Question

我想在两行不同的表之间减去：

我创建了一个名为view的{{1}}和名为leave_taken的{{1}}。

我想从这两个表中得到这个结果：

table

并按leave_balance分组

两个表的代码

-----------------查看Leave_Taken -----------

leave_taken.COUNT(*) - leave_balance.balance

----------------表leave_balance ----------

leave_type_id_leave_type

-------表格离开----------

CREATE ALGORITHM = UNDEFINED DEFINER=`1`@`localhost` SQL SECURITY DEFINER 
VIEW `leave_taken` 
AS 
   select 
     `leave`.`staff_leave_application_staff_id_staff` AS `staff_leave_application_staff_id_staff`,
     `leave`.`leave_type_id_leave_type` AS `leave_type_id_leave_type`,
     count(0) AS `COUNT(*)` 
   from 
   (
      `leave` 
      join `staff` on((`staff`.`id_staff` = `leave`.`staff_leave_application_staff_id_staff`))
   ) 
   where (`leave`.`active` = 1) 
   group by `leave`.`leave_type_id_leave_type`;

Answer 1

好吧，我仍然认为你没有提供足够的信息。获得一些样本数据和预期输出（以表格格式）会非常有用非常有用。也就是说，我可能会有一些你可以开始合作的东西。此查询查找所有工作人员，计算其当前休假（按类型分组），并通过休假类型确定该休假与其余额之间的差异。看看它，更重要的是（也许）我使用的sqlfiddle here中包含样本数据（对于确定这是否是数据的正确路径非常重要）。

SELECT
    staff.id_staff,
    staff.name,
    COUNT(`leave`.id_leave) AS leave_count,
    leave_balance.balance,
    (COUNT(`leave`.id_leave) - leave_balance.balance) AS leave_difference,
    `leave`.leave_type_id_leave_type AS leave_type
  FROM
    staff
    JOIN `leave` ON staff.id_staff = `leave`.staff_leave_application_staff_id_staff
    JOIN leave_balance ON
      (
        staff.id_staff = leave_balance.staff_id_staff
        AND `leave`.leave_type_id_leave_type = leave_balance.leave_type_id_leave_type
      )
  WHERE 
    `leave`.active = 1
  GROUP BY
    staff.id_staff, leave_type;

祝你好运！

mySql减去不同表的行

1 个答案: