Java Instant.parse cannot validate date

Question

I have an instant field and trying to validate the values in this field using the following approach:

Instant.parse("2021-09-29 09:35:07.531")

However, it throws the error:

java.time.format.DateTimeParseException: Text '2021-09-29 09:35:07.531' could not be parsed at index 10.

So, how can I test if the given Instant date in String format is a valid Instant date?

I am implementing an Import feature and I need to check the date cells. So, the date is normally kept in Instant format in db and for this reason I need to check if the cell date value is a valid Instant date. So, if there is a more proper way, of course I can follow that approach.

I do not use the value, I just need to check if the date is a valid date.

Answer 1

A reference to the time zone is mandatory if you are trying to use the Instant class. So, let's try this:

LocalDateTime.from(DateTimeFormatter.ofPattern("yyyy-MM-dd H:mm:ss.SSS").parse("2021-09-29 09:35:07.351")).atZone(ZoneId.systemDefault()).toInstant();

Answer 2

Your Date-Time string does not have timezone information and therefore you can not parse it into an Instant without introducing a timezone. I recommend you parse it into LocalDateTime and use the same for DB work if it is supposed to be used independent of timezones.

Demo:

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

public class Main {
    public static void main(String args[]) {
        DateTimeFormatter dtfInput = DateTimeFormatter.ofPattern("uuuu-MM-dd HH:mm:ss.SSS", Locale.ENGLISH);
        LocalDateTime ldt = LocalDateTime.parse("2021-09-29 09:35:07.531", dtfInput);
        System.out.println(ldt);
    }
}

Output:

2021-09-29T09:35:07.531

ONLINE DEMO

How to use LocalDateTime in JDBC?

Given below is a sample code to insert a LocalDateTime into columnfoo (which is of TIMESTAMP type):

PreparedStatement st = conn.prepareStatement("INSERT INTO mytable (columnfoo) VALUES (?)");
st.setObject(1, ldt);
st.executeUpdate();
st.close();

Given below is a sample code to retrieve a LocalDateTime from columnfoo:

Statement st = conn.createStatement();
ResultSet rs = st.executeQuery("SELECT * FROM mytable WHERE <some condition>");
while (rs.next()) {
    // Assuming the column index of columnfoo is 1
    LocalDateTime ldt = rs.getObject(1, LocalDateTime.class));
    System.out.println(ldt);
}
rs.close();
st.close();

In case you want to parse the given Date-Time string into Instant:

As described above, you need to introduce a timezone in order to parse it into an Instant.

Demo:

import java.time.Instant;
import java.time.ZoneId;
import java.time.format.DateTimeFormatter;
import java.util.Locale;

public class Main {
    public static void main(String args[]) {
        // Change the ZoneId to the applicable one e.g. ZoneId.of("Etc/UTC")
        DateTimeFormatter dtfInput = DateTimeFormatter.ofPattern("uuuu-MM-dd HH:mm:ss.SSS", Locale.ENGLISH)
                                        .withZone(ZoneId.systemDefault());

        Instant instant = Instant.from(dtfInput.parse("2021-09-29 09:35:07.531"));
        System.out.println(instant);
    }
}

Output in my timezone, Europe/London:

2021-09-29T08:35:07.531Z

ONLINE DEMO

Learn more about the modern Date-Time API^* from Trail: Date Time.

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^{* 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. Note that Android 8.0 Oreo already provides support for java.time.}

Answer 3

So, how can I test if the given Instant date in String format is a valid Instant date?

It is not.

An instant is a (unique) point in time. Your string holds a date and time of day. Without knowing the time zone this may denote some point within a range of 24 or 27 hours — so pretty far from being one point in time.

Edit: I understand that you are importing the string from somewhere, that you cannot decide the format or content of the string, and you need to validate it. You can validate it as a date and time. You basically cannot convert it to an Instant, or at least you only can under assumptions the validity of which I can’t know. For validation I suggest this formatter:

private static final DateTimeFormatter PARSER = new DateTimeFormatterBuilder()
        .append(DateTimeFormatter.ISO_LOCAL_DATE)
        .appendLiteral(' ')
        .append(DateTimeFormatter.ISO_LOCAL_TIME)
        .toFormatter(Locale.ROOT);

The formatter reuses built-in formatters and accepts a variable number of decimals in the seconds, which I think will make sense in many contexts. You know better whether it does in yours.

Validate like this:

    String importedDateTimeString = "2021-09-29 09:35:07.531";
    try {
        LocalDateTime.parse(importedDateTimeString, PARSER);
        System.out.format("Valid date and time: %s%n", importedDateTimeString);
    } catch (DateTimeParseException dtpe) {
        System.out.format("Not a valid date and time: %s. Validation error: %s%n",
                importedDateTimeString, dtpe);
    }

Output:

Valid date and time: 2021-09-29 09:35:07.531

Original recommendation: So instead use a format that includes offset from UTC. In particular the ISO 8601 format for an instant in UTC is a recommended option for several purposes, like 2021-09-29T01:35:07.531Z.

Link: Wikipedia article: ISO 8601