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在Javaservlet中分别从json对象获取值_Java_Json_Servlet 3.0 - Fatal编程技术网

在Javaservlet中分别从json对象获取值

在Javaservlet中分别从json对象获取值,java,json,servlet-3.0,Java,Json,Servlet 3.0,这是我将json对象发送到servlet的javascript代码 username = "Nash"; password = "619here"; type = "all"; data = '{ "username": "' + username + '", "password": "' + password + '", "type": "' + type + '" }'; request = JSON.parse(data); $.getJSON(url, request, functio

这是我将json对象发送到servlet的javascript代码

username = "Nash";
password = "619here";
type = "all";
data = '{ "username": "' + username + '", "password": "' + password + '", "type": "' + type + '" }';
request = JSON.parse(data);

$.getJSON(url, request, function (response) {
    $("#result").append("<h1>Success</h1>");
    $.each(response.vehicle, function (no, vehicle) {

        $.each(vehicle, function (key, value) {

            $("#result").append("<h2>" + key + " : " + value + "</h2>");

        });

        $("#result").append("<br>");
    });

})
我使用的是json-lib-2.4-jdk15.jar
请帮助我..

在servlet中,在
do[Get,Post]()
方法中执行以下操作

JSONObject jsonObject = JSONObject.fromObject( request.getQueryString() );  

String username= jsonObject.get( "username" );  
String password= jsonObject.get( "password" ); 
String type= jsonObject.get( "type" ); 
您还需要从JS中删除以下行

request = JSON.parse(data);

它给了这个例外。。net.sf.json.JSONException:JSONObject文本必须在callback=jquery19099722470087862343_1370721676614&username=Nash&password=619here&type=all&_=1370721676615的字符1处以“{”开头
request = JSON.parse(data);