Laravel 5.2 - 左连接 DB::Raw 不起作用?
Laravel 5.2 - Left Join DB::Raw not working?
我有以下查询,我在其中尝试使用 DB::Raw()
进行左连接,但出现错误:
Missing argument 2 for Illuminate\Database\Query\Builder::leftJoin()
这是我的查询:
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', '=', $userId)
->selectRaw("a.id
, a.name
, a.criteria
, GROUP_CONCAT(DISTINCT(i.name) SEPARATOR ', ') as 'Industries'
->groupBy('a.id')
->orderBy('a.created_at', 'desc');
leftJoin
函数声明如下:
public function leftJoin($table, $first, $operator = null, $second = null)
您想将原始函数作为第二列传递:
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', '=', $userId)
->selectRaw("a.id
, a.name
, a.criteria
, GROUP_CONCAT(DISTINCT(i.name) SEPARATOR ', ') as 'Industries'")
->groupBy('a.id')
->orderBy('a.created_at', 'desc');
find_in_set 建议来自 。
我不确定 '$.locationId'
是什么,但如果它是一个变量,您可以将其作为数组中的参数作为 DB::raw()
函数的第二个参数传递。
我有以下查询,我在其中尝试使用 DB::Raw()
进行左连接,但出现错误:
Missing argument 2 for Illuminate\Database\Query\Builder::leftJoin()
这是我的查询:
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', '=', $userId)
->selectRaw("a.id
, a.name
, a.criteria
, GROUP_CONCAT(DISTINCT(i.name) SEPARATOR ', ') as 'Industries'
->groupBy('a.id')
->orderBy('a.created_at', 'desc');
leftJoin
函数声明如下:
public function leftJoin($table, $first, $operator = null, $second = null)
您想将原始函数作为第二列传递:
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', '=', $userId)
->selectRaw("a.id
, a.name
, a.criteria
, GROUP_CONCAT(DISTINCT(i.name) SEPARATOR ', ') as 'Industries'")
->groupBy('a.id')
->orderBy('a.created_at', 'desc');
find_in_set 建议来自
我不确定 '$.locationId'
是什么,但如果它是一个变量,您可以将其作为数组中的参数作为 DB::raw()
函数的第二个参数传递。