Laravel：在Join Query中用戶定義的對象名稱

Question

我有成功地返回在Laravel中寫入的聯接查詢。我現在試圖將對象名添加到返回的值，就好像它從表中返回一樣。

Laravel查詢

$post = DB::table('follow') 
    ->join('posts', 'follow.user2', '=', 'posts.userid') 
    ->where('follow.user1',Auth::user()->id) 
    ->where('follow.user2','!=',Auth::user()->id) 
    ->where('posts.created_at','>',$update) 
    ->select('posts.created_at', 'posts.userid') 
    ->orderBy('posts.created_at','desc') 
    ->get(); 

上面的查詢返回以下

array (size=1) 
0 => 
    object(stdClass)[208] 
    public 'created_at' => int 1418466963 
    public 'userid' => int 5 

我想實現的是下面的輸出

array (size=1) 
0 => 
    object(stdClass)[208] 
    public 'created_at' => int 1418466963 
    public 'userid' => int 5 
    public 'oType' => string 'post' //This is user defined. 

我想什麼是（顯然是錯誤的，但只是暗示我在想什麼）

$post = DB::table('follow') 
    ->join('posts', 'follow.user2', '=', 'posts.userid') 
    ->where('follow.user1',Auth::user()->id) 
    ->where('follow.user2','!=',Auth::user()->id) 
    ->where('posts.created_at','>',$update) 
    ->select('posts.created_at', 'posts.userid', 'oType as post') //Compare this line with 1st query 
    ->orderBy('posts.created_at','desc') 
    ->get(); 

Answer 1

您可以通過使用DB :: raw（）來實現此目的。

->select('posts.created_at', 'posts.userid', DB::raw('\'post\' as oType'))