如何让Jackson根据简单的模式(如"dd-MM-yyyy")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!
发布于 2013-03-25 07:28:48
有了Jackson 2.0和Joda模块,这就变得非常容易了。
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/
发布于 2010-07-18 00:10:21
在要映射的对象中:
@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));
}
}
发布于 2014-09-16 17:17:03
正如@Kimble所说,在Jackson 2中,使用默认格式非常简单;只需在ObjectMapper
上注册JodaModule
即可。
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JodaModule());
对于自定义的DateTime
序列化/反序列化,您需要实现您自己的StdScalarSerializer
和StdScalarDeserializer
;这非常复杂,但不管怎样。
例如,下面是一个使用具有协调世界时时区的ISODateFormat
的DateTime
序列化器:
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);
}
}
和相应的反序列化程序:
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());
}
}
然后使用一个模块将这些绑定在一起:
public class DateTimeModule extends SimpleModule {
public DateTimeModule() {
super();
addSerializer(DateTime.class, new DateTimeSerializer());
addDeserializer(DateTime.class, new DateTimeDeserializer());
}
}
然后在您的ObjectMapper
上注册该模块
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new DateTimeModule());
https://stackoverflow.com/questions/3269459
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