Question

我有一个包含员工'拳头'（时钟输入/输出）的表格，每个拳击可以是'in'（punch_type = 1）或'out'（punch_type = 2）。

该表的格式如下：

    emp_num | report_date | punch_time             | punch_type
    -----------------------------------------------------------
    1       | 2018-04-20  |2018-04-20 04:46:00.000 | 1
    1       | 2018-04-20  |2018-04-20 06:58:00.000 | 2
    1       | 2018-04-20  |2018-04-20 08:10:00.000 | 1
    1       | 2018-04-20  |2018-04-20 12:00:00.000 | 2

我试图在同一行中获得第一个“打孔”（时钟输入）和随后的“打孔”（时钟输出）。然后，当然，任何以下都是一样的。

期望的输出：

    emp_num | report_date | punch_in               | punch_out
    -----------------------------------------------------------
    1       | 2018-04-20  |2018-04-20 04:46:00.000 | 2018-04-20 06:58:00.000
    1       | 2018-04-20  |2018-04-20 08:10:00.000 | 2018-04-20 12:00:00.000

请记住，如示例所示，一天内可能会有多个打卡/打卡组合。

任何帮助将不胜感激！

Answer 1

首先，您想知道哪个打卡时间属于哪个打卡时间。答：第n次打卡时间属于第n次打卡。为你的记录编号：

select
  p_in.emp_num,
  p_in.report_date,
  p_in.punch_time as punch_in,
  p_out.punch_time as punch_out
from
(
  select
    emp_num,
    report_date,
    punch_time,
    row_number() over (partition by emp_num, report_date order by punch_time) as rn
  from mytable
  where punch_type = 1
) p_in
left join
(
  select
    emp_num,
    report_date,
    punch_time,
    row_number() over (partition by emp_num, report_date order by punch_time) as rn
  from mytable
  where punch_type = 2
) p_out on p_out.emp_num = p_in.emp_num
        and p_out.report_date = p_in.report_date
        and p_out.rn = p_in.rn
order by p_in.emp_num, p_in.report_date, punch_in;

Answer 2

select emp_num, report_date, max(case when punch_type=1 then punch_time else null end) punch_in, 
max(case when punch_type=2 then punch_time else null end) punch_out 
from (select *, row_number() over(partition by emp_num, report_date, punch_type order by emp_num, report_date, punch_time) value from yourtable )a
group by emp_num, report_date, value

按日期透视DateTime字段

2 个答案: