JSON响应作为文件打开,但我无法使用JavaScript访问它

时间:2011-03-03 12:35:36

标签: javascript jquery json response

在下面的代码中,我向以这种方式回复的servlet发出POST请求:

response.setContentType("application/json");
json = "{success:true,sessionUid:\""+sessionUid+"\"}";
response.getWriter().write(json);

所以Firefox打开它就像一个文件,我可以看到它没问题。在这里你有JSON:

{success:true,sessionUid:"D07WC15R7LFRFRGPF4P5"}

问题是我无法检查JSON对象。它似乎不存在于我的回调函数中(也使用Firebug)。看看代码和警报。

<html>
    <head>
        <script type="text/javascript" src="http://code.jquery.com/jquery-1.5.1.min.js"></script>
        <script type="text/javascript">
            $(document).ready(function() {
                $("#loginForm").submit(function(response){
alert("response="+response);    //output: "response=[object Object]"
                    var obj = jQuery.parseJSON(response);
alert("obj.sessionUid="+obj.sessionUid);    //doesn't work, Firebug says "obj is null"
                    if (response.success == true){ //never true
                        document.location.href = 'http://localhost:8080/QuoteroClient/logged.jsp';
                    }else{
                        alert("Something went wrong in the login process.");
                    }
                    return false;
                });
            });
        </script>
    </head>
    <body>
        <form id="loginForm" action="http://localhost:8080/QuoteroClient/Main?servlet=Security" method="post">
            <fieldset><legend>Login to Quotero:</legend>
                <label>Action:</label><input type="text" name="action" value="login"/><br />
                <label>Username:</label><input type="text" name="login-quotero" value="admin"/><br />
                <label>Password:</label><input type="text" name="password-quotero" value="admin" /><br />
                <label>Domain:</label><input type="text" name="combo-domain" value="Quotero" /><br />
            </fieldset>
            <input type="submit" value="Submit" />
        </form>
    </body>
</html>

编辑:我也试图对AJAX请求做同样的事情,但是没有成功:

$("#ajaxSubmit").click(function () {
  $.ajax({
    type: "GET", //GET or POST is the same for this servlet
    url: "http://localhost:8080/QuoteroClient/Main?servlet=Security&action=login&login-quotero=admin&password-quotero=admin&combo-domain=Quotero",
    dataType: "json",
    success: function (response) {
alert("response=" + response);
      var obj = jQuery.parseJSON("" + response);
alert("obj.sessionUid=" + obj.sessionUid);
      if (response.success == true) {
        document.location.href = contextPath + 'http://localhost:8080/QuoteroClient/logged.jsp';
      } else {
        alert("Something went wrong in the login process.");
      }
    }
  });
  return false;
});

2 个答案:

答案 0 :(得分:3)

这是无效的JSON:

{success:true,sessionUid:"D07WC15R7LFRFRGPF4P5"}

这是有效的JSON:

{"success":true,"sessionUid":"D07WC15R7LFRFRGPF4P5"}

在JSON中,必须始终引用键。请参阅DEMO

答案 1 :(得分:0)

我认为你把ajax与提交混淆了。提交只是一个事件,当提交表单时,请执行以下操作。那么你可以

 $("#loginForm").submit(function(){
     var post_data = $(this).serialize();
     $.ajax({
        url: '',//url of the php file that handles the forms
        type: 'GET',
        dataType:'json',
        data:post_data,//this are the query strings, e.g. ?q=blabla&s=blabla
        success:function (data){//if page was 200 or successfully loaded
             alert(data.sessionUid);
             // do what ever you wish with the data here
        },
        error: function (){
            alert('Page failed to load');
        }
     })
     return false;
 });