Ajax请求不在Iphone上工作但在笔记本电脑上工作

时间:2016-07-13 18:38:48

标签: javascript jquery ios iphone ajax

所以我的简单的ajax请求不适用于使用Safari的iPhone IOS。它适用于我的笔记本电脑,适用于所有浏览器 - 包括safari。

当我说它不起作用时,我的意思是成功方法中的数据是空的,但事实并非如此。

            $.ajax({
                url: '/ajax-get-video-url-iphone',
                type: "post",
                data: {
                    'id' : id,   //this is correct, I alert(id) and it shows correct
                    '_token': $("meta[name=_token]").attr('content')  //this is also correct
                },
                success: function(data){
                    alert('hello'); // This pops up on the screen
                    alert(data);    // This pops up as nothing, empty but it should be a url
                }
            });

上面的代码在其他设备和浏览器上完全正常,所以我不知道为什么会发生这种情况。

修改::

这是我的后端代码,顺便提一下

public function ajaxGetVideoUrlIphone(Request $request)
{
    $videoId = $request['id'];

    if(Auth::check()){
        $user = Auth::user();
        $userOwnsVideo = 0;

        $shoot = Shoot::where('id',$videoId)->with('users')->get();

        if($user->subscribed('main')){
            $userOwnsVideo = 1;
        }else{

            if(!empty($shoot[0]['users']) && count($shoot[0]['users']) > 0){
                //this stops the loop once its found the users video
                $endForEach = 0;
                foreach($shoot[0]['users'] as $userVideo){
                    if($endForEach == 0){
                        if($userVideo['pivot']['user_id'] == $user['id']){
                            $endForEach++;
                            $userOwnsVideo = 1;
                        }
                    }else{
                        break;
                    }
                }
            }
        }
        if($userOwnsVideo){
            $shoot = Shoot::where('id', $videoId)->get();
            $url = $shoot[0]['iphone_full'];
        }else{
            $shoot = Shoot::where('id', $videoId)->get();
            $url = $shoot[0]['iphone_trailer'];
        }

    }else{
        $shoot = Shoot::where('id', $videoId)->get();
        $url = $shoot[0]['iphone_trailer'];
    }

    //return 'return from route';
    return $url;
}

0 个答案:

没有答案