来自原始查询的laravel查询生成器

时间:2017-08-24 06:03:05

标签: php laravel laravel-5.4 lumen

美好的一天! 我是Laravel的新手。我尝试了很多方法,但它仍然给我一个错误。我试图将这个SQL查询转换成雄辩的

 Select     t.employee_code, 
            CASE WHEN t.day = '2017-08-19' THEN t.PRESENT ELSE NULL END AS `2017-08-19`,
            CASE WHEN t.day = '2017-08-20' THEN t.PRESENT ELSE NULL END AS `2017-08-20`,
            CASE WHEN t.day = '2017-08-21' THEN t.PRESENT ELSE NULL END AS `2017-08-21`,
            CASE WHEN t.day = '2017-08-22' THEN t.PRESENT ELSE NULL END AS `2017-08-22`,
            CASE WHEN t.day = '2017-08-23' THEN t.PRESENT ELSE NULL END AS `2017-08-23`,
            CASE WHEN t.day = '2017-08-24' THEN t.PRESENT ELSE NULL END AS `2017-08-24`
FROM (
select e.employee_code, 
    Cast(e_l.time_in As date) As Day,
    Case 
        WHEN e_l.time_in IS NULL THEN 'A' 
        WHEN DAYOFWEEK(e_l.time_in) In(7, 1) Then 'W'
        ELSE 'P' 
    end as PRESENT
from employee As e
left join employees_logs As e_l on e.id = e_l.employee_id)
AS t

我尝试了这种雄辩的方式

public static function statusReport($data){
$start_date = $data['start_date'];
$end_date = $data['end_date'];
$date_array = self::getDatesFromRange($start_date, $end_date);

$query = DB::raw("(Select t.employee_code,
          CASE WHEN t.day = '$start_date' THEN t.Present ELSE NULL END AS '$start_date')");
$query->addSelect(
        DB::raw("(
            SELECT employee.id as emp_id, 
            CONCAT(employee.firstname, " ",employee.lastname) AS employee_name,
            CAST(employees_log.time_in as date) as date_given,
            CASE    WHEN employee_logs.time_in IS NULL THEN 'A'
                    WHEN leave.status_id = 4 AND leave_request.with_pay = 1  THEN 'L'
                    WHEN leave.status_id = 4 AND leave_request.with_pay = 0  THEN 'LOP'
                    WHEN DAYOFWEEK(employee_logs.time) In(7, 1) THEN 1 AS 'W' 
                    ELSE 'P' END as status
            LEFT JOIN employee_logs On employee.id = employees_logs.employee_id
            JOIN leave On employee.id = leave.emp_id
            JOIN leave_request On leave.id = leave_request.leave_id
            WHERE employee_logs.time_in BETWEEN '$start_date' AND '$end_date') As `t`"));
$data = $query->get();
return $data;       

}

this the error that I'm getting

预期输出是这个查询将告诉每天员工的状态。例如,如果这一天是周末,它将显示' W'在输出上。如果员工缺席' A'如果存在' P' 。我希望有人可以帮助我。提前致谢

1 个答案:

答案 0 :(得分:0)

首先看一下head -1000 alert_PROD.log > alert_tmp ; cat alert_tmp > alert_PROD.log ; rm alert_tmp 方法:

addSelect

这只是通过与现有选定列合并来添加列。

要获得更多说明,请访问https://laravel.com/docs/5.4/queries#selects

现在我猜你会在不使用addSelect方法的情况下解决问题。

我的参考this

相关问题