Question

我曾尝试计算累积金额列，以便在每个月找出当前工作员工，但是按照上个月的数字而不是现在的员工。

表员工：

id    date_started     date_terminated
1      01-Apr-14       NULL
2      21-Apr-14       NULL
3      11-Apr-14       NULL
4      01-Apr-14       NULL
5      01-Apr-14       NULL
6      05-Apr-14       NULL
7      01-Apr-14       NULL
8      01-Apr-14       NULL
9      01-Apr-14       NULL
10     29-Apr-14       NULL
11     21-Apr-14       NULL
12     01-Apr-14       NULL
13     01-Apr-14       NULL
14     01-Apr-14       NULL
15     05-Aug-14       NULL
16     01-Oct-1        NULL
17     13-Oct-14       NULL
18     22-Oct-14       NULL
19     25-Oct-14       NULL
10     29-Oct-14       NULL

表格日期：它包含date列，其中包含从2011-Jan-01到当前日期的数据。

从我的查询中获得结果表：

+--------------------------------------------------------------+
| date                  | employee_joined | present_employees  |
+--------------------------------------------------------------+
| 2014-01-01 00:00:00-7 |            NULL |              NULL  |
| 2014-02-01 00:00:00-7 |            NULL |              NULL  |
| 2014-03-01 00:00:00-7 |            NULL |              NULL  |
| 2014-04-01 00:00:00-7 |              14 |                14  |
| 2014-05-01 00:00:00-7 |            NULL |              NULL  |
| 2014-06-01 00:00:00-7 |            NULL |              NULL  |
| 2014-07-01 00:00:00-7 |            NULL |              NULL  |
| 2014-08-01 00:00:00-7 |               1 |                15  |
| 2014-09-01 00:00:00-7 |            NULL |              NULL  |
| 2014-10-01 00:00:00-7 |               5 |                20  |
+--------------------------------------------------------------+

我正在寻找结果表：

+--------------------------------------------------------------+
| date                  | employee_joined | present_employees  |
+--------------------------------------------------------------+
| 2014-01-01 00:00:00-7 |            NULL |              NULL  |
| 2014-02-01 00:00:00-7 |            NULL |              NULL  |
| 2014-03-01 00:00:00-7 |            NULL |              NULL  |
| 2014-04-01 00:00:00-7 |              14 |                14  |
| 2014-05-01 00:00:00-7 |            NULL |                14  |
| 2014-06-01 00:00:00-7 |            NULL |                14  |
| 2014-07-01 00:00:00-7 |            NULL |                14  |
| 2014-08-01 00:00:00-7 |               1 |                15  |
| 2014-09-01 00:00:00-7 |            NULL |                15  |
| 2014-10-01 00:00:00-7 |               5 |                20  |
+--------------------------------------------------------------+

我试图从以下查询中获取数据：

/*-----ONLY FOR PRESENT EMPLOYEES USING CUMULATIVE SUM--------*/
WITH fdates AS 
    (
        SELECT DATE_TRUNC('month', d.date) AS date
        FROM dates d
        WHERE d.date::DATE <= '10-01-2014' AND
        d.date::DATE >= '01-01-2014'
        group by DATE_TRUNC('month', d.date)
    ),  
employeeJoin AS
    (
        SELECT COALESCE( COUNT(e.id), 0 ) AS employee_joined, 
            DATE_TRUNC( 'month', e.date_started) AS date_started
        FROM employees e GROUP BY DATE_TRUNC( 'month', e.date_started)
    ),
employeeJoinRownum AS
    (   
        SELECT employee_joined, date_started, row_number() OVER (order by date_started) rownum
        FROM employeeJoin
    ) 
SELECT d.*, employee_joined AS employee_joined,
        (SELECT sum(employee_joined) FROM employeeJoinRownum eJ2 WHERE eJ2.rownum <= eJ1.rownum) AS Total_Joined_Employees
    FROM fdates d
    LEFT OUTER JOIN employeeJoinRownum eJ1 ON( eJ1.date_started = DATE_TRUNC('month', d.date) )
    ORDER BY d.date

Answer 1

以下查询计算每个日期加入的员工和员工的剩余时间，然后使用 window function 累积结果。

SELECT
  dates.date,
  COUNT(DISTINCT ej.id) AS employee_joined,
  COUNT(DISTINCT el.id) AS employee_left,
  SUM(COUNT(DISTINCT ej.id) - COUNT(DISTINCT el.id)) OVER (ORDER BY dates.date) AS present_employees
FROM
  dates LEFT JOIN employees ej
ON
  ej.date_started = dates.date LEFT JOIN employees el
ON
  el.date_terminated = dates.date
GROUP BY
  dates.date;

如果您没有预填充的dates表，则可以使用 generate_series 设置返回功能，然后将其加入。

SELECT
  ...
FROM
  GENERATE_SERIES('2014-01-01', '2014-01-10', '1 day'::interval) dates LEFT JOIN employees ej
ON
  ...

Answer 2

您可以通过为join和terminate事件创建一行来规范化表：

select  welcome as date
,       1 as size_change
from    emps
union all
select  bye
,       -1
from    emps
where   bye is not null

现在您可以使用运行总和来计算当前大小：

; with  events as
        (
        select  welcome as date
        ,       1 as size_change
        from    emps
        union all
        select  bye
        ,       -1
        from    emps
        where   bye is not null
        )
select  distinct to_char(date, 'YYYY-MM-DD') as date
,       sum(size_change) over (order by date) as family_size
from    events
order by
        date
;

Example at SQL Fiddle.

空值的累积总和

2 个答案: