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));
}

Westnorthwest

Answer 1

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);
    }
}

Forbore

How are you implementing this CustomerDateAndTimeDeserialize class? – Tifanytiff

Answer 2

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.

Gimbals

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

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

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

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 – Shelbyshelden

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

Answer 3

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.

Factotum

what if date is blank? – Alarise

Answer 4

1. If you want to bind a JSON string to date, this process is called deserialization, not serialization.

2. 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.

Douglas

Answer 5

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();
    }
}

Selvage

Answer 6

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
}

2. Write CustomDateAndTimeDeserialize.java Sample Code

Moonlighting

Answer 7

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;

Katharyn