原因
错误日志:
com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "xxx fieldName"
导致原因是因为Jackson将JSON转换成Java对象时,JSON中包含Java对象没有的属性。
解决方案:
方案一:单个类配置
在类上增加@JsonIgnoreProperties(ignoreUnknown = true)注解
@JsonIgnoreProperties(ignoreUnknown = true)
public class Person {
private String name;
private Integer age;
}
public class Main {
private final static ObjectMapper objectMapper = new ObjectMapper();
public static void main(String[] args) throws JsonProcessingException {
String jsonStr = "{\"name\":\"brian7788\",\"age\":18,\"sex\":\"1\"}";
Person person = objectMapper.readValue(jsonStr, Person.class);
System.out.println(person);
}
}
方案二:全局配置
使用 objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false); 实现全局禁用遇到未知属性抛出异常。
public class Main {
private final static ObjectMapper objectMapper = new ObjectMapper();
// Jackson个性化配置可在静态代码块中完成
static {
// 禁用遇到未知属性抛出异常
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
}
public static void main(String[] args) throws JsonProcessingException {
String jsonStr = "{\"name\":\"brian7788\",\"age\":18,\"sex\":\"1\"}";
Person person = objectMapper.readValue(jsonStr, Person.class);
System.out.println(person);
}
}
方案三:使用@JsonAnySetter注解将JSON中多余的属性反序列化到Java POJO
该方式需要在Java POJO 中增加一个Map的属性用来接收JSON中所有多余的属性,且对应的setter方法入参为key-value键值对。
使用@JsonAnySetter注解可以将JSON中多余出来的属性反序列化到person对象的other属性中去。
使用@JsonAnyGetter 注解将Person对象的other属性中的kay-value键值对以平级的方式序列化到JSON对象当中,注意是和其他属性是平级的。
public class Person {
private String name;
private Integer age;
private Map<String, Object> other = new HashMap<>();
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public Integer getAge() {
return age;
}
public void setAge(Integer age) {
this.age = age;
}
@JsonAnyGetter
public Map<String, Object> getOther() {
return other;
}
@JsonAnySetter
public void setOther(String key, Object value) {
this.other.put(key, value);
}
}
public class Main {
private final static ObjectMapper objectMapper = new ObjectMapper();
public static void main(String[] args) throws JsonProcessingException {
String jsonStr = "{\"name\":\"brian7788\",\"age\":18,\"sex\":\"1\"}";
Person person = objectMapper.readValue(jsonStr, Person.class);
System.out.println(person);
System.out.println(mapper.writeValueAsString(person));
}
}