Question

我有一个名为rjs_attendance的表，后面有四列

_________________________________________________
|attenedance_id | admin_id | note | created_date|
-------------------------------------------------

用户可以在一天内多次出席。奇数输入被假定为登录，甚至输入被假定为退出。我需要的输出看起来像这样。

_______________________________________________________________________________  
|admin_id   | time_in  | time_in_note  | time_out | time_out_note | date      |
-------------------------------------------------------------------------------
|1          |10:00     | none          | 11:00    | none          | 2015-12-24|
-------------------------------------------------------------------------------
|1          |11:30     |none           |12:15     |none           | 2015-12-24|
-------------------------------------------------------------------------------

我无法获取同一日期的所有记录，但我能够记录同一日期的一条记录。我运行的查询如下：

SELECT 
    `atd_in`.`admin_id` AS `admin_id`,
    CAST(MIN(`atd_in`.`created_date`) AS TIME) AS `time_in`,
    `atd_in`.`note` AS `time_in_note`,
    CAST(MAX(`atd_out`.`created_date`) AS TIME) AS `time_out`,
    `atd_out`.`note` AS `time_out_note`,
    CAST(`atd_in`.`created_date` AS DATE) AS `date_on` 
FROM 
    `zf2`.`rjs_attendance` `atd_in` 
    LEFT JOIN `zf2`.`rjs_attendance` `atd_out` 
        ON
            `atd_in`.`admin_id` = `atd_out`.`admin_id`
            AND CAST(`atd_in`.`created_date` AS DATE) = CAST(`atd_out`.`created_date` AS DATE) 
            AND `atd_in`.`attendance_id` <> `atd_out`.`attendance_id`
GROUP BY 
    CAST(`atd_in`.`created_date` AS DATE), `atd_in`.`admin_id`

非常感谢任何帮助。

Answer 1

试试这个：

SELECT A.admin_id, 
       MAX(IF(A.ID % 2 = 1, CAST(A.created_date AS TIME), NULL)) AS time_in, 
       MAX(IF(A.ID % 2 = 1, note, NULL)) AS time_in_note, 
       MAX(IF(A.ID % 2 = 0, CAST(A.created_date AS TIME), NULL)) AS time_out, 
       MAX(IF(A.ID % 2 = 0, note, NULL)) AS time_out_note, 
       CAST(A.created_date AS DATE) AS date_on 
FROM (SELECT IF(@adminId=@adminId:=A.admin_id, @id:=@id+1, @id:=1) AS ID, 
             A.admin_id, A.note, A.created_date 
      FROM zf2.rjs_attendance A, (SELECT @id:=1, @adminId:=0) AS B
      ORDER BY A.admin_id, A.attendance_id
    ) AS A
GROUP BY A.admin_id, CAST(A.created_date AS DATE), CEILING(A.ID / 2);

sqlfiddle

Answer 2

在看到任何代码之前 - 如果你假设奇数，甚至作为进入和退出=＆gt;加入这个条件 -

on A.id +1 = B.id

或

`atd_in`.`attendance_id` +1 = `atd_out`.`attendance_id`

您甚至可以将其保留为非退出条目（当然，ID应该是数字的）

如果您的记录没有按照这种方式排序，只需在使用order by排序后重新发出id列并创建新的id列

如何在MySql中查找具有相同表的别名的相同日期的多个记录（出勤）

2 个答案: