Question

员工表列在下面......

    Name          DOJ                  DOL
 ............   ...............      ............
    Ram            2014-01-12          2014-02-12
    Kiran          2014-02-05          2014-07-05
    Jhon           2014-01-25          2014-10-01

预期输出是......

    Month          Joining_count     Leaving_count
  ...........     ................  .................
   Jan-2014              2                 0
   Feb-2014              1                 1
      .                  .                 .
      .                  .                 .
      .                  .                 .

我尝试使用下面的Mysql Query但我无法获得预期的输出请帮帮我

        SELECT monthname(current_date) as month, 

        count( `DATE_OF_JOINING`)  as 'Joining_count' , 
        count( `DATE_OF_LEAVING`)  as 'leaving_count' , 

        group by year(`DATE_OF_JOINING`),month('DATE_OF_JOINING`),
        year(`DATE_OF_LEAVING),month(`DATE_OF_LEAVING)

Answer 1

您正在对同一数据执行两种不同的聚合 - 因此您需要两个汇总查询join。不幸的是，mysql没有一个完整的外部联接，因此处理有人离开但没有人加入的几个月以及几个月没有人加入的人都有点麻烦。我用三个连接解决了这个问题 - 一个查询获取所有可能的日期，另一个查询用于每个聚合，尽管there are other ways。

SELECT    my_table.d, COALESCE(join_count, 0), COALESCE(leave_count, 0)
FROM      (SELECT DISTINCT DATE_FORMAT(doj, '%b-%y') AS d
           FROM   my_table
           UNION
           SELECT DISTINCT DATE_FORMAT(dol, '%b-%y')
           FROM   my_table) dates
LEFT JOIN (SELECT   DATE_FORMAT(doj, '%b-%y') d, COUNT(*) AS join_count
           FROM     my_table
           GROUP BY DATE_FORMAT(doj, '%b-%y')
          ) joins ON dates.d = joins.d
LEFT JOIN (SELECT   DATE_FORMAT(dol, '%b-%y') d, COUNT(*) AS leave_count
           FROM     my_table
           GROUP BY DATE_FORMAT(dol, '%b-%y')
          ) leaves ON dates.d = leaves.d

Answer 2

基本上，您需要两个查询，一个用于分组加入日期，另一个用于离开日期。您可以使用UNION ALL合并两个结果，并将合并后的结果视为新表，然后选择（和SUM）值。

SELECT 
  s.month, 
  SUM(s.Joining_count) AS Joining_count, 
  SUM(s.Leaving_count) AS Leaving_count
FROM
  (
   SELECT 
     DATE_FORMAT(doj, '%b-%y') as `month`, 
     COUNT(ej.doj)  as 'Joining_count',
     0 AS 'Leaving_count'
   FROM employee ej
     GROUP BY `month`

   UNION ALL

   SELECT 
     DATE_FORMAT(dol, '%b-%y') as `month`, 
     0  as 'Joining_count',
     COUNT(el.dol)  as 'Leaving_count'  
   FROM employee el
     GROUP BY `month`
  ) s
GROUP BY s.`month`;

输出

month       Joining_count   Leaving_count
---------------------------------------------
Feb-2014    1               1
Jan-2014    2               0
Jul-2014    0               1
Oct-2014    0               1

在mysql中按2个列的月份和年份排序

2 个答案: