问题
I have a Date format coming from API like this:
"start_time": "2015-10-1 3:00 PM GMT+1:00"
Which is YYYY-DD-MM HH:MM am/pm GMT timestamp. I am mapping this value to a Date variable in POJO. Obviously, its showing conversion error.
I would like to know 2 things:
- What is the formatting I need to use to carry out conversion with Jackson? Is Date a good field type for this?
- In general, is there a way to process the variables before they get mapped to Object members by Jackson? Something like, changing the format, calculations, etc.
回答1:
What is the formatting I need to use to carry out conversion with Jackson? Is Date a good field type for this?
Date is a fine field type for this. You can make the JSON parse-able pretty easily by using ObjectMapper.setDateFormat:
DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm a z");
myObjectMapper.setDateFormat(df);
In general, is there a way to process the variables before they get mapped to Object members by Jackson? Something like, changing the format, calculations, etc.
Yes. You have a few options, including implementing a custom JsonDeserializer, e.g. extending JsonDeserializer<Date>. This is a good start.
回答2:
Since Jackson v2.0, you can use @JsonFormat annotation directly on Object members;
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm a z")
private Date date;
回答3:
Of course there is an automated way called serialization and deserialization and you can define it with specific annotations (@JsonSerialize,@JsonDeserialize) as mentioned by pb2q as well.
You can use both java.util.Date and java.util.Calendar ... and probably JodaTime as well.
The @JsonFormat annotations not worked for me as I wanted (it has adjusted the timezone to different value) during deserialization (the serialization worked perfect):
@JsonFormat(locale = "hu", shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm", timezone = "CET")
@JsonFormat(locale = "hu", shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm", timezone = "Europe/Budapest")
You need to use custom serializer and custom deserializer instead of the @JsonFormat annotation if you want predicted result. I have found real good tutorial and solution here http://www.baeldung.com/jackson-serialize-dates
There are examples for Date fields but I needed for Calendar fields so here is my implementation:
The serializer class:
public class CustomCalendarSerializer extends JsonSerializer<Calendar> {
public static final SimpleDateFormat FORMATTER = new SimpleDateFormat("yyyy-MM-dd HH:mm");
public static final Locale LOCALE_HUNGARIAN = new Locale("hu", "HU");
public static final TimeZone LOCAL_TIME_ZONE = TimeZone.getTimeZone("Europe/Budapest");
@Override
public void serialize(Calendar value, JsonGenerator gen, SerializerProvider arg2)
throws IOException, JsonProcessingException {
if (value == null) {
gen.writeNull();
} else {
gen.writeString(FORMATTER.format(value.getTime()));
}
}
}
The deserializer class:
public class CustomCalendarDeserializer extends JsonDeserializer<Calendar> {
@Override
public Calendar deserialize(JsonParser jsonparser, DeserializationContext context)
throws IOException, JsonProcessingException {
String dateAsString = jsonparser.getText();
try {
Date date = CustomCalendarSerializer.FORMATTER.parse(dateAsString);
Calendar calendar = Calendar.getInstance(
CustomCalendarSerializer.LOCAL_TIME_ZONE,
CustomCalendarSerializer.LOCALE_HUNGARIAN
);
calendar.setTime(date);
return calendar;
} catch (ParseException e) {
throw new RuntimeException(e);
}
}
}
and the usage of the above classes:
public class CalendarEntry {
@JsonSerialize(using = CustomCalendarSerializer.class)
@JsonDeserialize(using = CustomCalendarDeserializer.class)
private Calendar calendar;
// ... additional things ...
}
Using this implementation the execution of the serialization and deserialization process consecutively results the origin value.
Only using the @JsonFormat annotation the deserialization gives different result I think because of the library internal timezone default setup what you can not change with annotation parameters (that was my experience with Jackson library 2.5.3 and 2.6.3 version as well).
回答4:
Just a complete example for spring boot application with RFC3339 datetime format
package bj.demo;
import com.fasterxml.jackson.databind.ObjectMapper;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.context.event.ApplicationReadyEvent;
import org.springframework.context.ApplicationListener;
import java.text.SimpleDateFormat;
/**
* Created by BaiJiFeiLong@gmail.com at 2018/5/4 10:22
*/
@SpringBootApplication
public class BarApp implements ApplicationListener<ApplicationReadyEvent> {
public static void main(String[] args) {
SpringApplication.run(BarApp.class, args);
}
@Autowired
private ObjectMapper objectMapper;
@Override
public void onApplicationEvent(ApplicationReadyEvent applicationReadyEvent) {
objectMapper.setDateFormat(new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssXXX"));
}
}
回答5:
Building on @miklov-kriven's very helpful answer, I hope these two additional points of consideration prove helpful to someone:
(1) I find it a nice idea to include serializer and de-serializer as static inner classes in the same class. NB, using ThreadLocal for thread safety of SimpleDateFormat.
public class DateConverter {
private static final ThreadLocal<SimpleDateFormat> sdf =
ThreadLocal.<SimpleDateFormat>withInitial(
() -> {return new SimpleDateFormat("yyyy-MM-dd HH:mm a z");});
public static class Serialize extends JsonSerializer<Date> {
@Override
public void serialize(Date value, JsonGenerator jgen SerializerProvider provider) throws Exception {
if (value == null) {
jgen.writeNull();
}
else {
jgen.writeString(sdf.get().format(value));
}
}
}
public static class Deserialize extends JsonDeserializer<Date> {
@Overrride
public Date deserialize(JsonParser jp, DeserializationContext ctxt) throws Exception {
String dateAsString = jp.getText();
try {
if (Strings.isNullOrEmpty(dateAsString)) {
return null;
}
else {
return new Date(sdf.get().parse(dateAsString).getTime());
}
}
catch (ParseException pe) {
throw new RuntimeException(pe);
}
}
}
}
(2) As an alternative to using @JsonSerialize and @JsonDeserialize annotations on each individual class member you could also consider overriding Jackson's default serialization by applying the custom serialization at an application level, that is all class members of type Date will be serialized by Jackson using this custom serialization without explicit annotation on each field. If you are using Spring Boot for example one way to do this would as follows:
@SpringBootApplication
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
@Bean
public Module customModule() {
SimpleModule module = new SimpleModule();
module.addSerializer(Date.class, new DateConverter.Serialize());
module.addDeserializer(Date.class, new Dateconverter.Deserialize());
return module;
}
}
回答6:
Working for me. SpringBoot.
import com.alibaba.fastjson.annotation.JSONField;
@JSONField(format = "yyyy-MM-dd HH:mm:ss")
private Date createTime;
output:
{
"createTime": "2019-06-14 13:07:21"
}
回答7:
If anyone has problems with using a custom dateformat for java.sql.Date, this is the simplest solution:
ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addSerializer(java.sql.Date.class, new DateSerializer());
mapper.registerModule(module);
(This SO-answer saved me a lot of trouble: https://stackoverflow.com/a/35212795/3149048 )
Jackson uses the SqlDateSerializer by default for java.sql.Date, but currently, this serializer doesn't take the dateformat into account, see this issue: https://github.com/FasterXML/jackson-databind/issues/1407 . The workaround is to register a different serializer for java.sql.Date as shown in the code example.
回答8:
I want to point out that setting a SimpleDateFormat like described in the other answer only works for a java.util.Date which I assume is meant in the question.
But for java.sql.Date the formatter does not work.
In my case it was not very obvious why the formatter did not work because in the model which should be serialized the field was in fact a java.utl.Date but the actual object ended up beeing a java.sql.Date.
This is possible because
public class java.sql extends java.util.Date
So this is actually valid
java.util.Date date = new java.sql.Date(1542381115815L);
So if you are wondering why your Date field is not correctly formatted make sure that the object is really a java.util.Date.
Here is also mentioned why handling java.sql.Date will not be added.
This would then be breaking change, and I don't think that is warranted. If we were starting from scratch I would agree with the change, but as things are not so much.
回答9:
To add characters such as T and Z in your date
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss'Z'")
private Date currentTime;
output
{
"currentTime": "2019-12-11T11:40:49Z"
}
来源：https://stackoverflow.com/questions/12463049/date-format-mapping-to-json-jackson