How to bind date @ConfigurationProperties with time only using @DateTimeFormat?

Question

New to spring-boot.

I'm trying to parse the properties from the file with annotation @ConfigurationProperties. I'm able to parse the fields other than date field.

issue is My property file has only time without date. ie date=09:30:00.

I'm able to parse it with @DateTimeFormat(pattern = "HH:mm:ss"). But the issue is, it is giving date as date=Thu Jan 01 09:30:00 GST 1970.

I would like to get the date as todays date with time 09:30:00. Is it possible?

@ConfigurationProperties
public class Config {

    private int id;
    
    private int version;
        
    @DateTimeFormat(pattern = "HH:mm:ss")
    private Date date;
    
}

Property

id=12
version=2
date=09:30:00

Answer 1

Why not use a type that represents time only?

    @DateTimeFormat(pattern = "HH:mm:ss")
    private LocalTime time;

    public LocalDateTime getDate() {
        return LocalDateTime.of(LocalDate.now(), time);
    } 

Answer 2

The output you are getting is as expected because a string is parsed into java.util.Date using a SimpleDateFormat which defaults the date-time to January 1, 1970, 00:00:00 GMT .

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;

public class Main {
    public static void main(String[] args) throws ParseException {
        String text = "09:30:00";
        SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss");
        Date date = sdf.parse(text);
        System.out.println(date);
    }
}

The output in my timezone (Europe/London):

Thu Jan 01 09:30:00 GMT 1970

Note that the java.util.Date object is not a real date-time object like the modern date-time types ; rather, it represents the number of milliseconds since the standard base time known as "the epoch", namely January 1, 1970, 00:00:00 GMT (or UTC). When you print an object of java.util.Date , its toString method returns the date-time in the JVM's timezone, calculated from this milliseconds value. If you need to print the date-time in a different timezone, you will need to set the timezone to SimpleDateFormat and obtain the formatted string from it. The java.util date-time API and their formatting API, SimpleDateFormat are not only outdated but also error-prone because of many such things. It is recommended to stop using them completely and switch to the modern date-time API ¹ .

Given below are a couple of options:

1. Recommended: Use LocalTime which truly represents time.

    @DateTimeFormat(pattern = "HH:mm:ss")
    private LocalTime time;

2. Dirty way: Declare your field as a String and parse it in your business logic making it error-prone and dirty.

    private String time;

I strongly recommend NOT to go with the second option.

A quick demo using LocalTime :

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

public class Main {
    public static void main(String[] args) {
        String text = "9:30:00";

        // The optional part can be put inside square bracket
        DateTimeFormatter dtfInput = DateTimeFormatter.ofPattern("H:m[:s]", Locale.ENGLISH);
        LocalTime time = LocalTime.parse(text, dtfInput);

        // Default implementation of LocalTime#toString omits the seconds part if it is zero
        System.out.println(time);

        // Custom output
        DateTimeFormatter dtfOutput = DateTimeFormatter.ofPattern("HH:mm:ss", Locale.ENGLISH);
        String formatted = dtfOutput.format(time);
        System.out.println(formatted);
    }
}

Output:

09:30
09:30:00

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

---

^{1. For any reason, if you have to stick to Java 6 or Java 7, you can use ThreeTen-Backport which backports most of the java.time functionality to Java 6 & 7. If you are working for an Android project and your Android API level is still not compliant with Java-8, check Java 8+ APIs available through desugaring and How to use ThreeTenABP in Android Project .}

Answer 3

Don't use old and obsolete class Date. Look into package java.time and in particular in your case - class LocalTime .

change your code to: @ConfigurationProperties public class Config {

private int id;

private int version;
    
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "HH:mm:ss")
private LocalTime date;

}

This should work. You may need to add the following dependency:

<dependency>
    <groupId>com.fasterxml.jackson.datatype</groupId>
    <artifactId>jackson-datatype-jsr310</artifactId>
    <version>2.6.0</version>
</dependency>

This is modified answer from this question: Spring Data JPA - ZonedDateTime format for json serialization