语法错误，在Laravel

Question

意外“$ POST_ID”（T_VARIABLE）我有一个问题，我在laravel

我试图执行下面的代码功能：

public static function likePost($value) 
{ 
    $post_id = $value['post_id']; 
    $user_id = $value['user_id']; 
    if($value['liked'] == "1") 
    { 
     $model = new Like; 
     $model->fill($value); 
     $model->created = time(); 
     if($model->save()) 
     { 
     $post = Post::with('user')->where('id', '=', $value['post_id'])->get()->first(); 
     $text = $post['user']['name']." ".Notification::TEXT_LIKE_POST; 
     Push::push($text, $value['post_id']); 
     } 
    } 
    else 
    { 
     Like::where('user_id', '=', $user_id)->where('post_id', '=' $post_id)->get()->first()->delete(); 
    } 
} 

但我发现了出现以下错误：syntax error, unexpected '$post_id' (T_VARIABLE)

这个函数有什么问题？如果$ post_id工作正常，但在其他方面不起作用。

Answer 1

由于@andrewsi说你在'='之后缺少逗号。您也可以只使用此代码，这将做同样的工作：

Like::where('user_id', $user_id)->where('post_id', $post_id)->delete(); 

Answer 2

你缺少一个,在 Like::where('user_id', '=', $user_id)->where('post_id', '=' $post_id)->get()->first()->delete();

应该 Like::where('user_id', '=', $user_id)->where('post_id', '=', $post_id)->get()->first()->delete();

或 Like::where('user_id', '=', $user_id)->where('post_id', $post_id)->get()->first()->delete();