fastjson如何解析list类型的json字符串？

Question

jquery表单转换了一个列表：

$.fn.serializeObject = function() {
		var o = {};
		var a = this.serializeArray();
		$.each(a, function() {
			if (o[this.name]) {
				if (!o[this.name].push) {
					o[this.name] = [ o[this.name] ];
				}
				o[this.name].push(this.value || '');
			} else {
				o[this.name] = this.value || '';
			}
		});
		return o;
	};

	function onchang() {
		//var v = $("#form1").serializeObject();
		var v = $("#form1").serializeObject();
		var sjson = JSON.stringify(v);
		//alert(sjson);
		//$("p").append(sjson);
		$.ajax({
			type : "post",
			//data:{"username":"username","password":"password"},
			//data:{"jsonString":'{"username":"username","password":"password"}'},
			data : {
				"jsonString" : sjson
			},

			//data : {test:test,username:userDao5,password:userDao},

			//data: sjson,
			url : "Batch_batchChangAll.action",
			dataType : "JSON",
			success : alert("发送成功")
		});

	}

输出的json字符串是：

{"username":["1","2"],"password":["1","2"],"charaname":["1","2"]} 

那么请问怎么在java后台解析这样格式的json呢？

我已经定义好User.class，里面也有上面的三个值了。。

Answer 1

谢谢，知道怎么操作了！

Answer 2

大大，麻烦您有空帮我再看看这个问题：http://www.oschina.net/question/143998_2151958

Answer 3

String str = "{"username":["1","2"],"password":["1","2"],"charaname":["1","2"]} ";
	JSONObject obj = JSONObject.parseObject(str);
	if(obj.get("username")!=null){
		JSONArray username = obj.getJSONArray("username");
		JSONArray password = obj.getJSONArray("password");
		JSONArray charaname = obj.getJSONArray("charaname");
		for(int i=0;i<username.size();i++){
			User user = new User();
			user.setUsername(username.getString(i));
			//.....
		}
	}

Answer 4

JSONArray.parseArray(jsonstr,User.class)

Answer 5

是不是要将类型为 List<User> 的对象 打印到控制台？因为你使用了fastjson，可以使用JSON.toJSONString(list)