对于java.util.Date

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")  
  private Date dateOfBirth;

然后在JSON请求中发送

{ {"dateOfBirth":"01/01/2000"} }  

它的工作原理。

我应该如何为Java 8的LocalDate字段这样做??

我试着

@JsonDeserialize(using = LocalDateDeserializer.class)  
@JsonSerialize(using = LocalDateSerializer.class)  
private LocalDate dateOfBirth;  

但是没有成功。

谁能告诉我正确的方法是什么?

下面是依赖关系

<dependency>
    <groupId>org.jboss.resteasy</groupId>
    <artifactId>jaxrs-api</artifactId>
     <version>3.0.9.Final</version>
</dependency>
<dependency>
    <groupId>com.fasterxml.jackson.jaxrs</groupId>
    <artifactId>jackson-jaxrs-json-provider</artifactId>
    <version>2.4.2</version>
</dependency>
<dependency>
    <groupId>com.wordnik</groupId>
    <artifactId>swagger-annotations</artifactId>
    <version>1.3.10</version>
</dependency>

当前回答

到目前为止最简单和最短的:

@JsonFormat(pattern = "yyyy-MM-dd")
private LocalDate localDate;

@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
private LocalDateTime localDateTime;

Spring引导>= 2.2+不需要依赖

其他回答

ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JavaTimeModule());
mapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);

因为LocalDateSerializer默认将其转换为“[年,月,日]”(json数组)而不是“年-月-日”(json字符串),并且因为我不想需要任何特殊的ObjectMapper设置(如果禁用SerializationFeature,您可以使LocalDateSerializer生成字符串)。WRITE_DATES_AS_TIMESTAMPS,但这需要额外的设置到您的ObjectMapper),我使用以下:

进口:

import com.fasterxml.jackson.databind.ser.std.ToStringSerializer;
import com.fasterxml.jackson.datatype.jsr310.deser.LocalDateDeserializer;

代码:

// generates "yyyy-MM-dd" output
@JsonSerialize(using = ToStringSerializer.class)
// handles "yyyy-MM-dd" input just fine (note: "yyyy-M-d" format will not work)
@JsonDeserialize(using = LocalDateDeserializer.class)
private LocalDate localDate;

现在我可以使用new ObjectMapper()来读写我的对象,而不需要任何特殊的设置。

@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@JsonSerialize(using = LocalDateTimeSerializer.class)
@JsonDeserialize(using = LocalDateTimeDeserializer.class)
private LocalDateTime createdDate;

到目前为止最简单和最短的:

@JsonFormat(pattern = "yyyy-MM-dd")
private LocalDate localDate;

@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
private LocalDateTime localDateTime;

Spring引导>= 2.2+不需要依赖

在配置类中定义LocalDateSerializer和LocalDateDeserializer类,并通过JavaTimeModule将它们注册到ObjectMapper,如下所示:

@Configuration
public class AppConfig
{
@Bean
    public ObjectMapper objectMapper()
    {
        ObjectMapper mapper = new ObjectMapper();
        mapper.setSerializationInclusion(Include.NON_EMPTY);
        //other mapper configs
        // Customize de-serialization


        JavaTimeModule javaTimeModule = new JavaTimeModule();
        javaTimeModule.addSerializer(LocalDate.class, new LocalDateSerializer());
        javaTimeModule.addDeserializer(LocalDate.class, new LocalDateDeserializer());
        mapper.registerModule(javaTimeModule);

        return mapper;
    }

    public class LocalDateSerializer extends JsonSerializer<LocalDate> {
        @Override
        public void serialize(LocalDate value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
            gen.writeString(value.format(Constant.DATE_TIME_FORMATTER));
        }
    }

    public class LocalDateDeserializer extends JsonDeserializer<LocalDate> {

        @Override
        public LocalDate deserialize(JsonParser p, DeserializationContext ctxt) throws IOException {
            return LocalDate.parse(p.getValueAsString(), Constant.DATE_TIME_FORMATTER);
        }
    }
}