在Javaservlet中分别从json对象获取值
这是我将json对象发送到servlet的javascript代码在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
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);