Laravel 5.2 - Left Join Db: :Raw Not Working?
I Have the Following Query Where I'm Trying to Use Db: :Raw() for the Left Join but I'm Getting Error: Missing Argument 2 for Illuminate\Database\Query\Builder...
I have the following query where I'm trying to use DB::Raw() for the left join but I'm getting error:
Missing argument 2 for Illuminate\Database\Query\Builder::leftJoin()
This is my query:
return $this->model->from('alerts as a')
->leftJoin(DB::Raw("locations as l on l.id = JSON_UNQUOTE(JSON_EXTRACT(a.criteria, '$.locationId'))"))
->leftJoin(DB::Raw("industries as i on find_in_set(i.id, JSON_UNQUOTE(JSON_EXTRACT(a.criteria, '$.industries')))"))
->where('user_id', '=', $userId)
->selectRaw("a.id
, a.name
, a.criteria
, GROUP_CONCAT(DISTINCT(i.name) SEPARATOR ', ') as 'Industries'
->groupBy('a.id')
->orderBy('a.created_at', 'desc');
1 Answer
The leftJoin function is declared like this:
public function leftJoin($table, $first, $operator = null, $second = null)
You want to pass your raw functions in as the second column:
return $this->model->from('alerts as a')
->leftJoin('locations AS l', 'l.id', '=', DB::Raw("JSON_UNQUOTE(JSON_EXTRACT(a.criteria, '$.locationId'))"))
->leftJoin('industries as i', function($join){
$join->on(DB::raw("find_in_set(i.id, JSON_UNQUOTE(JSON_EXTRACT(a.criteria, '$.industries')))",DB::raw(''),DB::raw('')));
})
->where('user_id', '=', $userId)
->selectRaw("a.id
, a.name
, a.criteria
, GROUP_CONCAT(DISTINCT(i.name) SEPARATOR ', ') as 'Industries'")
->groupBy('a.id')
->orderBy('a.created_at', 'desc');
The find_in_set suggestion came from here.
I'm not sure what '$.locationId' is, but if it's a variable, you can pass that along as a parameter within an array as the second parameter on the DB::raw() function.
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