How to sanity check a date in Java [closed]

Answer 1

Assuming that both of those are Strings (otherwise they'd already be valid Dates), here's one way:

package cruft;

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

public class DateValidator
{
    private static final DateFormat DEFAULT_FORMATTER;

    static
    {
        DEFAULT_FORMATTER = new SimpleDateFormat("dd-MM-yyyy");
        DEFAULT_FORMATTER.setLenient(false);
    }

    public static void main(String[] args)
    {
        for (String dateString : args)
        {
            try
            {
                System.out.println("arg: " + dateString + " date: " + convertDateString(dateString));
            }
            catch (ParseException e)
            {
                System.out.println("could not parse " + dateString);
            }
        }
    }

    public static Date convertDateString(String dateString) throws ParseException
    {
        return DEFAULT_FORMATTER.parse(dateString);
    }
}

Here's the output I get:

java cruft.DateValidator 32-11-2010 31-02-2010 04-01-2011
could not parse 32-11-2010
could not parse 31-02-2010
arg: 04-01-2011 date: Tue Jan 04 00:00:00 EST 2011

Process finished with exit code 0

As you can see, it does handle both of your cases nicely.

Answer 2

This is working great for me. Approach suggested above by Ben.

private static boolean isDateValid(String s) {
    SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy");
    try {
        Date d = asDate(s);
        if (sdf.format(d).equals(s)) {
            return true;
        } else {
            return false;
        }
    } catch (ParseException e) {
        return false;
    }
}

Answer 3

I think the simpliest is just to convert a string into a date object and convert it back to a string. The given date string is fine if both strings still match.

public boolean isDateValid(String dateString, String pattern)
{   
    try
    {
        SimpleDateFormat sdf = new SimpleDateFormat(pattern);
        if (sdf.format(sdf.parse(dateString)).equals(dateString))
            return true;
    }
    catch (ParseException pe) {}

    return false;
}

Answer 4

        public static String detectDateFormat(String inputDate, String requiredFormat) {
        String tempDate = inputDate.replace("/", "").replace("-", "").replace(" ", "");
        String dateFormat;

        if (tempDate.matches("([0-12]{2})([0-31]{2})([0-9]{4})")) {
            dateFormat = "MMddyyyy";
        } else if (tempDate.matches("([0-31]{2})([0-12]{2})([0-9]{4})")) {
            dateFormat = "ddMMyyyy";
        } else if (tempDate.matches("([0-9]{4})([0-12]{2})([0-31]{2})")) {
            dateFormat = "yyyyMMdd";
        } else if (tempDate.matches("([0-9]{4})([0-31]{2})([0-12]{2})")) {
            dateFormat = "yyyyddMM";
        } else if (tempDate.matches("([0-31]{2})([a-z]{3})([0-9]{4})")) {
            dateFormat = "ddMMMyyyy";
        } else if (tempDate.matches("([a-z]{3})([0-31]{2})([0-9]{4})")) {
            dateFormat = "MMMddyyyy";
        } else if (tempDate.matches("([0-9]{4})([a-z]{3})([0-31]{2})")) {
            dateFormat = "yyyyMMMdd";
        } else if (tempDate.matches("([0-9]{4})([0-31]{2})([a-z]{3})")) {
            dateFormat = "yyyyddMMM";
        } else {
            return "Pattern Not Added";
//add your required regex
        }
        try {
            String formattedDate = new SimpleDateFormat(requiredFormat, Locale.ENGLISH).format(new SimpleDateFormat(dateFormat).parse(tempDate));

            return formattedDate;
        } catch (Exception e) {
            //
            return "";
        }

    }

Answer 5

looks like SimpleDateFormat is not checking the pattern strictly even after setLenient(false); method is applied on it, so i have used below method to validate if the date inputted is valid date or not as per supplied pattern.

import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeParseException;
public boolean isValidFormat(String dateString, String pattern) {
    boolean valid = true;
    DateTimeFormatter formatter = DateTimeFormatter.ofPattern(pattern);
    try {
        formatter.parse(dateString);
    } catch (DateTimeParseException e) {
        valid = false;
    }
    return valid;
}

Answer 6

// to return valid days of month, according to month and year
int returnDaysofMonth(int month, int year) {
    int daysInMonth;
    boolean leapYear;
    leapYear = checkLeap(year);
    if (month == 4 || month == 6 || month == 9 || month == 11)
        daysInMonth = 30;
    else if (month == 2)
        daysInMonth = (leapYear) ? 29 : 28;
    else
        daysInMonth = 31;
    return daysInMonth;
}

// to check a year is leap or not
private boolean checkLeap(int year) {
    Calendar cal = Calendar.getInstance();
    cal.set(Calendar.YEAR, year);
    return cal.getActualMaximum(Calendar.DAY_OF_YEAR) > 365;
}