN
        HIRED_TERMED_DIFF HTD
        ON HTD.hired_date BETWEEN DATEADD(day, -90, PFW.period) AND PFW.period
GROUP BY
    PFW.period
ORDER BY
    1 DESC
,   2

样本输出

period                    Terminated employees hired within past 90 days
2011-09-01 14:46:29.243   2
2011-08-25 14:46:29.243   3
2011-08-18 14:46:29.243   3
2011-08-11 14:46:29.243   3