您的位置:首页 > Web前端 > JavaScript

fastJson将json字符串转换为map

2017-03-24 16:22 429 查看
public class JsonUtil {

/**
* 将json转化成map
* @param jsonStr
* @return
*/
public static Map<String, Object> convertJsonStrToMap(String jsonStr){

Map<String, Object> map = JSON.parseObject(
jsonStr,new TypeReference<Map<String, Object>>(){} );

return map;
}
}

String jsonData = "{id:100,list:[{a:1},{a:2}]";
Map<String, Object> map = JsonUtil.convertJsonStrToMap(jsonData);
System.out.println(map);
Integer id = MapUtils.getInteger(map, "id");
Object list = MapUtils.getObject(map, "list");
List<JSONObject> ll = (List)list;
for (JSONObject s : ll) {
String eleVal = s.get("a").toString();
System.out.println(s+"---a的值是:"+eleVal);
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  json fastjson