JSON Serializing date in a custom format (Can not construct instance of java.util.Date from String value)

Question

could not read JSON: Can not construct instance of java.util.Date from String 
value '2012-07-21 12:11:12': not a valid representation("yyyy-MM-dd'T'HH:mm:ss.SSSZ", "yyyy-MM-dd'T'HH:mm:ss.SSS'Z'", "EEE, dd MMM yyyy HH:mm:ss zzz", "yyyy-MM-dd"))

passing json request to REST controller method in a POJO class.user should enter only in below datetime format other wise it should throw message.why DateSerializer is not calling?

add(@Valid @RequestBody User user)
{
}

json:

{
   "name":"ssss",
   "created_date": "2012-07-21 12:11:12"
}

pojo class variable

@JsonSerialize(using=DateSerializer.class)
@Column
@NotNull(message="Please enter a date")      
@Temporal(value=TemporalType.TIMESTAMP)
private Date created_date;

public void serialize(Date value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
    logger.info("serialize:"+value);
    DateFormat formatter = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
    logger.info("DateSerializer formatter:"+formatter.format(value));
    jgen.writeString(formatter.format(value));
}

Splaktar · Accepted Answer · 2018-12-03 00:36:57Z

54

Annotate your created_date field with the JsonFormat annotation to specify the output format.

@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss", timezone = TimeZone.getDefault(), locale = Locale.getDefault())

Note that you may need to pass in a different Locale and TimeZone if they should be based on something other than what the server uses.

You can find out more information in the docs.

edited Dec 3, 2018 at 0:36

answered Nov 25, 2014 at 16:12

Splaktar

5,9145 gold badges45 silver badges76 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Dariusz Over a year ago

Is it all you have to do? I tried this and it didn't work by itself. Maybe some other annotations are required somewhere?

Splaktar Over a year ago

You need to be using Jackson: fasterxml.github.io/jackson-annotations/javadoc/2.0.0/com/…

Anton Dozortsev Over a year ago

Be aware of using this annotation. Here is known issue with with one day less.

Pavlo Zvarych Over a year ago

Don't forget to add timezone = "Canonical_ID_of_your_timezone" parameter after "pattern" if you want to have a right time. Timezones names could be found here: joda-time.sourceforge.net/timezones.html

tibi Over a year ago

Or use: @Pattern(regexp = Constants.REGEX)

ansidev · Accepted Answer · 2018-01-19 05:36:10Z

21

I have the same problem, so I write a custom date deserialization with @JsonDeserialize(using=CustomerDateAndTimeDeserialize.class)

public class CustomerDateAndTimeDeserialize extends JsonDeserializer<Date> {

    private SimpleDateFormat dateFormat = new SimpleDateFormat(
            "yyyy-MM-dd HH:mm:ss");

    @Override
    public Date deserialize(JsonParser paramJsonParser,
            DeserializationContext paramDeserializationContext)
            throws IOException, JsonProcessingException {
        String str = paramJsonParser.getText().trim();
        try {
            return dateFormat.parse(str);
        } catch (ParseException e) {
            // Handle exception here
        }
        return paramDeserializationContext.parseDate(str);
    }
}

edited Jan 19, 2018 at 5:36

ansidev

3711 gold badge3 silver badges19 bronze badges

answered Nov 19, 2014 at 2:27

lbdfcgyjjr

2262 silver badges3 bronze badges

1 Comment

adamhurwitz.eth Over a year ago

How are you implementing this CustomerDateAndTimeDeserialize class?

Vikash Kumar · Accepted Answer · 2020-04-04 11:14:53Z

2

For someone ,If you are using DTO/VO/POJO to map your request you can simply annotate your date field

@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
private Date customerRegDate;

And json request should be:

{
"someDate":"2020-04-04 16:11:02"
}

You don't need to annotate Entity class variable.

answered Apr 4, 2020 at 11:14

Vikash Kumar

1,22613 silver badges10 bronze badges

1 Comment

Aadam Over a year ago

what if date is blank?

Ali Akbarpour · Accepted Answer · 2018-04-24 14:59:25Z

1

If you want to bind a JSON string to date, this process is called deserialization, not serialization.
To bind a JSON string to date, create a custom date deserialization, annotate created_date or its setter with

@JsonDeserialize(using=YourCustomDateDeserializer.class)

where you have to implement the method public Date deserialize(...) to tell Jackson how to convert a string to a date.

Enjoy.

edited Apr 24, 2018 at 14:59

Ali Akbarpour

9742 gold badges18 silver badges35 bronze badges

answered Aug 17, 2013 at 10:55

khoi nguyen

4093 silver badges12 bronze badges

Comments

Dariusz · Accepted Answer · 2015-03-26 11:17:50Z

Yet another way is to have a custom Date object which takes care of its own serialization.

While I don't really think extending simple objects like Date, Long, etc. is a good practice, in this particular case it makes the code easily readable, has a single point where the format is defined and is rather more than less compatible with normal Date object.

public class CustomFormatDate extends Date {

    private DateFormat myDateFormat = ...; // your date format

    public CustomFormatDate() {
        super();
    }

    public CustomFormatDate(long date) {
        super(date);
    }

    public CustomFormatDate(Date date) {
        super(date.getTime());
    }


    @JsonCreator
    public static CustomFormatDate forValue(String value) {
        try {
            return new CustomFormatDate(myDateFormat.parse(value));
        } catch (ParseException e) {
            return null;
        }
    }

    @JsonValue
    public String toValue() {
        return myDateFormat.format(this);
    }

    @Override
    public String toString() {
        return toValue();
    }
}

rahulnikhare · Accepted Answer · 2019-05-26 10:30:09Z

0

I solved this by using the below steps.

1.In entity class annote it using @JsonDeserialize

@Entity
@Table(name="table")
public class Table implements Serializable {
// Some code
@JsonDeserialize(using= CustomerDateAndTimeDeserialize.class)
@Temporal(TemporalType.TIMESTAMP)
@Column(name="created_ts")
private Date createdTs
}

Write CustomDateAndTimeDeserialize.java Sample Code

edited May 26, 2019 at 10:30

answered May 26, 2019 at 9:25

rahulnikhare

1,5461 gold badge19 silver badges29 bronze badges

Comments

Arefe · Accepted Answer · 2021-11-14 06:25:05Z

0

I needed to annotate with the @JsonFormat and afterwards, this is working well:

@Temporal(TemporalType.DATE)
@DateTimeFormat(pattern = "MM-dd-yyyy")
@JsonFormat(pattern = "MM-dd-yyyy")
private java.util.Date expDate;

answered Nov 14, 2021 at 6:25

Arefe

12.6k25 gold badges135 silver badges198 bronze badges

Collectives™ on Stack Overflow

JSON Serializing date in a custom format (Can not construct instance of java.util.Date from String value)

7 Answers 7

5 Comments

1 Comment

1 Comment

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

5 Comments

1 Comment

1 Comment

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related