Question

<script type="text/javascript" src="jquery.mobile/jquery-1.7.2.min"></script>
<script type="text/javascript" src="docs/assets/js/jquery.jsonp.js"></script>
<script> 
     $(document).ready(function(){
            var output = $('#output');
            $.ajax({
                url: 'http://musi.php/?oper=getds&dev_id=f587&cur_id=2',
                dataType: 'jsonp',
                jsonp: 'jsoncallback',
                timeout: 5000,
                success: function(data, status){
                    $.each(data, function(i,item){ 
                        var landmark ='<h1>'+item.Nick_Name+'</h1>';
                        output.append(landmark);
                    });
                },
                error: function(){
                    output.text('There was an error loading the data.');
                }
            });
        });
</script>

我有一个url（例如）和它的json值如下：

{"RetVal":"Ok","ValueRsp":[{"Feedback_Id":"22","Customer_Id":"543","Feedback_Type_Id":"1","First_Name":"Tester Tester","Last_Name":"NA","Nick_Name":"Xgcfyxfu"}]}

error: invalid label
Line 1 [Break On This Error] 
{"RetVal":"Ok","ValueRsp":[{"Feedback_Id":"22","Customer_Id":"543","Feedback_Typ...

我需要解析它并在我的应用中使用它。我是javascript的新手。请帮助

Answer 1

您可以使用jQuery.parseJSON

<强> Live Demo

jsonobj = $.parseJSON('{"RetVal":"Ok","ValueRsp":[{"Feedback_Id":"22","Customer_Id":"543","Feedback_Type_Id":"1","First_Name":"Tester Tester","Last_Name":"NA","Nick_Name":"Xgcfyxfu"}]}');
alert(jsonobj.RetVal);
alert("jsonobj.ValueRsp[0].Customer_Id: " + jsonobj.ValueRsp[0].Customer_Id);

Answer 2

有点谷歌搜索会做到这一点......

var obj = JSON.parse(yourString);
// now use obj.RetVal

Answer 3

你可以使用这个功能：

var json_obj = JSON.parse("correct_json_format_string");

现在您可以使用json_object访问所有属性：

var return_value = json_obj.RetVal;

希望有所帮助。

json数组到phonegap中的javascript

3 个答案: