无法使用Map,ObjectMapper解析Json

Question

{   
    "time":1352113682,
    "api_version":"1",
    "firstname":"abc",
    "lastname":"xyz",
    "company":"Cool Apps",
    "email":"[e-mail]"
}

public class AuthenticateUser implements Serializable{

    // Response when Successfully Login
    public String time;
    public String api_version;
    public String firstname;
    public String lastname;
    public String company;
    public String email;

}

Map<String, AuthenticateUser> map=null;
ObjectMapper mapper=new ObjectMapper();
try{
    map=mapper.readValue(result,new TypeReference<Map<String, AuthenticateUser>>(){});
    Set<String> keys=map.keySet();
    for (String key : keys) {
    System.out.println(map.get(key).time);
    System.out.println(map.get(key).api_version);
        System.out.println(map.get(key).firstname);
    System.out.println(map.get(key).lastname);
    System.out.println(map.get(key).company);
    System.out.println(map.get(key).email);
    }
}catch (Exception e) {
    e.printStackTrace();
}

JSONMappingException：无法从JSON整数实例化类型[simple type,class AuthenticateUser]的值；在com.fasterxml.jackson.databind.deser.std.stdvalueinstantiator.createfromint（stdvalueinstantiator.java:316)在com.fasterxml.jackson.databind.deser.beandeserializer.deserializefromnumber（beandeserializer.java:427)在com.fasterxml.jackson.databind.deser.std.mapdeserializer.deserializer.deserialize（beandeserializer.119)在

Answer 1

您在开头给出的示例是针对单个AuthenticateUser对象的。这是作为输入传递给这个程序的全部内容吗？这绝对不能正确解析。

单个AuthenticateUser不是类型（String,AuthenticateUser）的映射

我想也许你很困惑，你为什么要做地图？如果您只想解析单个对象，那么您只需要：

AuthenticateUser user =mapper.readValue(result, AuthenticateUser.class);

System.out.println(user.time);