Question

jquery ajax servlet取值

请教，如何从servlet中把jQuery的ajax中以json格式传入的数据接收到?

javaScript:
$('#but_json_json').click(function(){
var j ={"name":"王","password":123456};
$.ajax(
{
url:"servlet/JsonObject", //访问路径
type:"POST", //访问方式
data:j, //传入服务端的数据
dataType:"json",
contentType:"application/json;charset=utf-8",
success : function(data){
alert(data);
alert(data.name);

alert(data.password);
}

}
);
});

Servlet：

public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
//如接收客户端传入的对象j?
//此处接收到的都是null
String user = request.getParameter("j");
String name = request.getParameter("name");
String password = request.getParameter("password");
System.out.println(user);

//JSON对象
JSONObject jsonObject = new JSONObject();
jsonObject.accumulate("password", password).accumulate("name", "www");
response.setContentType("application/json");
response.getWriter().write(jsonObject.toString());
}

希望知道的朋友指点一二

Answer 1

url:"servlet/JsonObject", //访问路径
type:"POST", //访问方式
data:{ j: "j"}, //传入服务端的数据
dataType:"json",
data格式改下
然后 servlet 按普通的数据接收就行了
传到servlet的数据j
只是追加到了url上ervlet/JsonObject?j=j

Answer 2

String param1 = request.getParameter("param1");
json 改为 var p={}; p.param1:"abcd";或者 p={param1:"abcd",param2:"def"};试试
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException { init(request,response);}
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException { init(request,response);}
public void init(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException { 主程序内容}

Answer 3

data:{"name":"王","password":123456}

之后后台 String name= request.getParameter("name");

String password = request.getParameter("password");