In a spring-boot based project i have a simple DTO object:
public class ExpenseDTO {
#Min(value = 1, message = "expense.amount.negative")
private int amount;
#Past
private Calendar createdAt;
// setters/getters/constructor are omitted
}
and such rest controller:
public class ExpenseController {
private final ExpenseService expenseService;
#Autowired
public ExpenseController(ExpenseService expenseService) {
this.expenseService = expenseService;
}
#RequestMapping(value = ADD_EXPENSE, method = POST)
public ResponseEntity addExpense(#Valid #RequestBody ExpenseDTO expenseDTO, Principal principal) {
expenseService.addExpense(expenseDTO, principal.getName());
return ResponseEntity.ok().build();
}
}
From a client i'm gonna send a current date: {"createdAt": "2017-01-27T21:32:19.183Z"} but during validation on the back end the date will be parsed as "2017-01-28T01:30:00.000+0200" so the result is wrong and validation fails. I was trying to play around with #JsonFormat but with no result.
NOTE: i'm using H2 db and if remove #Past from the DTO object everything works just fine, but i have to disable future date.
So how can i validate the date without timezone, i mean i need the exactly the same date on the back end as it was send from a client?!
It might be because of the TimeZone set in your local machine. You can set it to UTC in the application startup, e.g.:
TimeZone.setDefault(TimeZone.getTimeZone("UTC"));
Also, you need to set the TimeZone in the SimpleDateFormat instance that is being configured inside ObjectMapper (if you are configuring ObjectMapper as a bean), e.g.:
#Bean
public ObjectMapper objectMapper(){
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
objectMapper.setSerializationInclusion(Include.NON_NULL);
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd'T'hh:mm:ss.SSS'Z'");
simpleDateFormat.setTimeZone(TimeZone.getTimeZone("UTC"));
objectMapper.setDateFormat(simpleDateFormat);
return objectMapper;
}
If it still does not work, I would recommend creating a custom deserializer for Calendar and manually setting TimeZone in new instance, e.g.:
#Component
public class CalendarDeserialiser extends JsonDeserializer<Calendar>{
TimeZone UTC = TimeZone.getTimeZone("UTC");
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd'T'hh:mm:ss.SSS'Z'");
#Override
public Calendar deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
Calendar calendar;
try{
calendar = Calendar.getInstance(UTC);
calendar.setTime(dateFormat.parse(p.getText()));
}catch(Exception e){
throw new IOException(e);
}
calendar.setTimeInMillis(p.getLongValue());
return calendar;
}
}
And annotate your Calendar field with #JsonDeserialize(using = CalendarDeserialiser.class).
Related
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 use the below code for serializing the response that get from an external service an return a json response back as part of my service. However when the external service return a time value along with timezone (10:30:00.000-05.00) , jackson is converting it to 15:30:00. How can I ignore the timezone value?
public interface DateFormatMixin {
#JsonFormat(shape=JsonFormat.Shape.STRING, pattern="HH:mm:ss")
public XMLGregorianCalendar getStartTime();
#JsonFormat(shape=JsonFormat.Shape.STRING, pattern="HH:mm:ss")
public XMLGregorianCalendar getEndTime();
}
public ObjectMapper objectMapper() {
com.fasterxml.jackson.databind.ObjectMapper responseMapper = new com.fasterxml.jackson.databind.ObjectMapper();
responseMapper.addMixIn(Time.class, DateFormatMixin.class);
return responseMapper;
}
You can create custom deserializer
public class CustomJsonTimeDeserializerWithoutTimeZone extends JsonDeserializer<Time>{
#Override
public Time deserialize(JsonParser jp, DeserializationContext ctxt)
throws IOException, JsonProcessingException {
DateFormat format = new SimpleDateFormat("hh:mm:ss.SSS");
Time time = null;
try{
Date dt = format.parse("10:30:00.000-05.00".substring(0,12)); // remove incorrect timezone format
return new Time(dt.getTime());
}catch (ParseException e){
e.printStackTrace();
}
}
}
tell jackson to use your custom deserializer
public class Model{
#JsonDeserialize(using = CustomJsonTimeDeserializerWithoutTimeZone.class)
private Time time;
}
and use it like this:
ObjectMapper mapper = new ObjectMapper();
String jsonString = ...// jsonString retrieve from external service
Model model = mapper.readValue(jsonString, Model.class);
You can use Jackson Custom Serialization to add timezone information for your service response
You can create deserializer as below:
public Calendar deserialize(JsonParser jsonParser, DeserializationContext context)
throws IOException, JsonProcessingException {
DateFormat formatter = new SimpleDateFormat(("yyyy-MM-dd'T'HH:mm:ss.SSS"));
String date = jsonParser.getText();
try {
Calendar cal = Calendar.getInstance();
cal.setTime(formatter.parse(date));
return cal;
} catch (ParseException e) {
throw new RuntimeException(e);
}
}
I have the below Code :
DTO :
Class MyDTO {
import java.util.Date;
private Date dateOfBirth;
public Date getDateOfBirth() {
return dateOfBirth;
}
public void setDateOfBirth(Date dateOfBirth) {
this.dateOfBirth = dateOfBirth;
}
}
Controller
public void saveDOB(#RequestBody MyDTO myDTO, HttpServletRequest httprequest, HttpServletResponse httpResponse) {
System.out.println("Inside Controller");
System.out.println(myDTO.getDateOfBirth());
}
JSON Request :
{
"dateOfBirth":"2014-09-04",
}
If I send the request as yyyy-mm-dd automatic conversion to date object happens.
output in controller:-
dateOfBirth= Thu Sep 04 05:30:00 IST 2014
But when I send DateofBirth in dd-mm-yyyy format It does not convert String to Date automatically.So how i can i handle this case.
JSON Request :
{
"dateOfBirth":"04-09-2014",
}
Output: No Output in console does not even reaches controller.
I have tried with #DateTimeFormat but its not working.
I am using Spring 4.02 Please suggest is there any annotation we can use.
#DateTimeFormat is for form backing (command) objects. Your JSON is processed (by default) by Jackson's ObjectMapper in Spring's MappingJackson2HttpMessageConverter (assuming the latest version of Jackson). This ObjectMapper has a number of default date formats it can handle. It seems yyyy-mm-dd is one of them, but dd-mm-yyyy is not.
You'll need to register your own date format with a ObjectMapper and register that ObjectMapper with the MappingJackson2HttpMessageConverter. Here are various ways to do that :
Configuring ObjectMapper in Spring
Spring, Jackson and Customization (e.g. CustomDeserializer)
Alternatively, you can use a JsonDeserializer on either your whole class or one of its fields (the date). Examples in the link below
Right way to write JSON deserializer in Spring or extend it
How to deserialize JS date using Jackson?
List itemCreate a class to extend JsonDeserializer
public class CustomJsonDateDeserializer extends JsonDeserializer<Date> {
#Override
public Date deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss");
String date = jsonParser.getText();
try {
return format.parse(date);
} catch (ParseException e) {
throw new RuntimeException(e);
}
}
}
Use #JsonDeserialize(using = CustomJsonDateDeserializer.class) annotation on setter methods.
Thanks #Varun Achar answer, url
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.