如何使用Jackson JSON处理器序列化Joda DateTime？

Question

如何根据一个简单的模式(如"dd-MM-yyyy")让杰克逊序列化我的Joda DateTime对象？

我试过了:

@JsonSerialize(using=DateTimeSerializer.class)
private final DateTime date;

我也尝试过:

ObjectMapper mapper = new ObjectMapper()
    .getSerializationConfig()
    .setDateFormat(df);

谢谢!

Answer 1

使用Jackson 2.0和Joda模块,这变得非常容易.

ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JodaModule());

Maven依赖:

<dependency>
  <groupId>com.fasterxml.jackson.datatype</groupId>
  <artifactId>jackson-datatype-joda</artifactId>
  <version>2.1.1</version>
</dependency>  

代码和文档:https: //github.com/FasterXML/jackson-datatype-joda

二进制文件:http: //repo1.maven.org/maven2/com/fasterxml/jackson/datatype/jackson-datatype-joda/

Answer 2

在您要映射的对象中:

@JsonSerialize(using = CustomDateSerializer.class)
public DateTime getDate() { ... }

在CustomDateSerializer中:

public class CustomDateSerializer extends JsonSerializer<DateTime> {

    private static DateTimeFormatter formatter = 
        DateTimeFormat.forPattern("dd-MM-yyyy");

    @Override
    public void serialize(DateTime value, JsonGenerator gen, 
                          SerializerProvider arg2)
        throws IOException, JsonProcessingException {

        gen.writeString(formatter.print(value));
    }
}

Answer 3

正如@Kimble所说,使用Jackson 2,使用默认格式非常容易; 只需注册JodaModule你的ObjectMapper.

ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JodaModule());

对于自定义序列化/反序列化DateTime,则需要实现自己的StdScalarSerializer和StdScalarDeserializer; 它很复杂,但无论如何.

例如,这是一个DateTime使用ISODateFormatUTC时区的序列化程序:

public class DateTimeSerializer extends StdScalarSerializer<DateTime> {

    public DateTimeSerializer() {
        super(DateTime.class);
    }

    @Override
    public void serialize(DateTime dateTime,
                          JsonGenerator jsonGenerator,
                          SerializerProvider provider) throws IOException, JsonGenerationException {
        String dateTimeAsString = ISODateTimeFormat.withZoneUTC().print(dateTime);
        jsonGenerator.writeString(dateTimeAsString);
    }
}

和相应的反序列化器:

public class DateTimeDesrializer extends StdScalarDeserializer<DateTime> {

    public DateTimeDesrializer() {
        super(DateTime.class);
    }

    @Override
    public DateTime deserialize(JsonParser jsonParser,
                                DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
        try {
            JsonToken currentToken = jsonParser.getCurrentToken();
            if (currentToken == JsonToken.VALUE_STRING) {
                String dateTimeAsString = jsonParser.getText().trim();
                return ISODateTimeFormat.withZoneUTC().parseDateTime(dateTimeAsString);
            }
        } finally {
            throw deserializationContext.mappingException(getValueClass());
        }
    }

然后将这些与模块结合在一起:

public class DateTimeModule extends SimpleModule {

    public DateTimeModule() {
        super();
        addSerializer(DateTime.class, new DateTimeSerializer());
        addDeserializer(DateTime.class, new DateTimeDeserializer());
    }
}

然后在您的注册模块ObjectMapper:

ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new DateTimeModule());

Answer 4

/sf/answers/758458011/

虽然您可以为每个日期字段添加注释,但最好为对象映射器执行全局配置.如果您使用jackson,您可以按如下方式配置弹簧:

<bean id="jacksonObjectMapper" class="com.company.CustomObjectMapper" />

<bean id="jacksonSerializationConfig" class="org.codehaus.jackson.map.SerializationConfig"
    factory-bean="jacksonObjectMapper" factory-method="getSerializationConfig" >
</bean>

对于CustomObjectMapper:

public class CustomObjectMapper extends ObjectMapper {

    public CustomObjectMapper() {
        super();
        configure(Feature.WRITE_DATES_AS_TIMESTAMPS, false);
        setDateFormat(new SimpleDateFormat("EEE MMM dd yyyy HH:mm:ss 'GMT'ZZZ (z)"));
    }
}

当然,SimpleDateFormat可以使用您需要的任何格式.

Answer 5

简单的解决方案

我遇到过类似的问题,我的解决方案比上面清楚得多.

我只是在 注释中使用了模式@JsonFormat

基本上我的班级有一个DateTime字段,所以我在getter周围添加了一个注释:

@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
public DateTime getDate() {
    return date;
}

我用这个类序列化了 ObjectMapper

    ObjectMapper mapper = new ObjectMapper();
    mapper.registerModule(new JodaModule());
    mapper.disable(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS);
    ObjectWriter ow = mapper.writer();
    try {
        String logStr = ow.writeValueAsString(log);
        outLogger.info(logStr);
    } catch (IOException e) {
        logger.warn("JSON mapping exception", e);
    }

我们使用Jackson 2.5.4

Answer 6

同时,当JodaModule在classpath中时,Jackson会自动注册Joda模块.我刚刚将Jackson-datatype-joda添加到了Maven,它立刻起作用了.

<dependency>
  <groupId>com.fasterxml.jackson.datatype</groupId>
  <artifactId>jackson-datatype-joda</artifactId>
  <version>2.8.7</version>
</dependency>

JSON输出:

{"created" : "2017-03-28T05:59:27.258Z"}

Answer 7

对于使用Spring Boot的用户，您必须将模块添加到上下文中，并且它将像这样添加到您的配置中。

@Bean
public Module jodaTimeModule() {
    return new JodaModule();
}

如果要使用新的java8时间模块jsr-310。

@Bean
public Module jodaTimeModule() {
    return new JavaTimeModule();
}