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2020年和Jackson 2.10.1,不需要任何特殊的代码,只需要告诉Jackson你想要什么:

ObjectMapper objectMapper = new ObjectMapper();

// Register module that knows how to serialize java.time objects
// Provided by jackson-datatype-jsr310
objectMapper.registerModule(new JavaTimeModule());

// Ask Jackson to serialize dates as String (ISO-8601 by default)
objectMapper.disable(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS);

这已经在这个答案中提到了,我添加了一个单元测试来验证功能:

import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.datatype.jsr310.JavaTimeModule;
import lombok.Data;
import org.junit.jupiter.api.Test;

import java.time.LocalDate;

import static org.junit.jupiter.api.Assertions.assertEquals;

public class LocalDateSerializationTest {

    @Data
    static class TestBean {
        // Accept default ISO-8601 format
        LocalDate birthDate;
        // Use custom format
        @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
        LocalDate birthDateWithCustomFormat;
    }

    @Test
    void serializeDeserializeTest() throws JsonProcessingException {
        ObjectMapper objectMapper = new ObjectMapper();

        // Register module that knows how to serialize java.time objects
        objectMapper.registerModule(new JavaTimeModule());

        // Ask Jackson to serialize dates as String (ISO-8601 by default)
        objectMapper.disable(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS);

        // The JSON string after serialization
        String json = "{\"birthDate\":\"2000-01-02\",\"birthDateWithCustomFormat\":\"03/02/2001\"}";

        // The object after deserialization
        TestBean object = new TestBean();
        object.setBirthDate(LocalDate.of(2000, 1, 2));
        object.setBirthDateWithCustomFormat(LocalDate.of(2001, 2, 3));

        // Assert serialization
        assertEquals(json, objectMapper.writeValueAsString(object));

        // Assert deserialization
        assertEquals(object, objectMapper.readValue(json, TestBean.class));
    }
}

TestBean使用Lombok为bean生成样板。

其他回答

在Spring Boot web应用程序中,使用Jackson和JSR 310版本“2.8.5”

compile "com.fasterxml.jackson.core:jackson-databind:2.8.5"
runtime "com.fasterxml.jackson.datatype:jackson-datatype-jsr310:2.8.5"

@JsonFormat的工作原理:

import com.fasterxml.jackson.annotation.JsonFormat;

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd")
private LocalDate birthDate;

最简单的解决方案(也支持反序列化和序列化)是

import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.databind.annotation.JsonDeserialize;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import com.fasterxml.jackson.datatype.jsr310.deser.LocalDateDeserializer;
import com.fasterxml.jackson.datatype.jsr310.ser.LocalDateSerializer;

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
@JsonDeserialize(using = LocalDateDeserializer.class)
@JsonSerialize(using = LocalDateSerializer.class)
private LocalDate dateOfBirth;

在项目中使用以下依赖项时。

Maven

<dependency>
   <groupId>com.fasterxml.jackson.core</groupId>
   <artifactId>jackson-databind</artifactId>
   <version>2.9.7</version>
</dependency>
<dependency>
   <groupId>com.fasterxml.jackson.datatype</groupId>
   <artifactId>jackson-datatype-jsr310</artifactId>
   <version>2.9.7</version>
</dependency>

Gradle

compile "com.fasterxml.jackson.core:jackson-databind:2.9.7"
compile "com.fasterxml.jackson.datatype:jackson-datatype-jsr310:2.9.7"

不需要额外的ContextResolver、Serializer或Deserializer实现。

只是更新了克里斯托弗的回答。

从2.6.0版本开始

<dependency>
    <groupId>com.fasterxml.jackson.datatype</groupId>
    <artifactId>jackson-datatype-jsr310</artifactId>
    <version>2.9.0</version>
</dependency>

使用JavaTimeModule而不是JSR310Module(已弃用)。

@Provider
public class ObjectMapperContextResolver implements ContextResolver<ObjectMapper> {  
    private final ObjectMapper MAPPER;

    public ObjectMapperContextResolver() {
        MAPPER = new ObjectMapper();
        MAPPER.registerModule(new JavaTimeModule());
        MAPPER.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
    }

    @Override
    public ObjectMapper getContext(Class<?> type) {
        return MAPPER;
    }  
}

根据文档,新的JavaTimeModule使用相同的标准设置默认为序列化,不使用Timezone id,而只使用ISO-8601兼容的Timezone偏移量。

可以使用SerializationFeature改变行为。WRITE_DATES_WITH_ZONE_ID

春天更容易些:

///...

@Configuration
public class ApplicationCtxBeans {
//....
    @Bean
    public ObjectMapper objectMapper() {
        ObjectMapper MAPPER = new ObjectMapper(); 
        MAPPER.registerModule(new JavaTimeModule()); // to handle LocalDateTime etc
        return MAPPER;
    }
//...
}

用法:

@Service
public class SomeService {
    
//...
    @Autowired
    ObjectMapper jsonMapper;
//...
  JsonNode node = jsonMapper.readTree(
    jsonMapper.writeValueAsString(instance_Of_Class_With_LocalDate_Fields)
  );
//...
}

春季启动2.3.9。RELEASE,我刚刚注册了java时间模块,在POJO类中没有显式注释,具有LocalDate字段&它工作。

var objectMapper = new ObjectMapper();
objectMapper.registerModule(new JavaTimeModule());