Question

我试图想出最好的方法来增加员工的总时间。

INOUT类型是S＆amp;即

我认为这样的事情可能是最好的解决方案，但我不知道它是否实用甚至可能：

例如：

No  EnNo    INOUT   DateTime
1   12      S       2017-06-02 08:35
2   28      S       2017-06-02 10:10
3   28      E       2017-06-02 13:00
4   12      E       2017-06-02 14:02
5   12      S       2017-06-02 15:03
6   12      E       2017-06-02 18:09

所以在这个例子中。我可以尝试计算工作时间的总时间。再说一遍，我不知道这是否可能。

我真的很感谢有关如何完成此任务的任何意见/建议，或任何其他可能更实际的方法！

像这样查询：

$query = "SELECT EnNo, 
            sum(case 
            when INOUT = 'E' then -DATEDIFF(MINUTE,2017-06-02,DateTime)
            when INOUT = 'S' then  DATEDIFF(MINUTE,2017-06-02,DateTime) 
            end 
            )/60 as Working Hours 
        From `dataimport`
        Group By dataimport.EnNo";

但结果不会产生..

Answer 1

可行的方法是选择两名员工之间的TIMESTAMPDIFF。您必须以这样的方式加入表格，使得您在同一天具有S和E的两个员工记录之间存在差异。

我尝试了自己并在未经测试的情况下提出了以下内容，但它应该指向正确的方向：

SELECT 
A.EnNo, TIMESTAMPDIFF(MINUTE,A.DateTime,B.DateTime)/60 AS WorkingHours
FROM dataimport A INNER JOIN dataimport B ON A.EnNo = B.EnNo AND 
(A.INOUT = 'E' AND B.INOUT = 'S') OR  (A.INOUT = 'S' AND B.INOUT = 'E') 
AND TIMESTAMPDIFF(DAY,A.DateTime,B.DateTime) = 0

注意：您也可以更改TIMESTAMPDIFF以一次性查找小时数差异。

参考文献：Calculate the time difference between of two rows https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_timestampdiff

Answer 2

另外一种在php中计算的简单方法!!!

$ datetime1 = new DateTime（'2017-06-02 08:35'）;

$ datetime2 = new DateTime（'2017-06-02 14:02'）;

$ duration = $ datetime1-＆gt; diff（$ datetime2）;

echo $ duration-＆gt;格式（'％h'）。“小时”。$ duration-＆gt;格式（'％i'）。“分钟”;

Answer 3

Please check the query twice and try with your code ...i have done same scenario which you have

$query->select("a.*,a.DATE_FORMAT(a.date_time,'%Y-%m-%d') as `new_date`,
                            DATE_FORMAT(a.date_time,'%Y-%m-%d 23:59:00') as `last_datetime`,
                            DATE_FORMAT(a.date_time,'%Y-%m-%d 00:01:00') as `first_datetime`,
                    @last_value:=(SELECT date_time FROM master a2
                    WHERE  a2.enroll=a.enroll
                    and (a2.INOUT = 'E' )
                    and (a2.`date_time` <= last_datetime and a2.`date_time`>first_datetime)
                    group by a2.date_time
                    ORDER BY a2.date_time DESC LIMIT 1) as last_value,



            ");
        $query->from('master as a')

        ->andWhere('a.enroll IS NOT NULL and a.INOUT="S"')
        ->groupBy('a.enroll,new_date')
        ->orderBy('a.enroll,a.date_time');
        $attendence_db_data = $query->all();

You will get user's first IN And last OUT of particular one day in one record so that you can easily calculate his/her work hour from that value.

查询mysql以计算一天工作总时间的可能方法

3 个答案: