2015-02-10 64 views
0

我想通过让用户选择合适的图像将用户转发到其他页面。用户选择两张图像时出现问题。我不知道如何设置转发,当用户选择这两个图像,任何想法? Here's我用的工具%windir%\Temp文件夹。谢谢你们的帮助!如何在此JavaScript代码(imagepicker)中获得另一个值?

<html> 
<head> 
     <!-- scripts for imagepicker --> 
    <link rel="stylesheet" href="assets/css/image-picker.css"> 
    <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script> 
    <script type="text/javascript" src="assets/js/image-picker.js"></script> 
    <script type="text/javascript"> 


     "use strict"; 
     //wait for the page to be fully loaded 
     window.onload = function() { 
      initialize(); 
     } 
    </script> 
    <title></title> 
</head> 
<body> 
    <div class="container"> 
     <br> <br> <br> <br> 
     <div class="jumbotron"> 

        <!-- 2 obrazki --> 

        <div class="row"> 
      <div class="col-xs-6 col-md-3"></div> 
      <center> 
      <select multiple="multiple" class="image-picker show-html" id="selectImg"> 

      <div class="col-xs-6 col-md-3 flags container" > 
      <option class="col-xs-6 col-md-3 flags" data-img-src="assets/media/img/german.png" value="1"></option> 
      </div> 

      <div class="col-xs-6 col-md-3 flags container" > 
      <option class="col-xs-6 col-md-3 flags" data-img-src="assets/media/img/spain.png" value="2"></option> 
      </div> 

     </div> 
     </select> 
     </center> 
      <!-- 2 obrazki --> 
       </div> 
       <div class="form-group container"> 
       <center> 
        <p>Kliknij tu i <br><a id="forwardButton" class="btn btn-lg btn-success" type="submit" href="#" role="button">Dołącz do Poliglotów</a></p> 
       </center> 
       </div> 
      </form> 
     </div> 
    </div> 


    <!--SCRIPTS imgpickr DON'T TOUCH--> 

    <script> 

    $("select").imagepicker(); 
    function initialize() { 

     var choice; //= document.getElementById("selectImg").value; 


     //changes destination when you set or change your choice 
     document.getElementById("selectImg").onchange = function() { 

      choice = document.getElementById("selectImg").value; 


     } 
     //when button is clicked 
     document.getElementById("forwardButton").onclick = function() { 


      if(choice == 1){ 
       window.location = "https://www.google.com"; 
      } 
      else if(choice == 2){ 
       window.location = "https://www.facebook.com"; 

      } 
      else if(choice == 1 && choice == 2){ 
       window.location = "https://www.twitter.com"; 

      } 

     } 
    } 


    </script> 




</body> 
</html> 

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