php – 如何使用sql查询获取时间总和值？

2023年7月23日 209次阅读

我有一个ReportBuilder类用于生成文档(Excel).我想要的是选择员工和日期,然后计算该日期的总工作时间.这是我的php文件

<?php
class EmployeeSummaryTimesheet extends ReportBuilder{

    public function getMainQuery(){
        $query = "SELECT
(SELECT `employee_id` from Employees where id = at.employee) as 'Employee',
(SELECT concat(`first_name`,' ',`middle_name`,' ', `last_name`) from Employees where id = at.employee) as 'Employee',

TIMEDIFF(out_time, in_time) as 'Working Hours'
FROM Attendance at";

        return $query;
}

public function getWhereQuery($request){
        if(($request['employee'] != "NULL" && !empty($request['employee']))){
            $query = "where employee = ? and in_time >= ? and out_time <= ? group by employee;";
            $params = array(
                $request['employee'],
                $request['date_start']." 00:00:00",
                $request['date_end']." 23:59:59",
            );
        }else{
            $query = "where in_time >= ? and out_time <= ? group by employee;";
            $params = array(
                $request['date_start']." 00:00:00",
                $request['date_end']." 23:59:59",
            );
        }
 return array($query, $params);
    }}

现在我无法得到那个特定日期的总和我第一天获得TIMEDIFF(out_time,in_time)值只是没有工作时间总和帮助我

编辑 – 这里我有两个表Employee和Attendance(id,employee(这是Employee表中的员工ID),in_time,out_time)

编辑：我将DATE_FORMAT(in_time,’％Y-％m-％d’)取消为’Date’,我不需要此列

最佳答案 您希望每位员工和日期有一条记录,并显示该员工当天的总工作时间.因此,按员工和日期加入和分组,并将工作时间相加.

select
  e.employee_id,
  concat(e.first_name, ' ', e.middle_name, ' ', e.last_name) as employee_name 
  date_format(at.in_time, '%Y-%m-%d') as 'Date',
  sum(timediff(at.out_time, at.in_time)) as 'Working Hours'
from attendance at
join employee e on e.id = at.employee
where ...
group by e.employee_id, date_format(at.in_time, '%Y-%m-%d');
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