Reputation: 301
I use three table spareparts, categories and parameter and YajraDatatables.
My controller action is:
public function anyData()
{
$spareparts_param_list = Sparepart::all();
$list='';
foreach ($spareparts_param_list as $value) {
foreach ($value->category->parameter as $par_list) {
$list .= $par_list->Name.',';
}
}
$spareparts = Sparepart::
join('cars', 'spareparts.car_id', '=', 'cars.id')
->select(['spareparts.id', 'cars.Brend', 'spareparts.Model', $list]);
$datatables = app('datatables')->of($spareparts);
return $datatables->make();
}
My array list $list print parameters such as color,type,tires,. How to pass $list array in select query?
Upvotes: 0
Views: 2535
Reputation: 688
You can use whereIn
:
public function anyData()
{
$spareparts_param_list = Sparepart::all();
$list = [];
foreach ($spareparts_param_list as $value) {
foreach ($value->category->parameter as $par_list) {
$list[] = $par_list->Name;
}
}
$spareparts = Sparepart::
join('cars', 'spareparts.car_id', '=', 'cars.id')
->whereIn('Name', $list)
->select(['spareparts.id', 'cars.Brend', 'spareparts.Model']);
$datatables = app('datatables')->of($spareparts);
return $datatables->make();
}
You can run this raw Mysql
command:
select sp.id as sp_id, sp.model as sp_model, c.brend as car_brend, json_arrayagg(p.name) as p_name
from spareparts as sp
join cars as c on sp.car_id=c.id
join categories as cat on sp.category_id=cat.id
join parameters as p on cat.id=p.category_id
group by sp.model;
Above command will give you a result like the following:
1 | Audi A6 door | Audi A6 | ["color", "window"]
The last column will be a json column of all parameters related to each category.
Upvotes: 1
Reputation: 3268
You can use whereIn(), something like this:
$users = DB::table('users')
->whereIn('id', [1, 2, 3])
->get();
Docs - https://laravel.com/docs/5.7/queries#where-clauses
Upvotes: 1