Question

我尝试按以下顺序设置员工出勤日期。

row data

emp_no  date                time
001   2019-08-07 00:00:00    1899-12-30 08:29:28
001   2019-08-07 00:00:00    1899-12-30 17:29:28

我想准备像样并保存在其他表格中

emp_no    date      time_in   time_out
001     2019-08-07 08:29:28   17:29:28

time in 03:00 am - 12:15 pm
time out 12:15 pm to 03:00 am

行数据通过指纹机获取。每天记录2000名员工详细信息。如何解决此问题？我尝试根据emp_no和date分组，但输出错误我正在使用php和MySQL进行开发

Answer 1

这可能有帮助。

如果您需要保存在新表中，请首先创建new_table
在主表的new_table中插入格式化的数据
从new_table中选择所有数据

    create table new_table(emp_no number,
                           day_date date,
                           time_in varchar2(15),
                           time_out varchar2(15));

    insert into new_table 
    select emp_no,
           date,
           extract(hour from min(time)) || ':' || extract(minute from min(time)) || ':' || extract(second from min(time)),
           extract(hour from max(time)) || ':' || extract(minute from max(time)) || ':' || extract(second from max(time))
      from employee
    group by emp_no, date;

    select * from new_table;

尽管我不确定您是从数据中得到什么

time in 03:00 am - 12:15 pm
time out 12:15 pm to 03:00 am

Answer 2

这是一个主意。获取每个emp_no的行号（按empno，日期和时间排序）。然后将time_out（第二个条目）加入到您的第一个表中。

SELECT t1.emp_no, t1.date, t1.time as time_in, IFNULL(t2.type, 'no time out') as time_out FROM (SELECT 
      @row_number:=CASE
          WHEN @id = emp_no THEN @row_number + 1
          ELSE 1
      END AS num,
      @id:=emp_no as emp_no,
      date,
      time
  FROM
      table1,(SELECT @id:=0,@row_number:=0) as t
  ORDER BY emp_no, date, time) as t1
LEFT JOIN
    (SELECT 
      @row_number:=CASE
          WHEN @id = emp_no THEN @row_number + 1
          ELSE 1
      END AS num,
      @id:=emp_no as emp_no,
      date,
      time
  FROM
      table1,(SELECT @id:=0,@row_number:=0) as t
  ORDER BY emp_no, date, time) as t2 on t2.num - 1 = t1.num and t2.emp_no = t1.emp_no
WHERE mod(t1.num, 2) = 1

如何在不覆盖sql中的数据的情况下对数据集进行分组

2 个答案: