laravel查询生成器从原始查询

美好的一天！我是Laravel的新手。我尝试了很多方法来做到这一点，但它仍然给我一个错误。我想这个SQL查询转换成雄辩laravel查询生成器从原始查询

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

我想这雄辩的语言

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;

}

预期输出是这个查询会在每一天告诉员工的状态。例如，如果这一天是周末，它将在输出中显示'W'。如果员工缺席'A'并且如果出现'P'。我希望有人能帮助我。在此先感谢

来源

2017-08-24 Alyssa Andrea

你得到了什么错误，把它添加到你的问题。 –

语法错误，意外'''在C：\ xampp \ htdocs \ hrpayroll \ api \ app \ Http \ Models \ AttendanceReportModel.php在线178 –

为什么你有两个'DB :: raw'语句， t连接结果，我会说错误来自那里 –

在先来看看addSelect方法：

public function addSelect($column) 
{ 
    $column = is_array($column) ? $column : func_get_args(); 
    $this->columns = array_merge((array) $this->columns, $column); 
    return $this; 
}

这只是与现有的选定列合并将此列（S）。

更多的澄清访问https://laravel.com/docs/5.4/queries#selects

现在，我想你会解决你的问题，而无需使用addSelect方法。

我的参考this

来源

2017-08-24 06:42:39

感谢的最终代码。许多 –

laravel查询生成器从原始查询

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