Question

我想写一个查询来生成员工的出勤报告。首先，我将告诉您员工的存在如何存储在我的数据库中。

我有以下表格。

Employee Table with Columns

    emp_id  emp_Name   Joining_Date
     1      john         11-01-2012
     2      Scott        12-01-2012


Holiday Table

Holiday_Name         Date 
Chrismas          25-12-2012  
Dushera           08-03-2012
Independance Day  15-08-2012

Leave Table

Subject  from_Date     to_Date      Emp_Id     status 
PL       02-01-2012    04-01-2012      1       Approved
CL       11-01-2012    12-01-2012      2       Declined      


Doctor Table

Subject   Call_Date    call_Done_By(emp_id)
 Call     15-01-2012      1
 CA       21-02-2012      2

Chemist Table
Subject   Call_Date    call_Done_By(emp_id)
Chemist   1-02-2012     2
Texo      21-03-2012    1

如果员工被医生或化学家访问，该特定日期将与employee_id一起存储在该特定医生或化学家表中

现在，人们将选择年份和月份，他应该能够以下列格式获得出勤报告示例：假设用户选择年份为“2011”，月份为“Dec”，则输出应为

Employee   year  Month    1  2   3   4   5   6    7.... 
 John      2011   Nov     Y  Y   Y   Y   Y   L    S....
 Scott     2011   Nov     Y  Y   L   M   Y   L    S

这里输出1,2,3 ....是0-30个月的日子，我们可以使用'case'来写

考虑员工是否在当天出现，其状态为“Y”，否则为L else 如果他去了医生，化学家这样的顾客，那就换成'S'。

那么我该怎么写一个查询来实现这个输出呢？任何建议对我都有帮助....

Answer 1

这是一个应该按预期工作的漫长道路：

SELECT
    Employee.emp_Name,
    '2011' AS `Year`,
    'Dec' AS `Month`,
    CASE (
        IF(
            DATE('1-12-2011') < DATE(Employee.Joining_Date)),
            '0' --Not joined yet
            IF (
                (SELECT COUNT(*) FROM Holiday WHERE DATE('1-12-2011') = DATE(Holiday.date)) = 1,
                '1', --National Holiday
                IF (
                    (SELECT COUNT(*) FROM Leave WHERE DATE('1-12-2011') > DATE(Leave.to_Date) AND DATE('1-12-2011') < DATE(Leave.from_Date) AND Leave.Emp_Id = Employee.emp_id) = 1,
                    '2', --On Leave
                    IF(
                        (SELECT COUNT(*) FROM Doctor WHERE DATE('1-12-2011') > DATE(Doctor.Call_Date) AND Doctor.call_Done_By = Employee.emp_id) = 1 OR 
                        (SELECT COUNT(*) FROM Chemist WHERE DATE('1-12-2011') > DATE(Chemist.Call_Date) AND Chemist.call_Done_By = Employee.emp_id) = 1,
                        '3' --Visit Doctor or Chemist
                        '4' --Employee was at work
                    )
                )
            )
        )
    )
        WHEN 0 THEN 'N/A' --Not joined yet
        WHEN 1 THEN 'L' --National Holiday
        WHEN 2 THEN 'L' --On Leave
        WHEN 3 THEN 'S' --Visit Doctor or Chemist
        ELSE 'Y' --Employee was at work
    END AS `1`, --first day of month
    ... AS `2`, --repeat for second day of the month till max day of current month replace '1-12-2011' with each different day of month
    ...
    ... AS `30`
FROM
    Employee

我的建议是创建一个视图，为每个员工执行if语句，使代码更容易维护。请记住，这是伪代码，可能需要一些更改才能运行。

希望这有帮助。

MySql中的考勤报告

1 个答案: