Spring Boot - 如何为“java.time.LocalDateTime”

Question

我用 Spring Boot 开发了一个休息服务。我想以毫秒为单位返回用户生日的 json 响应。如何将java.time.LocalDateTime对象序列化为毫秒？我的模型类：    @Entity(name = "users")    public class User implements Serializable {        @Id        @GeneratedValue        @Column(name = "user_id")        private Long id;        @Column(name = "first_name")        private String firstName;        @Column(name = "last_name")        private String lastName;        @Column(name = "date_of_birth")        private LocalDateTime dateOfBirth;   . . .    }目前的回应：{ . . ."dateOfBirth":[2018,7,25,7,0],. . . }首选回复：{ . . ."dateOfBirth": 1532786354419,. . . }

Answer 1

用 @JsonSerialize(using = CustomSerializer.class)

@Entity(name = "users")

public class User implements Serializable {

    @Id

    @GeneratedValue

    @Column(name = "user_id")

    private Long id;

    @Column(name = "first_name")

    private String firstName;

    @Column(name = "last_name")

    private String lastName;

    @JsonSerialize(using = CustomSerializer.class)

    @Column(name = "date_of_birth")

    private LocalDateTime dateOfBirth;  

    . . .

}

自定义序列化程序类：

public class CustomSerializer extends JsonSerializer<LocalDateTime> {

@Override

public void serialize(LocalDateTime value, JsonGenerator gen, SerializerProvider serializers) throws IOException {

    //add your custom date parser

    gen.writeString(value.atZone(ZoneId.systemDefault()).toInstant().toEpochMilli()+"");

}

}

Answer 2

如果您不想装饰所有对象，@JsonSerialize可以将对象映射器配置为始终返回 long for LocalDateTime.

@Bean

public ObjectMapper objectMapper() {

    ObjectMapper objectMapper = new ObjectMapper();

    JavaTimeModule javaTimeModule = new JavaTimeModule();

    javaTimeModule.addSerializer(LocalDateTime.class, new LocalDateSerializer());

    javaTimeModule.addDeserializer(LocalDateTime.class, new LocalDateDeserializer());

    objectMapper.registerModule(javaTimeModule);

    return objectMapper;

}

和 de- 和序列化器。

public class LocalDateSerializer extends JsonSerializer<LocalDateTime> {

    @Override

    public void serialize(LocalDateTime value, JsonGenerator gen, SerializerProvider serializers) throws IOException {

        gen.writeString(String.valueOf(value.atZone(ZoneId.systemDefault()).toEpochSecond()));

    }

}

public class LocalDateDeserializer extends JsonDeserializer<LocalDateTime> {

    @Override

    public LocalDateTime deserialize(JsonParser p, DeserializationContext ctxt) throws IOException {

        return LocalDateTime.ofInstant(Instant.ofEpochSecond(Long.parseLong(p.getValueAsString())), ZoneId.systemDefault());

    }

}

在您的示例中，您使用的出生日期可能是LocalDate.

Answer 3

为什么不按如下方式存储日期属性：

@Temporal(TemporalType.TIMESTAMP)

private Date date = new Date();

它以毫秒为单位给出日期。将其格式化为所需的输出是应用层的责任。不要在实体本身上做一些注释。