Laravel Ajax 分页

Question

我正在尝试基于复选框单击来获取 ajax 搜索过滤器的分页。每个复选框单击时我的控制器功能：

public function ajaxGetServices(){

    $a_data = Input::all();

    Session::put('a_data',$a_data);
    $i=0;
    if(count($a_data) == 0){

        $temp = User::has('editor')->paginate(4);
        $this->data['editors'] = $temp->toArray();

        if(Auth::user()){
            $this->data['favorited_editors'] = Favorite::where('member_id','=',Auth::user()->id)->get()->toArray();
            $i=1;
        }


    }else{


        $editors = User::with(['editor.credentials','editor.specialties','editor.ratings']);
        $temp=$editors->whereHas('editor', function($q) use ($a_data){
            $q->whereHas('specialties',function($sq) use($a_data){
            $sq->whereIn('specialty',$a_data);
            });
        })->paginate(2);

        //var_dump($temp->toJson()); die();

        $this->data['editors'] = $temp->toArray();
        if(Auth::user()){
            $this->data['favorited_editors'] = Favorite::where('member_id','=',Auth::user()->id)->get()->toArray();
            $i=1;
        }

    }//endelse

    if($i==1){

        return Response::json(array('editors' => $this->data['editors'], 'favorited_editors' => $this->data['favorited_editors']));

    }
    else{
        $this->data['favorited_editors'] = array();
        return Response::json(array('editors' => $this->data['editors'], 'favorited_editors' => $this->data['favorited_editors']));
    }


}

这给了我总页数、当前页、每页、从和到分页字段。现在我需要重新生成分页链接以反映上述结果。请指导我如何去做。此外，如果需要，我可以提供更多代码来澄清我的问题。感谢您的建议。

Answer 1

也许这篇文章会有所帮助 http://www.laravel-tricks.com/tricks/easy-ajax-pagination https://gist.github.com/tobysteward/6163902（完整示例所以我没有提供任何解释）