javascript
  • css
  • ajax
  • laravel
  • modal-dialog
  • 2017-08-04 18 views 1 likes 
    1

    我試圖創建圖像廚房,當圖像點擊模式打開,但是當我加載我的網頁一會兒勳章的內容顯示,我不想要它如何可以修復它嗎?我在view.blade.php代碼是:創建圖像庫莫塔爾與ajax在Laravel

    <script> 
        $(".li-img").click(function() { 
        $.ajax({ 
         method: 'GET', 
         url: './comment?media_id=' + this.id, 
         success: function (data) { 
          /*$("#comments").html(data);**/ 
          $("#getCodeModal").modal("toggle"); 
          $("#getCode").html(data); 
    
         } 
        }); 
        }); 
    </script> 
        <body> 
        @foreach($array as $img) 
        <div class=" portfolio-container text-center"> 
        <ul class="portfolio-list" style="margin:0 auto"> 
         <li id = "{{$img['id']}}" class="li li-img" style="margin:0 
          auto"> 
          <img id="{{$img['id']}}" src="{{$img['image']}}"> 
         </li> 
        </ul> 
        </div> 
        @endforeach 
        <div class="modal fade" id="getCodeModal" tabindex="-1" role="dialog" 
        aria- 
        labelledby="myModalLabel" aria-hidden="true"> 
        <div class="modal-dialog modal-lg"> 
         <div class="modal-content"> 
         <div class="modal-header"> 
          <button type="button" class="close" data-dismiss="modal" aria- 
          label="Close"><span aria-hidden="true">&times;</span> 
          </button> 
         <h4 class="modal-title" id="myModalLabel"> API CODE </h4> 
         </div> 
         <div class="modal-body" id="getCode" style="overflow-x: scroll;"> 
         //ajax success content here. 
         </div> 
        </div> 
    </div> 
    

    我也想從控制器的模式我的控制器發送JSON信息有這樣的代碼:

    public function comment(Request $request) 
        { 
        $media_id = $request->get('media_id'); 
        if ($request->session()->has("access_token")) { 
        $access_token = $request->session()->get("access_token"); 
        $client = new Client(); 
        $res = $client->request('GET', 'https://api.instagram.com/v1/media/' 
         . $media_id . '/comments?access_token=' . $access_token); 
        /*echo /*$res->getBody();*/ 
        } else { 
         echo "there is no access token in session"; 
        } 
    } 
    

    我想送$水庫> getbody()模態view.blade.php

    回答

    0

    因爲你沒有添加引導作爲依賴。

    將它添加到head標籤:

    <link rel="stylesheet" type="text/css" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css"> 
    

    ,我發現,你失去了一位親密的標籤</div>這div有一類modal

    +0

    我加了這個鏈接,但片刻的模式內容顯示在頁 – Honey

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