2015-04-02 50 views
1

我正在開發一個驚悚電影網站。我想從IMDB獲取數據。我需要使用Ajax來異步獲取數據。我的代碼不能正常工作。如何使用PHP中的Ajax從網站(例如IMDB)獲取數據

電影信息沒有出現在carousal中。 ,其取HTML文件數據:

<?php 
     require_once('functions.php'); 
     $triangle = getMovieInfo('triangle'); 
     $Predestination = getMovieInfo('Predestination'); 
     $silence_of_the_lambs = getMovieInfo('The Silence of the Lambs'); 
     $shutter_island = getMovieInfo('Shutter Island'); 
    ?> 
    <br> 
    <!-- Text which has greater value --> 
    <div> 
     <div class = "row"> 
     <div class="alert alert-success text-center" role="alert"><h3>If you are looking for a site, where you can see the list of all verified thrillers, then you are at the right place.</h3></div> 
     </div> 
    </div> 
    <!-- slider for the home page --> 

     <div class="container"> 

     <div class="row"> 
     <div class = "col-md-6 col-lg-6 col-sm-6"> 
      <div id="myCarousel" class="carousel slide" data-ride="carousel"> 
      <!-- Indicators --> 
      <ol class="carousel-indicators"> 
       <li data-target="#myCarousel" data-slide-to="0" class="active"></li> 
       <li data-target="#myCarousel" data-slide-to="1"></li> 
       <li data-target="#myCarousel" data-slide-to="2"></li> 
       <li data-target="#myCarousel" data-slide-to="3"></li> 
      </ol> 

      <!-- Wrapper for slides --> 
      <div class="carousel-inner" role="listbox"> 

       <div class="item active"> 
       <?php echo "<img src=\"$triangle->Poster\">"; ?> 
       <!-- <span> <?php echo "Plot : ".$triangle->Plot; ?> </span> --> 
       </div> 
       <div class="item"> 
       <?php echo "<img src=\"$silence_of_the_lambs->Poster\">"; ?> 
       <!-- <span> <?php echo "Plot : ".$silence_of_the_lambs->Plot; ?> </span> --> 
       </div> 
       <div class="item"> 
       <?php echo "<img src=\"$Predestination->Poster\">"; ?> 
       <!-- <span> <?php echo "Plot : ".$Predestination->Plot; ?> </span> --> 
       </div> 
       <div class="item"> 
       <?php echo "<img src=\"$shutter_island->Poster\">"; ?> 
       <!-- <span> <?php echo "Plot : ".$shutter_island->Plot; ?> </span> --> 
       </div> 
      </div> 

      <!-- Left and right controls --> 
      <a class="left carousel-control" href="#myCarousel" role="button" data-slide="prev"> 
       <span class="glyphicon glyphicon-chevron-left" aria-hidden="true"></span> 
       <span class="sr-only">Previous</span> 
      </a> 
      <a class="right carousel-control" href="#myCarousel" role="button" data-slide="next"> 
       <span class="glyphicon glyphicon-chevron-right" aria-hidden="true"></span> 
       <span class="sr-only">Next</span> 
      </a> 
      </div> 
      </div> 
      <div class ="col-md-6 col-lg-6 col-sm-6"> 
      <div class="well"> 
       <div class="thriller"> 
        <h3>What is a thriller movie?</h3> 
        <h4>If the genre is to be defined strictly, a genuine thriller is a film that rentlessly pursues a single-minded goal - to provide thrills and keep the audience cliff-hanging at the 'edge of their seats' as the plot builds towards a climax.</h4> 
        <h4> 
        Believe me, I love watching thriller movies more than anything. These movies are real movies, where you can't guess the last scene. I enjoyed a lot of thriller movies and then thought to make a site, where you can easily watch a movie. 
        </h4> 
       </div> 
      </div> 
      </div> 
     </div> 

    </div> 

的getMovieInfo功能是在這裏: 功能getMovieInfo($標題) { $標題=用urlencode($標題);

我的Java腳本文件是:

$(document).ready(function() 
{ 
    $("#home").click(function(event) 
    { 
     alert("home is clicked"); 
     $('#contents').load('home.html'); 
    }); 


}); 

The movies info is not printed. 請幫我解決這個問題。

+0

「我的代碼無法正常工作。」真的?如何處理一些特定的問題,錯誤信息等? – zdeniiik 2015-04-02 12:13:36

+0

我正在添加一個屏幕截圖.. – khan 2015-04-02 12:14:00

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