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Spring 使用 Jackson 的 InstantSerializer 来写出我的 Instant 字段，如下所示：

now: "2021-04-07T10:51:53.043320Z"

不过，我不想要纳秒 - 只是毫秒。我猜想设置应用程序属性

spring.jackson.serialization.write-date-timestamps-as-nanoseconds=false

会实现这一点，但这没有什么区别。

我如何告诉 Spring/Jackson 在序列化 Instant 时省略纳秒？

（我使用的是 Spring Boot 2.2.11.RELEASE）

更新

基于这个答案，我最终得到了它的工作。我不得不使用已弃用的 JSR310Module 而不是 JavaTimeModule，并重写 createContextual(...) 方法以强制它始终使用我的序列化程序。

@Bean
public ObjectMapper objectMapper(Jackson2ObjectMapperBuilder builder) {
    ObjectMapper objectMapper = builder.createXmlMapper(false).build();
    JSR310Module jsr310Module = new JSR310Module();
    jsr310Module.addSerializer(Instant.class, new MyInstantSerializer());
    objectMapper.registerModule(jsr310Module);
    return objectMapper;
}

private static class MyInstantSerializer extends InstantSerializer {
    public MyInstantSerializer() {
        super(InstantSerializer.INSTANCE, false, false, 
                new DateTimeFormatterBuilder().appendInstant(3).toFormatter());
    }

    @Override
    public JsonSerializer<?> createContextual(SerializerProvider prov, BeanProperty property) {
        return this;
    }
}

这也有效（基于下面 Volodya 的回答）：

@Bean
public Jackson2ObjectMapperBuilderCustomizer addCustomTimeSerialization() {
    return jacksonObjectMapperBuilder -> 
            jacksonObjectMapperBuilder.serializerByType(Instant.class, new JsonSerializer<Instant>() {

        private final DateTimeFormatter formatter = 
                new DateTimeFormatterBuilder().appendInstant(3).toFormatter();

        @Override
        public void serialize(
                Instant instant, JsonGenerator generator, SerializerProvider provider) throws IOException {
            generator.writeString(formatter.format(instant));
        }
    });
}
4

2 回答 2

2

为此，您可以使用@JsonFormat(pattern = "yyyy-MM-dd'T'HH:mm:ss[.SSS]")

完整示例：

    import com.fasterxml.jackson.annotation.JsonFormat;
    import java.time.LocalDateTime;
    import java.time.ZonedDateTime;
    
    public class Message {
    
        @JsonFormat(pattern = "yyyy-MM-dd'T'HH:mm:ss[.SSS]")
        private final LocalDateTime dateTime;
    
        @JsonFormat(pattern = "yyyy-MM-dd'T'HH:mm:ss[.SSS]")
        private final ZonedDateTime zonedDateTime;
    
        public Message(ZonedDateTime zonedDateTime) {
            this(zonedDateTime.toLocalDateTime(), zonedDateTime);
        }
    
        public Message(LocalDateTime dateTime, ZonedDateTime zonedDateTime) {
            this.dateTime = dateTime;
            this.zonedDateTime = zonedDateTime;
        }
    
        public LocalDateTime getDateTime() {
            return dateTime;
        }
    
        public ZonedDateTime getZonedDateTime() {
            return zonedDateTime;
        }
    }

测试：

    import com.fasterxml.jackson.core.JsonProcessingException;
    import com.fasterxml.jackson.databind.ObjectMapper;
    import org.junit.jupiter.api.Test;
    import org.springframework.beans.factory.annotation.Autowired;
    import org.springframework.boot.test.autoconfigure.json.JsonTest;
    import java.time.*;
    import java.time.temporal.ChronoUnit;
    import static org.junit.jupiter.api.Assertions.*;
    
    @JsonTest
    class MessageTest {
    
        @Autowired
        ObjectMapper mapper;
    
        @Test
        public void serializationTest() throws JsonProcessingException {
            final LocalDate date = LocalDate.of(2000, Month.JANUARY, 1);
            final LocalTime time = LocalTime.of(12, 20, 10).plus(123, ChronoUnit.MILLIS);
            final Message message = new Message(ZonedDateTime.of(date, time, ZoneId.systemDefault()));
    
            final String res = mapper.writeValueAsString(message);
    
            assertEquals("{\"dateTime\":\"2000-01-01T12:20:10.123\",\"zonedDateTime\":\"2000-01-01T12:20:10.123\"}", res);
        }
    
    }

更新：

如果您想集中配置它，您可以：

  1. 尝试将日期格式设置为您的 ObjectMapper，如此处所述
mapper.setDateFormat(new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS"));
  1. 像这里描述的那样自定义您的映射器
@SpringBootApplication
public class InstantSerializerApplication {

    public static void main(String[] args) {
        SpringApplication.run(InstantSerializerApplication.class, args);
    }

    @Bean
    public Jackson2ObjectMapperBuilderCustomizer addCustomTimeSerialization() {
        return jacksonObjectMapperBuilder -> jacksonObjectMapperBuilder.serializerByType(ZonedDateTime.class, new JsonSerializer<ZonedDateTime>() {
            @Override
            public void serialize(ZonedDateTime zonedDateTime, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException {
                final DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ss.SSS");
                jsonGenerator.writeString(formatter.format(zonedDateTime));
            }
        });
    }
}
于 2021-04-07T11:39:27.300 回答
0

更好的方法是

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss.SSSZ")
private final ZonedDateTime zonedDateTime;

有关更多信息，请参阅此问题的已接受答案

于 2021-05-20T10:44:12.837 回答