Laravel Eloquent inner join with multiple conditions

PhpLaravelEloquent

Php Problem Overview


I have a question about inner joins with multiple on values. I did build my code like this in laravel.

public function scopeShops($query) {
    return $query->join('kg_shops', function($join)
    {
        $join->on('kg_shops.id', '=', 'kg_feeds.shop_id');
        // $join->on('kg_shops.active', '=', "1"); // WRONG
        // EDITED ON 28-04-2014
        $join->on('kg_shops.active', '=', DB::raw("1"));

    });
}

Only problem is, it gives this outcome:

Column not found: 1054 Unknown column '1' in 'on clause' (SQL: select `kg_feeds`.* from `kg_feeds` inner join `kg_shops` on `kg_shops`.`id` = `kg_  
  feeds`.`shop_id` and `kg_shops`.`active` = `1`) (Bindings: array (                                                                                        )) 

As you can see, the multiple conditions in the join go fine, but it thinks the 1 is a column instead of a string. Is this even possible, or do I have to fix it in the where.

Thanks in advance!

Php Solutions


Solution 1 - Php

return $query->join('kg_shops', function($join)
 {
   $join->on('kg_shops.id', '=', 'kg_feeds.shop_id');
    
 })
 ->select('required column names') 
 ->where('kg_shops.active', 1)
 ->get();

Solution 2 - Php

You can see the following code to solved the problem

return $query->join('kg_shops', function($join)
{
    $join->on('kg_shops.id', '=', 'kg_feeds.shop_id');
    $join->where('kg_shops.active','=', 1);
});

Or another way to solved it

 return $query->join('kg_shops', function($join)
{
    $join->on('kg_shops.id', '=', 'kg_feeds.shop_id');
    $join->on('kg_shops.active','=', DB::raw('1'));
});

Solution 3 - Php

//You may use this example. Might be help you...

$user = User::select("users.*","items.id as itemId","jobs.id as jobId")
        ->join("items","items.user_id","=","users.id")
        ->join("jobs",function($join){
            $join->on("jobs.user_id","=","users.id")
                ->on("jobs.item_id","=","items.id");
        })
        ->get();
print_r($user);

Solution 4 - Php

Because you did it in such a way that it thinks both are join conditions in your code given below:

public function scopeShops($query) {
    return $query->join('kg_shops', function($join)
    {
        $join->on('kg_shops.id', '=', 'kg_feeds.shop_id');
        $join->on('kg_shops.active', '=', "1");
    });
}

So,you should remove the second line:

return $query->join('kg_shops', function($join)
{
    $join->on('kg_shops.id', '=', 'kg_feeds.shop_id');
});

Now, you should add a where clause and it should be like this:

return $query->join('kg_shops', function($join)
{
  $join->on('kg_shops.id', '=', 'kg_feeds.shop_id')->where('kg_shops.active', 1);
})->get();

Solution 5 - Php

You can simply add multiple conditions by adding them as where() inside the join closure

->leftJoin('table2 AS b', function($join){
        $join->on('a.field1', '=', 'b.field2')
        ->where('b.field3', '=', true)
        ->where('b.field4', '=', '1');
})

Solution 6 - Php

More with where in (list_of_items):

    $linkIds = $user->links()->pluck('id')->toArray();

    $tags = Tag::query()
        ->join('link_tag', function (JoinClause $join) use ($linkIds) {
            $joinClause = $join->on('tags.id', '=', 'link_tag.tag_id');
            $joinClause->on('link_tag.link_id', 'in', $linkIds ?: [-1], 'and', true);
        })
        ->groupBy('link_tag.tag_id')
        ->get();

    return $tags;

Hope it helpful ;)

Solution 7 - Php

This is not politically correct but works

   ->leftJoin("players as p","n.item_id", "=", DB::raw("p.id_player and n.type='player'"))

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionWouter NeuteboomView Question on Stackoverflow
Solution 1 - PhpramaView Answer on Stackoverflow
Solution 2 - PhpMajbah HabibView Answer on Stackoverflow
Solution 3 - PhpYagnesh bhalalaView Answer on Stackoverflow
Solution 4 - PhpThe AlphaView Answer on Stackoverflow
Solution 5 - PhpKamleshView Answer on Stackoverflow
Solution 6 - PhpThangTDView Answer on Stackoverflow
Solution 7 - PhpUncokeView Answer on Stackoverflow