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PHP-Laravel : Raw Sql Query showing Error in controller

转载 作者:行者123 更新时间:2023-11-29 07:17:52 24 4
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我正在构建一个 Laravel 应用程序,我必须在其中生成一个报告,因为我已经构建了一个 Controller ,并且我已经编写了一个查询来生成该报告,但它显示了以下错误。

FatalErrorException in RoomController.php : syntax error, unexpected 'R' (T_STRING)

问题在于 R No,它不是数据库,但为了在报告中显示,我必须将其与我连接的行值一起使用。我该如何解决这个问题?

public function ajax_view_schedule(Request $request)
{
$dept_id=$request->get(['dept_id']);
$schedule= DB::select(DB::raw('SELECT courses.code as c_code, courses.name as c_name,COALESCE( CONCAT('R. No',':',rooms.room_number,', ',days.name ,', ', allocate_rooms.start,' - ',allocate_rooms.end),"Not Scheduled Yet") AS Schedule
FROM departments join courses on departments.id = courses.department_id
left join allocate_rooms on allocate_rooms.course_id=courses.id
left join rooms on allocate_rooms.room_id=rooms.id
left join days on allocate_rooms.day_id=days.id WHERE departments.id='.$dept_id.''));
return \Response::json($course);
}

最佳答案

尝试在 SQL 查询中使用双引号和单引号引用 PHP 字符串。

因此,您的查询字符串将如下所示:

$schedule= DB::select(DB::raw("SELECT courses.code as c_code, courses.name as c_name,COALESCE( CONCAT('R. No',':',rooms.room_number,', ',days.name ,', ', allocate_rooms.start,' - ',allocate_rooms.end),'Not Scheduled Yet') AS Schedule 
FROM departments join courses on departments.id = courses.department_id
left join allocate_rooms on allocate_rooms.course_id=courses.id
left join rooms on allocate_rooms.room_id=rooms.id
left join days on allocate_rooms.day_id=days.id WHERE departments.id='.$dept_id.'"));

关于PHP-Laravel : Raw Sql Query showing Error in controller,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37236015/

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