Sql leftJoin query using laravel syntax showing error - mysql

Question

This is my query that i used for left join, it is working fine but something is missing when I convert it to laravel syntax.
Query to convert is
$result = DB::select("select amenities.name as
name,amenities.type_id,amenities.id as id, amenities.icon, rooms.id as status
from amenities left join rooms on find_in_set(amenities.id, rooms.amenities)
and rooms.id = $room_id and type_id !=4");
and the I am doing this
$result = DB::table('amenities')
->select('amenities.name as name', 'amenities.type_id' , 'amenities.id as id'
, 'amenities.icon', 'rooms.id as status' )
->leftJoin('rooms', function ($join) {
$join->on('FIND_IN_SET(amenities.id, rooms.amenities)')
->where('rooms.id' , '=', '$room_id')
->where('type_id','!=', 4);
})->get();
The error is
InvalidArgumentException in
F:\xampp\htdocs\arheb\Arheb\vendor\laravel\framework\src\Illuminate\Database\Query\JoinClause.php
line 79: Not enough arguments for the on clause.

Answer 1

Your query is wrong. I assume that amenities.id and rooms.amenities are attributes of amenities and rooms table respectively.
MySQL FIND_IN_SET() returns the position of a string if it is present (as a substring) within a list of strings.
You need to pass column names in first and second parameter of on() function.
$result = DB::table('amenities')
->select('amenities.name as name', 'amenities.type_id' , 'amenities.id as id'
, 'amenities.icon', 'rooms.id as status' )
->leftJoin('rooms', function ($join) {
$join->on('amenities.id', '=', 'rooms.amenities')
->where('rooms.id' , '=', '$room_id')
->where('type_id','!=', 4);
})->get();

Answer 2

I think you can try this:
$result = DB::table('amenities')
->select('amenities.name as name', 'amenities.type_id' , 'amenities.id as id'
, 'amenities.icon', 'rooms.id as status' )
->leftJoin('rooms', function ($join) {
$join->on(DB::raw("find_in_set(amenities.id, rooms.amenities)"))
->where('rooms.id' , '=', '$room_id')
->where('type_id','!=', 4);
})->get();
Hope this work for you!