2014-03-04 42 views
0

我使用jquery代碼來調用ashx文件。它完美的工作,而頁面加載。當我從另一個功能調用該功能意味着不能正常工作。我在這裏放置了兩個代碼。無法在函數內調用ashx

請檢查並幫助我。

<script type="text/javascript"> 
     $(document).ready(function() { 
        try { 
         $("#submit").click(function() { 
          var user = $("#login").val(); 
          var pass = $("#password").val();      
          CallLoginHandler(user, pass); 
         } 
        }catch(ex){ 
     } 
</script> 

<script type="text/javascript"> 
     function CallLoginHandler(user, pass) {   
      $.ajax({ 
       type: "POST", 
       url: "../handler/JQGridHandler.ashx?MainPage=GetUserDetails&Type=2&user=" + user + "&pass=" + pass + "", 
       contentType: "application/json; charset=utf-8", 
       dataType: "json", 
       success: function (response) { 
        if (response.length > 0) { 

         alert(response[0]["FLD_ID"]); 
         //successCallback(response); 
        } else { 
         alert(response[0]["FLD_ID"]); 
        } 
       } 
      }); 

     } 

</script> 

回答

0

我想你錯過了正確的jQuery語法。嘗試以下修改代碼: -

<script type="text/javascript"> 
    $(document).ready(function() { 

     try { 
      $("#submit").click(function() { 

       var user = $("#login").val(); 
       var pass = $("#password").val();      
       CallLoginHandler(user, pass); 
      }); 

     } catch (ex) { 

     } 

    }); 


    function CallLoginHandler(user, pass) { 
     $.ajax({ 
      type: "POST", 
      url: "../handler/JQGridHandler.ashx?MainPage=GetUserDetails&Type=2&user=" + user + "&pass=" + pass + "", 
      contentType: "application/json; charset=utf-8", 
      dataType: "json", 
      success: function (response) { 
       if (response.length > 0) { 

        alert(response[0]["FLD_ID"]); 
        //successCallback(response); 
       } else { 
        alert(response[0]["FLD_ID"]); 
       } 
      } 
     }); 

    } 

</script> 
+0

不工作Dude .. – user3085540