Can not deserialize value of type java.util.Date from String "Aug 29, 2023 5:48:51 PM": expected format "yyyy-MM-dd HH:mm:ss.SSS"

Question

I am trying to parse this data "Aug 29, 2023 5:48:51 PM" which is in String format to JSON but getting the below error

Can not deserialize value of type java.util.Date from String "Aug 29, 2023 5:48:51 PM": expected format "yyyy-MM-dd HH:mm:ss.SSS"

The POJO declaration of the corresponding field is given below:

@JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd HH:mm:ss.SSS")
    @Column(name = "createdOn")
    private Date createdOn;

Pls help

I have tried to modify the pojo declaration of the field to this

@JsonFormat(pattern="yyyy-MM-dd HH:mm:ss.SSS")
@Column(name = "createdOn")
private Date createdOn;

Answer 1

java.time

The java.util date-time API and their corresponding parsing/formatting type, SimpleDateFormat are outdated and error-prone. In March 2014, the modern Date-Time API supplanted the legacy date-time API. Since then, it has been strongly recommended to switch to java.time, the modern date-time API.

The format corresponding to your date-time string is MMM dd, uuuu h:m:s a in the en_US locale.

Demo:

import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;
import java.util.Locale;

class Main {
    public static void main(String[] args) {
        DateTimeFormatter parser = DateTimeFormatter.ofPattern("MMM dd, uuuu h:m:s a", new Locale("en_US"));
        System.out.println(LocalDateTime.parse("Aug 29, 2023 5:48:51 PM", parser));
    }
}

Output:

2023-08-29T17:48:51

Online Demo

Based on this description, you can make the following changes in your code:

@JsonFormat(pattern = "MMM dd, uuuu h:m:s a", shape = JsonFormat.Shape.STRING, locale = "en_US")
@Column(name = "createdOn")
private LocalDateTime createdOn;

Learn about the modern date-time API from Trail: Date Time