Jackson ignore unrecognized field 2016-11-17 20:14
You will get UnrecognizedPropertyException if you convert json string to object with unknown field. There are 2 ways to avoid this problem. Use JsonIgnoreProperties annotation or add DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES to ObjectMapper configuration. Examples are in the following.
Use JsonIgnoreProperties
public class ConvertJson2ObjectUnknownField {
public static void main(String[] args) throws IOException {
ObjectMapper objectMapper = new ObjectMapper();
String userJsonStr = "{\"name\": \"Henry\",\"age\": 28,\"address\": \"Beijing\"}";
User user = objectMapper.readValue(userJsonStr, User.class);// throw UnrecognizedPropertyException
IgnoreFieldUser ignoreFieldUser = objectMapper.readValue(userJsonStr, IgnoreFieldUser.class);
System.out.println(ignoreFieldUser);
}
}
@JsonIgnoreProperties(ignoreUnknown = true)
class IgnoreFieldUser {
private String name;
private int age;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
@Override
public String toString() {
return "IgnoreFieldUser{" +
"name='" + name + '\'' +
", age=" + age +
'}';
}
}
Use DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES
public class ConvertJson2ObjectUnknownField {
public static void main(String[] args) throws IOException {
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
String userJsonStr = "{\"name\": \"Henry\",\"age\": 28,\"address\": \"Beijing\"}";
User user = objectMapper.readValue(userJsonStr, User.class);// won't throw UnrecognizedPropertyException
System.out.println(user);
}
}