杰克逊禁用@JsonFormat注解

Question

我正在使用 jackson 库，并且遇到了一种情况，我想在序列化/反序列化时使用 objectmapper 禁用 @JsonFormat 注释。

我的 Api 代码在第 3 方库中，因此我无法删除/添加任何注释，因此 objectMapper 是唯一的选择。

API 类：

public class ApiClass {

  @JsonFormat(pattern = "yyyy-MM-dd'T'HH:mm:ss'Z'")
  private DateTime time;

}

我的代码：

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
objectMapper.configure(Feature.ALLOW_COMMENTS, true);
objectMapper.configure(MapperFeature.AUTO_DETECT_IS_GETTERS, true);
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
objectMapper.configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false);
objectMapper.setSerializationInclusion(Include.NON_ABSENT);
objectMapper.registerModule(new JodaModule());
objectMapper.registerModule(new JavaTimeModule());

String str = " {\"time\": \"2012-05-01\"}";

ApiClass msg = objectMapper.readValue(str, ApiClass.class);

我希望此转换成功发生。

目前我得到：com.fasterxml.jackson.databind.JsonMappingException: Invalid format: "2012-05-01" is too short

请帮帮我。

提前致谢

Answer 1

下面是专门禁用 JsonFormat 的代码：

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
objectMapper.configure(Feature.ALLOW_COMMENTS, true);
objectMapper.configure(MapperFeature.AUTO_DETECT_IS_GETTERS, true);
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
objectMapper.configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false);
objectMapper.setSerializationInclusion(Include.NON_ABSENT);
objectMapper.registerModule(new JodaModule());
objectMapper.registerModule(new JavaTimeModule());

objectMapper.setAnnotationIntrospector(new JacksonAnnotationIntrospector() {
    @Override
    protected <A extends Annotation> A _findAnnotation(final Annotated annotated,
        final Class<A> annoClass) {
      if (!annotated.hasAnnotation(JsonFormat.class)) {    //since we need to disable JsonFormat annotation.
        return super._findAnnotation(annotated, annoClass);
      }
      return null;
    }
  });


String str = " {\"time\": \"2012-05-01\"}";

ApiClass msg = objectMapper.readValue(str, ApiClass.class);
System.out.println(objectMapper.writeValueAsString(msg ));

如果我们需要禁用多个注释（JsonFormat，JsonUnWrapped），那么：

替换：

if (!annotated.hasAnnotation(JsonFormat.class)) {

与：

if (!annotated.hasOneOf(new Class[] {JsonFormat.class, JsonUnwrapped.class})) {

谢谢大家。

Answer 2

在使用readValue方法之前尝试下面的代码

mapper.setDateFormat(new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss'Z'"));

Answer 3

尝试使用自定义反序列化器：

public class MyTest {
    @Test
    public void myTest() throws IOException {
        ObjectMapper objectMapper = new ObjectMapper();
        objectMapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
        objectMapper.configure(JsonParser.Feature.ALLOW_COMMENTS, true);
        objectMapper.configure(MapperFeature.AUTO_DETECT_IS_GETTERS, true);
        objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
        objectMapper.configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false);
        objectMapper.setSerializationInclusion(JsonInclude.Include.NON_ABSENT);
        objectMapper.registerModule(new JavaTimeModule());

        String str = " {\"time\": \"2012-05-01\"}";


        SimpleModule module = new SimpleModule();
        module.addDeserializer(DateTime.class, new DateTimeDeserializer());
        objectMapper.registerModule(module);


        ApiClass msg = objectMapper.readValue(str, ApiClass.class);
        System.out.println(msg.getTime());
    }
}

class ApiClass {

    @JsonFormat(pattern = "yyyy-MM-dd'T'HH:mm:ss'Z'")
    private DateTime time;

    public DateTime getTime() {
        return time;
    }

    public void setTime(DateTime time) {
        this.time = time;
    }
}

class DateTimeDeserializer extends StdDeserializer<DateTime> {

    public DateTimeDeserializer() {
        this(null);
    }

    public DateTimeDeserializer(Class<?> vc) {
        super(vc);
    }

    @Override
    public DateTime deserialize(JsonParser jp, DeserializationContext ctxt)
        throws IOException {
        JsonNode node = jp.getCodec().readTree(jp);
        String time = node.asText();
        return new DateTime(time);
    }
}

Answer 4

这个问题有一个解决方案。

请查看：https://www.baeldung.com/jackson-annotations 点号。 9 禁用杰克逊注解

objectMapper.disable(MapperFeature.USE_ANNOTATIONS);
ApiClass msg = objectMapper.readValue(str, ApiClass.class);

sout(msg.getTime()) //2012-05-01T00:00:00.000Z

但这会禁用所有注释。我只想禁用 @JsonFormat 注释。请提出建议。

Answer 5

您可以使用 SimpleDateFormat 将字符串类型日期转换为日期类型，如下所示

SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss'Z'");
    try {
        Date parse = simpleDateFormat.parse(String date);
        return parse;
    } catch (ParseException e) {
        e.printStackTrace();
    }