I am working on custom JSON deserialiser and have the below class
public class yyyy_MM_dd_DateDeserializer extends StdDeserializer <LocalDate> {
public yyyy_MM_dd_DateDeserializer() {
this(null);
}
public yyyy_MM_dd_DateDeserializer(Class t) {
super(t);
}
#Override
public LocalDate deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException {
String dateString = jsonParser.getText();
LocalDate localDate = null;
try {
localDate = LocalDate.parse(dateString, "yyyy-MM-dd");
} catch (DateTimeParseException ex) {
throw new RuntimeException("Unparsable date: " + dateString);
}
return localDate;
}
}
and in my request class
#Valid
#JsonDeserialize(using = LocalDateDeserializer.class)
#JsonSerialize(using = LocalDateSerializer.class)
private LocalDate endDate;
It works fine but I am wondering if I can pass the date format dynamically. Instead of hardcoding in yyyy_MM_dd_DateDeserializer.
I want to pass the date format from my request class so that my deserialiser is more generic any anyone can use it by sending the required format.
I think you working too hard to get what you want. There is a simpler way without writing your own deserializer. Look at this question. Essentially it looks like
#JsonFormat(shape= JsonFormat.Shape.STRING, pattern="EEE MMM dd HH:mm:ss Z yyyy")
#JsonProperty("created_at")
ZonedDateTime created_at;
And you just put your own mask. Also, I once had a task of parsing date with unknown format, essentially I needed to parse any valid date. Here is an article describing the idea of how to implement it: Java 8 java.time package: parsing any string to date. You might find it useful
Not when using a binder library (The very point of binding is that it is not dynamic.).
But you could when using a simple parsing library such as org.json
When you are working with java.time.* classes and Jackson is good to start from registering JavaTimeModule which comes from jackson-datatype-jsr310 module. We can extend it and register serialiser with provided pattern like in below example:
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.datatype.jsr310.JavaTimeModule;
import com.fasterxml.jackson.datatype.jsr310.ser.LocalDateSerializer;
import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
public class JsonApp {
public static void main(String[] args) throws Exception {
ObjectMapper mapperIso = createObjectMapper("yyyy-MM-dd");
ObjectMapper mapperCustom0 = createObjectMapper("yyyy/MM/dd");
ObjectMapper mapperCustom1 = createObjectMapper("MM-dd-yyyy");
System.out.println(mapperIso.writeValueAsString(new Time()));
System.out.println(mapperCustom0.writeValueAsString(new Time()));
System.out.println(mapperCustom1.writeValueAsString(new Time()));
}
private static ObjectMapper createObjectMapper(String pattern) {
JavaTimeModule javaTimeModule = new JavaTimeModule();
javaTimeModule.addSerializer(LocalDate.class, new LocalDateSerializer(DateTimeFormatter.ofPattern(pattern)));
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(javaTimeModule);
return mapper;
}
}
class Time {
private LocalDate now = LocalDate.now();
public LocalDate getNow() {
return now;
}
public void setNow(LocalDate now) {
this.now = now;
}
#Override
public String toString() {
return "Time{" +
"now=" + now +
'}';
}
}
Aboce code prints:
{"now":"2019-02-24"}
{"now":"2019/02/24"}
{"now":"02-24-2019"}
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.
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;
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.
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).
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"
}
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"));
}
}
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;
}
}
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.
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.
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"
}
If we are having the spring boot application, then one more option thats simple to implement for app wide configuration is to use below in application properties file. You can customize the format as needed.
spring.jackson.date-format=yyyy-MM-dd'T'HH:mm:ss.SSS
NOTE: If using this solution use the Spring Dependency Injection to get the reference of the ObjectMapper class.
Cons of not using explicit format is sometimes while upgrading the libraries for jackson code breaks because of change in the format for some versions.
I have a simple Spring Boot Application having a simple MyDateTime model class only having a java.util.Date instance variable with private access, getters/setters and default constructor.
A controller just instantiates this object and returns back.
In the output, I see that default representation of Date object is done as an Integer (maybe millis from Epoch)
Is there any way I can change the default jsonification of Date Object into ISO-String or any other String?
EDIT:
Some Clarification:
I'm very new to Spring and Spring Boot. I'm using the template from a sample application on spring's website. JSONification is done through Jackson. Rest, I don't know much about Spring in general.
You can set the default format Jackson uses when serializing dates in your application.properties file:
spring.jackson.date-format=yyyy-MM-dd HH:mm:ss
http://docs.spring.io/spring-boot/docs/1.2.3.RELEASE/reference/htmlsingle/#common-application-properties
You can also specify a specific format to use for a specific date using the #JsonFormat annotation, example below:
Example POJO:
public class Demo {
private Date timestamp1;
private Date timestamp2;
public Date getTimestamp1() {
return timestamp1;
}
public void setTimestamp1(Date timestamp1) {
this.timestamp1 = timestamp1;
}
#JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd'T'HH:mm:ss.SSSZ")
public Date getTimestamp2() {
return timestamp2;
}
public void setTimestamp2(Date timestamp2) {
this.timestamp2 = timestamp2;
}
}
Example Controller:
#RestController
public class DemoController {
#RequestMapping(value="/demo", method = RequestMethod.GET)
Demo start() {
Demo demo = new Demo();
Date timestamp = new Date();
demo.setTimestamp1(timestamp);
demo.setTimestamp2(timestamp);
return demo;
}
}
https://github.com/FasterXML/jackson-annotations/wiki/Jackson-Annotations
Date.toString() method returns epoch timestamp by default. What you want to do is about Jackson internals. To accomplish that, change your MyDateTime class to something like this:
public class MyDateTime {
private final Date date;
public MyDateTime(Date date) {
this.date = date;
}
public Date date() { //this is not read by Jackson
return date;
}
public String getDate() { //this is read by Jackson
SimpleDateFormat formatter = new SimpleDateFormat("dd/MM/yyyy");
return formatter.format(date);
}
}
When serializing, Jackson look for getter methods, the methods that start with get. If you don't want something to be serialized, remove get from the name. This is the simplest solution, you can check Jackson API for #JsonIgnore annotation for further solutions. If you want to see something other date in the field name in JSON string, look for #JsonProperty.
To print human readable date, you need SimpleDateFormat.
The examples works fine on my setup, I did not do anything further than returning custom class from HelloController.
The problem I am having is that I have some consumers that are Java and some that are browsers. My target browsers are IE7+ (json3 for IE7 only) & Chrome.
For a browser I wish to have the date deserialize to a Date JavaScript object (using the JSON.parse() method. For a Java consumer I wish to deserialize to a java.util.Date Java object.
Given that I can't change anything on the browser side. I have to do serialize the messages to something like this:
{ myDate: new Date(<EPOCH HERE>) }
Which of course will cause a problem for Java deserializer. However, I am hoping there is something I can do with Gson to make this work...amy ideas?
Or should I take a different strategy altogether?
I usually use the annotation #JsonSerialize and #JsonDeserialize to deal with this problem. I also use ISO8601 format as a standard for our REST API dates.
#JsonSerialize(using=JsonDateSerializer.class)
#JsonDeserialize(using=JsonDateDeserializer.class)
private Date expiryDate;
JsonDateSerializer class
#Component
public class JsonDateSerializer extends JsonSerializer<Date>
{
// ISO 8601
private static final DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd");
#Override
public void serialize(Date date, JsonGenerator gen, SerializerProvider provider)
throws IOException, JsonProcessingException
{
String formattedDate = dateFormat.format(date);
gen.writeString(formattedDate);
}
}
JsonDateDeserializer class
#Component
public class JsonDateDeserializer extends JsonDeserializer<Date>
{
// ISO 8601
private static final DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd");
#Override
public Date deserialize(JsonParser jsonParser, DeserializationContext deserializationContext)
throws IOException, JsonProcessingException
{
try
{
return dateFormat.parse(jsonParser.getText());
}
catch (ParseException e)
{
throw new JsonParseException("Could not parse date", jsonParser.getCurrentLocation(), e);
}
}
}
I am using Spring WebMVC, JodaTime and Jackson to build a RESTful webservice.
Every user who performs actions on this webservice has his default timezone saved in the database.
I need to provide all timestamp in the users timezone. I am able to convert every timestamp in a response object to the correspondending timezone of the user, but jackson deserializes every timestamp to a specific timezone, for example UTC.
How do I prevent jackson from doing this? I want a datetime field to be serialized in its timezone, not the timezone set for jackson.
I am using full ISO6801 format.
Edit:
For anyone who stumbles upon this question, there is currently a discussion going on at Github about this topic:
https://github.com/FasterXML/jackson-datatype-joda/issues/43
You can consider customizing the standard Joda time deserializer to read the timezone information from a thread local variable set by every request.
Here is an example:
public class JacksonTimezone {
public static class DataTimeDeserializerTimeZone extends DateTimeDeserializer {
public static final ThreadLocal<DateTimeZone> TIME_ZONE_THREAD_LOCAL = new ThreadLocal<DateTimeZone>() {
#Override
protected DateTimeZone initialValue() {
return DateTimeZone.getDefault();
}
};
public DataTimeDeserializerTimeZone() {
super(DateTime.class);
}
#Override
public ReadableDateTime deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException {
return super.deserialize(jp, ctxt).toDateTime().withZone(TIME_ZONE_THREAD_LOCAL.get());
}
}
public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();
JodaModule module = new JodaModule();
module.addDeserializer(DateTime.class,
(JsonDeserializer) new DataTimeDeserializerTimeZone());
mapper.registerModule(module);
DateTime dateTime1 = DateTime.parse("2014-02-03T10:00");
String json = mapper.writeValueAsString(dateTime1);
System.out.println(json + " " + TIME_ZONE_THREAD_LOCAL.get());
System.out.println(mapper.readValue(json, DateTime.class));
TIME_ZONE_THREAD_LOCAL.set(DateTimeZone.forID("US/Hawaii"));
System.out.println(mapper.readValue(json, DateTime.class));
}
}
Output:
1391418000000 Europe/Oslo
2014-02-03T10:00:00.000+01:00
2014-02-02T23:00:00.000-10:00
The value of the TIME_ZONE_THREAD_LOCAL static variable should be set to the correct timezone before JSON data reaches Jackson.