Develop Reference

Deserialize JSON date format to ZonedDateTime using objectMapper

Background
I have the following JSON (message from Kafka)
{
"markdownPercentage": 20,
"currency": "SEK",
"startDate": "2019-07-25"
}
I have the following (JSON schema generated) POJO (I cannot change the POJO as it is shared resource in the company)
public class Markdown {
#JsonProperty("markdownPercentage")
#NotNull
private Integer markdownPercentage = 0;
#JsonProperty("currency")
#NotNull
private String currency = "";
#JsonFormat(
shape = Shape.STRING,
pattern = "yyyy-MM-dd"
)
#JsonProperty("startDate")
#NotNull
private ZonedDateTime startDate;
// Constructors, Getters, Setters etc.
}
Our application is a Spring Boot application which reads the JSON message (1) from Kafka using Spring Cloud Stream and uses the POJO (2) and then does stuff with it.
Problem
When the application tries to deserialize the message to the object it throws the following exception
com.fasterxml.jackson.databind.exc.InvalidFormatException: Cannot deserialize value of type `java.time.ZonedDateTime` from String "2019-07-25": Failed to deserialize java.time.ZonedDateTime: (java.time.DateTimeException) Unable to obtain ZonedDateTime from TemporalAccessor: {},ISO resolved to 2019-07-25 of type java.time.format.Parsed
at [Source: (String)"{"styleOption":"so2_GreyMelange_1563966403695_1361997740","markdowns":[{"markdownPercentage":20,"currency":"SEK","startDate":"2019-07-25"},{"markdownPercentage":20,"currency":"NOK","startDate":"2019-07-25"},{"markdownPercentage":20,"currency":"CHF","startDate":"2019-07-25"}]}"; line: 1, column: 126] (through reference chain: com.bestseller.generated.interfacecontracts.kafkamessages.pojos.markdownScheduled.MarkdownScheduled["markdowns"]->java.util.ArrayList[0]->com.bestseller.generated.interfacecontracts.kafkamessages.pojos.markdownScheduled.Markdown["startDate"])
at com.fasterxml.jackson.databind.exc.InvalidFormatException.from(InvalidFormatException.java:67)
at com.fasterxml.jackson.databind.DeserializationContext.weirdStringException(DeserializationContext.java:1549)
at com.fasterxml.jackson.databind.DeserializationContext.handleWeirdStringValue(DeserializationContext.java:911)
at com.fasterxml.jackson.datatype.jsr310.deser.JSR310DeserializerBase._handleDateTimeException(JSR310DeserializerBase.java:80)
at com.fasterxml.jackson.datatype.jsr310.deser.InstantDeserializer.deserialize(InstantDeserializer.java:212)
at com.fasterxml.jackson.datatype.jsr310.deser.InstantDeserializer.deserialize(InstantDeserializer.java:50)
at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:127)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:288)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:151)
at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:286)
at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:245)
at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:27)
at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:127)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:288)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:151)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4013)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3004)
at com.bestseller.mps.functional.TestingConfiguration.test(TestingConfiguration.java:42)
at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at java.base/jdk.internal.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.base/java.lang.reflect.Method.invoke(Method.java:566)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:50)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:325)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:58)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:268)
at org.junit.runners.ParentRunner.run(ParentRunner.java:363)
at org.junit.runner.JUnitCore.run(JUnitCore.java:137)
at com.intellij.junit4.JUnit4IdeaTestRunner.startRunnerWithArgs(JUnit4IdeaTestRunner.java:68)
at com.intellij.rt.execution.junit.IdeaTestRunner$Repeater.startRunnerWithArgs(IdeaTestRunner.java:47)
at com.intellij.rt.execution.junit.JUnitStarter.prepareStreamsAndStart(JUnitStarter.java:242)
at com.intellij.rt.execution.junit.JUnitStarter.main(JUnitStarter.java:70)
Caused by: java.time.DateTimeException: Unable to obtain ZonedDateTime from TemporalAccessor: {},ISO resolved to 2019-07-25 of type java.time.format.Parsed
at java.base/java.time.ZonedDateTime.from(ZonedDateTime.java:566)
at com.fasterxml.jackson.datatype.jsr310.deser.InstantDeserializer.deserialize(InstantDeserializer.java:207)
... 35 more
Caused by: java.time.DateTimeException: Unable to obtain ZoneId from TemporalAccessor: {},ISO resolved to 2019-07-25 of type java.time.format.Parsed
at java.base/java.time.ZoneId.from(ZoneId.java:463)
at java.base/java.time.ZonedDateTime.from(ZonedDateTime.java:554)
... 36 more
Current Code
I have the following objectMapper defined
/**
* Date mapper.
*
* #return the {#link ObjectMapper}
*/
#Bean
public ObjectMapper objectMapper() {
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JavaTimeModule());
mapper.setDateFormat(new SimpleDateFormat("yyyy-MM-dd"));
return mapper;
}
Question
I understand that the resulting ZonedDateTime in the POJO needs a 'time' element which is not present in the source message. I have control only over the objectMapper. Is there any possible configuration that can make this work ?
Note
I am fine if the time element in the deserialised POJO is "assumed" to be startOfDay i.e. "00.00.00.000Z"

I have control only over the ObjectMapper. Is there any possible configuration that can make this work?
As long as you are happy with default values for the time and for the timezone, you could work around it with a custom deserializer:
public class ZonedDateTimeDeserializer extends JsonDeserializer<ZonedDateTime> {
#Override
public ZonedDateTime deserialize(JsonParser jsonParser,
DeserializationContext deserializationContext)
throws IOException {
LocalDate localDate = LocalDate.parse(
jsonParser.getText(),
DateTimeFormatter.ISO_LOCAL_DATE);
return localDate.atStartOfDay(ZoneOffset.UTC);
}
}
Then add it to a module and register the module to your ObjectMapper instance:
SimpleModule module = new SimpleModule();
module.addDeserializer(ZonedDateTime.class, new ZonedDateTimeDeserializer());
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(module);
If adding the deserializer to a module doesn't suit you (in the sense this configuration will be applied to other ZonedDateTime instances), then you could rely on mix-ins to define which fields the deserializer will be applied to. First define a mix-in interface, as shown below:
public interface MarkdownMixIn {
#JsonDeserialize(using = ZonedDateTimeDeserializer.class)
ZonedDateTime getDate();
}
And then bind the mix-in interface to the desired class:
ObjectMapper mapper = new ObjectMapper();
mapper.addMixIn(Markdown.class, MarkdownMixIn.class);

Problem: I would like to parse dates from json to java LocalDateTime/ZonedDateTime objects. ZonedDateTimeSerializer exists but the ZonedDateTimeDeserializer doesn't exist. Hence why I created a custom ZonedDateTimeDeserializer.
public static final String ZONED_DATE_TIME_FORMAT = "yyyy-MM-dd'T'HH:mm:ss.SSSz";
#Getter
#Setter
#JsonSerialize(using = ZonedDateTimeSerializer.class)
#JsonDeserialize(using = ZonedDateTimeDeserializer.class) // Doesn't exist, So I created a custom ZonedDateDeserializer utility class.
#JsonFormat(pattern = ZONED_DATE_TIME_FORMAT)
#JsonProperty("lastUpdated")
private ZonedDateTime lastUpdated;
Solution: I ended up with simpler & fewer lines of code.
The Utility Class for deserialising the ZonedDateTime:
/**
* Custom {#link ZonedDateTime} deserializer.
*
* #param jsonParser for extracting the date in {#link String} format.
* #param deserializationContext for the process of deserialization a single root-level value.
* #return {#link ZonedDateTime} object of the date.
* #throws IOException throws I/O exceptions.
*/
public class ZonedDateTimeDeserializer extends JsonDeserializer<ZonedDateTime> {
#Override
public ZonedDateTime deserialize(JsonParser jsonParser, DeserializationContext deserializationContext)
throws IOException {
return ZonedDateTime.parse(jsonParser.getText(), DateTimeFormatter.ofPattern(ZONED_DATE_TIME_FORMAT));
}
}
What if you're using LocalDateTime instead. In that case it's even easier, both the deserializer & the serializer classes are already provided to us. No need for custom utility classes as defined above:
#Getter
#Setter
#JsonSerialize(using = LocalDateSerializer.class)
#JsonDeserialize(using = LocalDateTimeDeserializer.class)
#JsonFormat(pattern = ZONED_DATE_TIME_FORMAT) //Specify the format you want: "yyyy-MM-dd'T'HH:mm:ss.SSS"
#JsonProperty("created")
private LocalDateTime created;
Other links that also helped with the research:
ideas that led to this solution
Keywords: json format localDateTime zonedDateTime

unfortunately, you can not deserialize String Object to ZonedDateTime format by default. But you can overcome this problem in two ways.
way 01
To change ZonedDateTime type to LocalDate type in your POJO class and passing value as a String 03-06-2012
way 02
But If you have to store the date and time with timezone then you have to do the following way to overcome
Step 01
Create a class for ZonedDateTime Deserialization with DateTimeFormat
public class ZonedDateTimeDeserializer extends JsonDeserializer<ZonedDateTime> {
#Override
public ZonedDateTime deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
DateTimeFormatter dateTimeFormatter=DateTimeFormatter.ofPattern("dd-MM-yyyy HH:mm:ss z");
LocalDate localDate = LocalDate.parse(p.getText(),dateTimeFormatter);
return localDate.atStartOfDay(ZoneOffset.UTC);
}
}
Step 02
You have to use that Deserialize class with the affected field in your POJO class with the support of #JsonDeserialize method level annotation.
#JsonDeserialize(using = ZonedDateTimeDeserializer.class)
private ZonedDateTime startDate;
Step 03
Passing the value as String in the above format which is given at the ZonedDateTimeDeserializer class
"startDate" : "09-03-2003 10:15:00 Europe/Paris"

Sadly, without changing the type of the POJO to LocalDate, that would be difficult.
The nearest solution that I can think of is to write a custom JsonDeserializer for Jackson, which is absolutely not a good practice for that kind of thing.
see https://fasterxml.github.io/jackson-databind/javadoc/2.3.0/com/fasterxml/jackson/databind/JsonDeserializer.html

You can write your own deserialiser like is shown in #cassiomolin answer. But also there is another option. On stack trace we have DeserializationContext.weirdStringException method which allows us to provide our DeserializationProblemHandler with handling weird strings values. See below example:
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.annotation.JsonFormat.Shape;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.deser.DeserializationProblemHandler;
import com.fasterxml.jackson.datatype.jsr310.JavaTimeModule;
import java.io.IOException;
import java.time.LocalDate;
import java.time.ZonedDateTime;
import java.time.format.DateTimeFormatter;
import java.util.TimeZone;
public class AppJson {
public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();
// override default time zone if needed
mapper.setTimeZone(TimeZone.getTimeZone("America/Los_Angeles"));
mapper.registerModule(new JavaTimeModule());
mapper.addHandler(new DeserializationProblemHandler() {
#Override
public Object handleWeirdStringValue(DeserializationContext ctxt, Class<?> targetType,
String valueToConvert, String failureMsg) {
LocalDate date = LocalDate.parse(valueToConvert, DateTimeFormatter.ISO_DATE);
return date.atStartOfDay(ctxt.getTimeZone().toZoneId());
}
});
String json = "{\"startDate\": \"2019-07-25\"}";
Markdown markdown = mapper.readValue(json, Markdown.class);
System.out.println(markdown);
}
}
Above code prints:
Markdown{startDate=2019-07-25T00:00-07:00[America/Los_Angeles]}

You need to convert your String variable startDate to ZonedDateTimne then it will be converted and saved to your DB or whatever.
As in coming json startDate is in string format and you have said that you can't change the POJO then, you need to convert it before assigning it to private ZonedDateTime startDate;
You can do so like the example below:
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss a z");
ZonedDateTime dateTime = ZonedDateTime.parse("2019-03-27 10:15:30 AM +05:30", formatter);
System.out.println(dateTime);

How to dynamically customise deserialiser for date format?

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"}

how to properly convert a string representation to java Date inline with jackson ObjectMapper readValue? [duplicate]

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.

Error "gson.JsonSyntaxException" when I try to retrieve a document with a date in MongoDB

In my MongoDB databse there are documents with a date field. By making a console find the result looks like:
"data" : ISODate("2015-03-01T00:40:45Z")
But when GSON try to retrieve the object appears this error:
javax.servlet.ServletException: com.google.gson.JsonSyntaxException: 03-01-2015-01-40-45-000
I tried to use the GSONBuilder as described below but the error persists:
Gson gson= new GsonBuilder().setDateFormat("yyyy-MM-dd'T'HH:mm:ssX").create();
How can I fix this?

I receive from server a Json Object with a long value (Timestamp) for Date.
I am creating my Gson object:
final Gson gson = new GsonBuilder().registerTypeAdapter(Date.class, new JsonDateDeserializer()).create();
This is code for my JsonDateDeserializer:
import com.google.gson.JsonDeserializationContext;
import com.google.gson.JsonDeserializer;
import com.google.gson.JsonElement;
import com.google.gson.JsonParseException;
import java.lang.reflect.Type;
import java.util.Date;
public class JsonDateDeserializer implements JsonDeserializer<Date> {
public Date deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
long time = json.getAsJsonPrimitive().getAsLong();
Date d = new Date(time);
return d;
}
}

Updated long time = json.getAsJsonObject().getAsJsonPrimitive("$date").getAsLong();
import com.google.gson.JsonDeserializationContext;
import com.google.gson.JsonDeserializer;
import com.google.gson.JsonElement;
import com.google.gson.JsonParseException;
import java.lang.reflect.Type;
import java.util.Date;
public class JsonDateDeserializer implements JsonDeserializer<Date> {
public Date deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
long time = json.getAsJsonObject().getAsJsonPrimitive("$date").getAsLong();
Date d = new Date(time);
return d;
}
}
http://www.techiesinfo.com/performance

I'm also using Gson to deserialize JSON string. For me, none of the following formats worked:
setDateFormat("yyyy-MM-dd'T'HH:mm:ssX").create(); //Didn't work
setDateFormat("yyyy-MM-dd'T'HH:mm:ssz").create(); //Didn't work
setDateFormat("yyyy-MM-dd'T'HH:mm:ssZ").create(); //Didn't work
Each of them was failing to parse the date. So I just removed the last character from the format. And it worked.
setDateFormat("yyyy-MM-dd'T'HH:mm:ss").create(); //worked
I know this is not the correct way, but at least it allows you to parse the date for the time being.
Edit: There is a better solution at this link Gson: JsonSyntaxException on date. Basically, this is the gist of the answer there:
I created a custom JsonDeserializer and registered it for the Date type. By doing so, Gson will use my deserializer for the Date type instead of its default. The same can be done for any other type if you want to serialize/deserialize it in a custom way.
public class JsonDateDeserializer implements JsonDeserializer<Date> {
public Date deserialize(JsonElement json, Type typeOfT,JsonDeserializationContext context) throws JsonParseException {
String s = json.getAsJsonPrimitive().getAsString();
long l = Long.parseLong(s.substring(6, s.length() - 2));
Date d = new Date(l);
return d;
}
}
Then, when I am creating my Gson object:
Gson gson = new GsonBuilder().registerTypeAdapter(Date.class, new JsonDateDeserializer()).create();"
EDIT 2: Turns out that the logic in the deserialize() function above did not work properly. So decided to use Joda-Time library. Here is the modified function:
public class JsonDateDeserializer implements JsonDeserializer<Date> {
public Date deserialize(JsonElement json, Type typeOfT,JsonDeserializationContext context) throws JsonParseException {
String s = json.getAsJsonPrimitive().getAsString();
DateTimeFormatter parser = ISODateTimeFormat.dateTimeParser();
DateTime dateTime = parser.parseDateTime(s);
return dateTime.toDate();
}
}
This is working. :)
So basically, either use the very first option, or use the last option here.

Employee employee = new GsonBuilder()
.registerTypeAdapter(LocalTime.class, new MyDateTypeAdapter())
.create().fromJson(json, Employee.class);
Change Locatime to Date orDatetime as per your requirement.
public class MyDateTypeAdapter implements JsonDeserializer<LocalTime> {
#Override
public synchronized LocalTime deserialize(JsonElement jsonElement, Type type,
JsonDeserializationContext jsonDeserializationContext) {
try {
JsonPrimitive dateStr= jsonElement.getAsJsonObject().getAsJsonPrimitive("$date") ;
Instant instant = Instant.parse(dateStr.getAsString());
LocalTime localTime = LocalTime.ofInstant(instant, ZoneId.of(ZoneOffset.UTC.getId()));
return localTime;
} catch (Exception e) {
log.error("error calling deserialize {}", e);
throw new JsonParseException(e);
}
}
}

Date format Mapping to JSON Jackson

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.