How to build MySQL query with Laravel

Question

I have a Laravel query like this but I got error like this

Undefined variable: keyword

$bookings = DB::table('bookings')
                    ->leftJoin('users', 'bookings.manager_id', '=', 'users.id')
                    ->leftJoin('projects', 'bookings.pr_project_id', '=', 'projects.id')
                    ->leftJoin('resources', 'bookings.pr_resource_id', '=', 'resources.id')
                    ->Where('unavailable','=','0')
                    ->orWhere(function ($query) {
                        $query->Where('users.name', 'LIKE', "%$keyword%")
                            ->Where('projects.project_name', 'LIKE', "%$keyword%")
                           ->Where('resources.r_name', 'LIKE', "%$keyword%");
                    })
                    ->orderBy('bookings.pr_working_date', 'desc')
                    ->paginate($perPage);

Now below query how to convert in Laravel

   select * from bookings left join users on bookings.manager_id=users.id left join projects on bookings.pr_project_id=projects.id left join resources on bookings.pr_resource_id=resources.id where unavailable='0' AND (users.name like '%One Web Series%' or projects.project_name like '%One Web Series%' or resources.r_name like '%One Web Series%')

Alexey Mezenin · Accepted Answer · 2018-01-23 09:44:24Z

4

You need to add the use() like this to pass the variable into the closure scope:

function ($query) use($keyword) {

answered Jan 23, 2018 at 9:44

Alexey Mezenin

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Sohel0415 · Accepted Answer · 2018-01-23 09:44:31Z

use use()-

$bookings = DB::table('bookings')
                ->leftJoin('users', 'bookings.manager_id', '=', 'users.id')
                ->leftJoin('projects', 'bookings.pr_project_id', '=', 'projects.id')
                ->leftJoin('resources', 'bookings.pr_resource_id', '=', 'resources.id')
                ->Where('unavailable','=','0')
                ->orWhere(function ($query) use($keyword){
                    $query->Where('users.name', 'LIKE', "%$keyword%")
                        ->Where('projects.project_name', 'LIKE', "%$keyword%")
                       ->Where('resources.r_name', 'LIKE', "%$keyword%");
                })
                ->orderBy('bookings.pr_working_date', 'desc')
                ->paginate($perPage);

Sohel0415 · Accepted Answer · 2018-01-23 10:10:43Z

1

You can pass the necessary variables from the parent scope into the closure with the use keyword . (i.e)

Where(function ($query) use($keyword)

edited Jan 23, 2018 at 10:10

Sohel0415

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answered Jan 23, 2018 at 10:08

AmalJo

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