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:

1. What is the formatting I need to use to carry out conversion with Jackson? Is Date a good field type for this?
2. 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.

Solution 1:

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;

Solution 2:

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.

Solution 3:

Of course there is an automated way called serialization and deserialization and you can define it with specific annotations (@JsonSerialize,@JsonDeserialize) as mentioned by pb2q as well.

You can use both java.util.Date and java.util.Calendar ... and probably JodaTime as well.

The @JsonFormat annotations not worked for me as I wanted (it has adjusted the timezone to different value) during deserialization (the serialization worked perfect):

@JsonFormat(locale = "hu", shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm", timezone = "CET")

@JsonFormat(locale = "hu", shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm", timezone = "Europe/Budapest")

You need to use custom serializer and custom deserializer instead of the @JsonFormat annotation if you want predicted result. I have found real good tutorial and solution here http://www.baeldung.com/jackson-serialize-dates

There are examples for Date fields but I needed for Calendar fields so here is my implementation:

The serializer class:

public class CustomCalendarSerializer extends JsonSerializer<Calendar> {

    public static final SimpleDateFormat FORMATTER = new SimpleDateFormat("yyyy-MM-dd HH:mm");
    public static final Locale LOCALE_HUNGARIAN = new Locale("hu", "HU");
    public static final TimeZone LOCAL_TIME_ZONE = TimeZone.getTimeZone("Europe/Budapest");

    @Override
    public void serialize(Calendar value, JsonGenerator gen, SerializerProvider arg2)
            throws IOException, JsonProcessingException {
        if (value == null) {
            gen.writeNull();
        } else {
            gen.writeString(FORMATTER.format(value.getTime()));
        }
    }
}

The deserializer class:

public class CustomCalendarDeserializer extends JsonDeserializer<Calendar> {

    @Override
    public Calendar deserialize(JsonParser jsonparser, DeserializationContext context)
            throws IOException, JsonProcessingException {
        String dateAsString = jsonparser.getText();
        try {
            Date date = CustomCalendarSerializer.FORMATTER.parse(dateAsString);
            Calendar calendar = Calendar.getInstance(
                CustomCalendarSerializer.LOCAL_TIME_ZONE, 
                CustomCalendarSerializer.LOCALE_HUNGARIAN
            );
            calendar.setTime(date);
            return calendar;
        } catch (ParseException e) {
            throw new RuntimeException(e);
        }
    }
}

and the usage of the above classes:

public class CalendarEntry {

    @JsonSerialize(using = CustomCalendarSerializer.class)
    @JsonDeserialize(using = CustomCalendarDeserializer.class)
    private Calendar calendar;

    // ... additional things ...
}

Using this implementation the execution of the serialization and deserialization process consecutively results the origin value.

Only using the @JsonFormat annotation the deserialization gives different result I think because of the library internal timezone default setup what you can not change with annotation parameters (that was my experience with Jackson library 2.5.3 and 2.6.3 version as well).

Solution 4:

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"
}

Solution 5:

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 [email protected] 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"));
    }
}