How to write laravel model based in this query with pagination? - laravel

Question

for ($i = 0; $i < count($keyskill); $i++) {
if ($i == count($keyskill) - 1) {
$key .= "skills like '%" . $keyskill[$i] . "%' ";
} else {
$key .= "skills like '%" . $keyskill[$i] . "%' or ";
}
}
Hear my Query is
$jobs = DB::select("SELECT * from post where ( $key ) and status=1 order by id desc");
How to write laravel model based in this query with pagination???

Answer 1

Post::where('skills' , 'like' , '%php%')->orWhere('skills' , 'like' , '%javascript%')->where('status' , 1)->orderBy('id' , 'desc')->paginate(10);

Answer 2

try this for dynamic array:
Post::where(function ($query) use ($keyskill) {
for ($i = 0; $i < count($keyskill); $i++) {
if ($i == 0) {
$query->where('skills' , 'like' ,"%$keyskill[$i]%");
} else {
$query->orWhere('skills' , 'like' , "%$keyskill[$i]%");
}
}
})->where('status' , 1)->orderBy('id' , 'desc')->paginate(10);

Answer 3

Below is the Eloquent Query considering Post as your Model Name and Page Size 50.
$posts = \App\Post::where('skills','like','%javascript%')
->orWhere('skills','like','%php%')
->where('status',1)
->orderBy('id','desc')
->paginate(50);
Loading records
$posts->items();
Rendering Pagination on Blade
{!! $posts->render() !!}
Just a recommendation
Your table name should be posts instead or post then Laravel can manage it implicitly but in your case currently you will have to mention the table name in your model with code protected $table = 'post'; in your app/post.php file, read more here.