Question

确定。这有点复杂。我一直在努力实现它，到目前为止还没有。

我有三张桌子。 leave_approval leave_applications ml_leave_type

这是leave_approval表 enter image description here

然后从leave_application_id链接到leave_application enter image description here

现在出现了棘手的问题，我想从表中获得两条记录。

我想要员工在2014年的总休假

我想要员工在2014年11月份采取的总叶子数

所以意味着需要每年和每月两个不同方法的两列。

到目前为止我有查询。

SELECT 
  LA.employee_id,
  DATEDIFF(
    LA.approved_to,
    LA.approved_from
  ) AS TotalLeavesTaken,
  LAPP.`entitlement_id`,
  LAPP.`ml_leave_type_id`,
  LA.leave_application_id,
  LA.approved_from AS LeaveFrom,
  LA.approved_to AS LeaveTo 
FROM
  leave_approval LA 
  INNER JOIN leave_application LAPP 
    ON LAPP.application_id = LA.leave_application_id 
WHERE YEAR(LA.approval_date) = 2014
  AND LA.`employee_id` = 1

这是我得到的查询结果.. enter image description here

= - = - = - = - = - = - = - = - = - = - = - = - = - = - = - = - = -

更新

我可能无法清楚地解释我的问题。

我想要实现的是添加两个额外的列而不删除我的列。我的意思是我使用大多数/我所有的选择字段

另外，我希望2014年的总叶数在一个栏目中，而在一年中不同领域的总叶数将为该年份。

所以这里是我想要的一个例子。＆gt;

ml_leave_type_id | Total Leaves Taken in Year (e-g 2014) | Total Leaves taken in Month (e-g November)
     1                          10                                 2
     2                           6                                 0
     3                           4                                 1

Answer 1

Edit2：澄清要求后更新的select语句。更新包括使用JOIN而不是UNION ALL重新组织

select a.*, b.*
from (
    SELECT 
      LA.employee_id,
      DATEDIFF(
        LA.approved_to,
        LA.approved_from
      ) AS TotalLeavesTaken,
      LAPP.`entitlement_id`,
      LAPP.`ml_leave_type_id`,
      LA.leave_application_id,
      LA.approved_from AS LeaveFrom,
      LA.approved_to AS LeaveTo
    FROM
      leave_approval LA 
      INNER JOIN leave_application LAPP 
        ON LAPP.application_id = LA.leave_application_id 
    WHERE YEAR(LA.approval_date) = 2014
      AND LA.`employee_id` = 1 
) a

JOIN (
    SELECT 
      LA.employee_id,
      DATEDIFF(
        LA.approved_to,
        LA.approved_from
      ) AS TotalLeavesTaken,
      LAPP.`entitlement_id`,
      LAPP.`ml_leave_type_id`,
      LA.leave_application_id,
      LA.approved_from AS LeaveFrom,
      LA.approved_to AS LeaveTo 
    FROM
      leave_approval LA 
      INNER JOIN leave_application LAPP 
        ON LAPP.application_id = LA.leave_application_id 
    WHERE MONTHNAME(LA.approval_date) = 'November'
      AND LA.`employee_id` = 1
) b
ON a.employee_id = b.employee_id

Answer 2

会这样吗？

employee_id = 1，2014年的总叶数

SELECT 
  LA.employee_id,
  SUM(
     DATEDIFF(
      LA.approved_to,
      LA.approved_from
     )
  ) AS TotalLeavesTaken,
  LAPP.`entitlement_id`,
  LAPP.`ml_leave_type_id`,
  LA.leave_application_id,
  LA.approved_from AS LeaveFrom,
  LA.approved_to AS LeaveTo 
FROM
  leave_approval LA 
  INNER JOIN leave_application LAPP 
    ON LAPP.application_id = LA.leave_application_id 
WHERE YEAR(LA.approval_date) = 2014
  AND LA.`employee_id` = 1
GROUP BY LA.`employee_id` = 1

employee_id的总休假= 1，年= 2014年和月= 11

SELECT 
  LA.employee_id,
  SUM(
     DATEDIFF(
      LA.approved_to,
      LA.approved_from
     )
  ) AS TotalLeavesTaken,
  LAPP.`entitlement_id`,
  LAPP.`ml_leave_type_id`,
  LA.leave_application_id,
  LA.approved_from AS LeaveFrom,
  LA.approved_to AS LeaveTo 
FROM
  leave_approval LA 
  INNER JOIN leave_application LAPP 
    ON LAPP.application_id = LA.leave_application_id 
WHERE YEAR(LA.approval_date) = 2014
  AND MONTH(LA.approval_date) = 11
  AND LA.`employee_id` = 1
GROUP BY LA.`employee_id` = 1

我确实注意到你的数据集中有些日期过于紧张。例如：

employee_id = 1正在休假2014/11/01 - 2014/11/06和2014/11/05 - 2014/11/08。什么是2014/11/05和2014/11/06日期的交易。系统将如何处理这种情况

MYSQL |从单个表中获取两个不同的记录

2 个答案: