如何在服务器上使用AJAX发布JSON数据

Question

我在黑莓开发上做了一个应用程序我在服务器上发布了一些数据。我能够做到这一点。但现在我正在学习jQuery和jQuery的移动。现在我需要发布数据在服务器上使用query.I使我的Ui在jQuery Mobile中。我需要一些帮助在服务器上发布数据。 我不喜欢在黑莓

Name gg 
Depot Barrow 
Turn No 1 
Date/Time:28:07:2013 02:22 
origin Ardwick 
Dest barrow 
Headcode: hnh 
Status :No 1st class Impact 

{"headcode":"hnh","destination":"Barrow","origin":"Ardwick","time":"28:07:2013 02:22","turnNumber":"1","depot":"Barrow","conductorName":"gg","devicePin":"123456.78.364813.8","noFirstClassImpact":"true","customersInvited":"false","customersUninvited":"false","customersLeft":"false"} 

这里是我的网址

public static String fsReportUrl = "http://50.57.145.165:8180/FTPEReport/ftpereports/fsreport?fsReport="; 

现在我需要发布此使用查询。这里是我的小提琴。 http://jsfiddle.net/ravi1989/NKBUF/2/

<div data-role="page" id="Home" > 
    <div data-role="content"> 

      <label for="name" style="text-align:top;margin-left: 0px;" >conductorName:</label> 
         <input name="name" id="name" value="" type="text" class="Name_h" autocorrect="off"> 

    <label for="deport" style="text-align:top;margin-left: 0px;" >Deport:</label> 
         <input name="deport" id="deport" value="" type="text" class="deport_h" autocorrect="off"> 

     <label for="dateandTime" style="text-align:top;margin-left: 0px;" >dateTime:</label> 
     <input name="dateandTime" id="dateandTime" value="" type="date" class=""> 

        <label for="origin" style="text-align:top;margin-left: 0px;" >origin:</label> 
     <input name="origin" id="origin" value="" type="text" class=""> 
           <label for="dest" style="text-align:top;margin-left: 0px;" >Destination:</label> 
     <input name="dest" id="dest" value="" type="text" class=""> 
     <label for="headcode" style="text-align:top;margin-left: 0px;" >Headcode:</label> 
     <input name="headcode" id="headcode" value="" type="text" class=""> 
       <label for="devicepin" style="text-align:top;margin-left: 0px;" >Devicepin:</label> 
     <input name="devicepin" id="devicepin" value="" type="text" class=""> 
      <div data-role="fieldcontain"> 
    <fieldset data-role="controlgroup"> 

     <input type="checkbox" name="checkbox-1" id="checkbox-1" class="custom" /> 
     <label for="checkbox-1">customersInvited</label> 
     <input type="checkbox" name="checkbox-2" id="checkbox-2" class="custom" /> 
     <label for="checkbox-2">customersunInvited</label> 
     <input type="checkbox" name="checkbox-3" id="checkbox-3" class="custom" /> 
     <label for="checkbox-3">customerLeft</label> 
     <input type="checkbox" name="checkbox-4" id="checkbox-4" class="custom" /> 
     <label for="checkbox-4">noFirstClassImpact</label> 
    </fieldset> 
</div> 
           <a href="#" data-role="button" data-corners="false" id="callJsonFunfunction">Call webservice</a> 

      </div> 

</div> 

我想用ajax我们会做essily？

我正在使用像BB那样。

public static String postJson(JSONObject jsonObj, String url) { 
     String response = ""; // this variable used for the server response 

      JSONObject postData = jsonObj; 
      String valueObj = ""; 

      try { 

       UrlImpl fullUrl = new UrlImpl(); 
       fullUrl.setBaseUrl(url); 
       valueObj = fullUrl.getFullUrl(); 

       HttpConnection connection = (HttpConnection) Connector 
         .open(valueObj); 
       // set the header property 
       connection.setRequestMethod(HttpConnection.POST); 
       connection.setRequestProperty("Content-Length", 
         Integer.toString(postData.length())); 
       connection.setRequestProperty("Content-Type", 
         "application/json;charset=UTF-8"); 
       byte[] postDataByte = postData.toString().getBytes("UTF-8"); 
       OutputStream out = connection.openOutputStream(); 
       out.write(postDataByte); 
       out.flush(); 
       out.close(); 
       int responseCode = connection.getResponseCode(); 
       if (responseCode == HttpConnection.HTTP_OK) { 

        InputStream in = connection.openInputStream(); 
        StringBuffer buf = new StringBuffer(); 
        int read = -1; 
        while ((read = in.read()) != -1) 
         buf.append((char) read); 

        response = buf.toString(); 
       } 

       connection.close(); 
      } catch (Exception e) { 
       XLogger.error(DataAccess.class, " Error in post reports -- " + e); 
      } 

     return response; 

    } 

Answer 1

这也许应该是这个样子：

$(document).ready(function() { 

    $('#callJsonFunfunction').on('click', function() { 
     var 
      data = $(this).closest('[data-role="content"]').find('input').serialize(); 

     $.post("http://50.57.145.165:8180/FTPEReport/ftpereports/fsreport?fsReport=", data, function(resp) { 
      alert("success"); 
      console.log(resp); 
     }) 
     .done(function() { alert("second success"); }) 
     .fail(function() { alert("error"); }) 
     .always(function() { alert("finished"); }); 
    }); 

});