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.SS         


        

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);
    }
}

Answer 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.

Answer 3

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

Answer 4

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();
    }
}

Answer 5

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.

Answer 6

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.