来自原始查询的 laravel 查询构建器答案

【问题标题】：laravel query builder from raw query来自原始查询的 laravel 查询构建器
【发布时间】：2018-02-01 21:20:37
【问题描述】：

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。我正在尝试将此sql查询转换为雄辩的

 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;

}

预期的输出是这个查询将告诉员工每天的状态。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，则为“A”，如果存在，则为“P”。我希望有人可以帮助我。提前致谢

【问题讨论】：

你遇到什么错误，添加你的问题。
语法错误，第 178 行 C:\xampp\htdocs\hrpayroll\api\app\Http\Models\AttendanceReportModel.php 中的意外 '"'
为什么你有两个DB::raw 语句，你也没有连接结果。我会说错误来自那里。
我不小心删除了addselect。谢谢@alex。现在编辑我的代码。
@404BrainNotFound 我为这个烂摊子道歉。谢谢你告诉我。现在这是我工作的最终代码

标签： php laravel laravel-5.4 lumen

【解决方案1】：

首先看一下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;
}

这只是通过与现有的选定列合并来添加列。

如需了解更多信息，请访问https://laravel.com/docs/5.4/queries#selects

现在我猜你会在不使用 addSelect 方法的情况下解决你的问题。

我的参考this

【讨论】：