Comment vérifier la validité d'une date en Java

StackOverflow https://stackoverflow.com/questions/226910

Question

Je trouve curieux que le moyen le plus évident de créer Date des objets en Java soit obsolète et semble avoir été & "substitué &"; avec un calendrier pas si évident d'utiliser un calendrier indulgent.

Comment vérifiez-vous qu'une date, donnée sous la forme d'une combinaison de jour, mois et année, est une date valide?

Par exemple, 2008-02-31 (comme dans aaaa-mm-jj) serait une date invalide.

Était-ce utile?

La solution

La méthode actuelle consiste à utiliser la classe calendar. Il possède le setLenient méthode qui validera la date et levera et exception si elle est hors limites comme dans votre exemple.

Oublié d'ajouter: Si vous obtenez une instance de calendrier et définissez l'heure à l'aide de votre date, voici comment vous obtenez la validation. 

Calendar cal = Calendar.getInstance();
cal.setLenient(false);
cal.setTime(yourDate);
try {
    cal.getTime();
}
catch (Exception e) {
  System.out.println("Invalid date");
}

Autres conseils

La clé est df.setLenient (false); . C'est plus que suffisant pour les cas simples. Si vous recherchez une bibliothèque plus robuste (je doute) et / ou alternative comme joda-time, consultez la réponse de l'utilisateur < !> quot; retarder " 

final static String DATE_FORMAT = "dd-MM-yyyy";

public static boolean isDateValid(String date) 
{
        try {
            DateFormat df = new SimpleDateFormat(DATE_FORMAT);
            df.setLenient(false);
            df.parse(date);
            return true;
        } catch (ParseException e) {
            return false;
        }
}

Comme l'a montré @Maglob, l'approche de base consiste à tester la conversion de chaîne en date à l'aide de SimpleDateFormat.parse . Cela capturera les combinaisons jour / mois invalides telles que 2008-02-31.

Cependant, dans la pratique, cela suffit rarement, car SimpleDateFormat.parse est extrêmement libéral. Vous pourriez être concerné par deux comportements:

Caractères non valides dans la chaîne de date Étonnamment, 2008-02-2x & Quot; passera & Quot; en tant que date valide avec le format de locale = " aaaa-MM-jj " par exemple. Même quand isLenient == false.

Années: 2, 3 ou 4 chiffres? Vous voudrez peut-être aussi appliquer des années à 4 chiffres plutôt que de permettre le comportement SimpleDateFormat par défaut (qui interprétera & 12-12-31 & Quot; différemment selon que votre format était & Quot; yyyy- MM-jj & "Ou &" Yy-MM-jj & ";)

Une solution stricte avec la bibliothèque standard

Une chaîne complète jusqu'au test de date pourrait donc ressembler à ceci: une combinaison de correspondance regex, puis une conversion de date forcée. Le truc avec l'expression rationnelle est de le rendre convivial pour les paramètres régionaux. 

  Date parseDate(String maybeDate, String format, boolean lenient) {
    Date date = null;

    // test date string matches format structure using regex
    // - weed out illegal characters and enforce 4-digit year
    // - create the regex based on the local format string
    String reFormat = Pattern.compile("d+|M+").matcher(Matcher.quoteReplacement(format)).replaceAll("\\\\d{1,2}");
    reFormat = Pattern.compile("y+").matcher(reFormat).replaceAll("\\\\d{4}");
    if ( Pattern.compile(reFormat).matcher(maybeDate).matches() ) {

      // date string matches format structure, 
      // - now test it can be converted to a valid date
      SimpleDateFormat sdf = (SimpleDateFormat)DateFormat.getDateInstance();
      sdf.applyPattern(format);
      sdf.setLenient(lenient);
      try { date = sdf.parse(maybeDate); } catch (ParseException e) { }
    } 
    return date;
  } 

  // used like this:
  Date date = parseDate( "21/5/2009", "d/M/yyyy", false);

Notez que l'expression rationnelle suppose que la chaîne de format ne contient que des caractères de jour, de mois, d'année et de séparation. En plus de cela, le format peut être dans n'importe quel format de lieu: & «D / MM / aa &», & «Aaaa-MM-jj &», Etc. La chaîne de format pour les paramètres régionaux actuels peut être obtenue comme suit: 

Locale locale = Locale.getDefault();
SimpleDateFormat sdf = (SimpleDateFormat)DateFormat.getDateInstance(DateFormat.SHORT, locale );
String format = sdf.toPattern();

Joda Time - Meilleure alternative?

J'ai entendu parler du temps joda récemment et j'ai pensé comparer. Deux points:

  1. Il semble préférable d’être strict sur les caractères non valides dans la chaîne de date, contrairement à SimpleDateFormat
  2. Vous ne pouvez pas encore trouver le moyen d'appliquer des années à 4 chiffres (mais je suppose que vous pourriez créer votre propre DateTimeFormatter à cet effet)

C'est assez simple à utiliser: 

import org.joda.time.format.*;
import org.joda.time.DateTime;

org.joda.time.DateTime parseDate(String maybeDate, String format) {
  org.joda.time.DateTime date = null;
  try {
    DateTimeFormatter fmt = DateTimeFormat.forPattern(format);
    date =  fmt.parseDateTime(maybeDate);
  } catch (Exception e) { }
  return date;
}

Vous pouvez utiliser SimpleDateFormat .

Par exemple, quelque chose comme: 

boolean isLegalDate(String s) {
    SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
    sdf.setLenient(false);
    return sdf.parse(s, new ParsePosition(0)) != null;
}

tl;dr

Use the strict mode on java.time.DateTimeFormatter to parse a LocalDate. Trap for the DateTimeParseException.

LocalDate.parse(                   // Represent a date-only value, without time-of-day and without time zone.
    "31/02/2000" ,                 // Input string.
    DateTimeFormatter              // Define a formatting pattern to match your input string.
    .ofPattern ( "dd/MM/uuuu" )
    .withResolverStyle ( ResolverStyle.STRICT )  // Specify leniency in tolerating questionable inputs.
)

After parsing, you might check for reasonable value. For example, a birth date within last one hundred years.

birthDate.isAfter( LocalDate.now().minusYears( 100 ) )

Avoid legacy date-time classes

Avoid using the troublesome old date-time classes shipped with the earliest versions of Java. Now supplanted by the java.time classes.

LocalDate & DateTimeFormatter & ResolverStyle

The LocalDate class represents a date-only value without time-of-day and without time zone.

String input = "31/02/2000";
DateTimeFormatter f = DateTimeFormatter.ofPattern ( "dd/MM/uuuu" );
try {
    LocalDate ld = LocalDate.parse ( input , f );
    System.out.println ( "ld: " + ld );
} catch ( DateTimeParseException e ) {
    System.out.println ( "ERROR: " + e );
}

The java.time.DateTimeFormatter class can be set to parse strings with any of three leniency modes defined in the ResolverStyle enum. We insert a line into the above code to try each of the modes.

f = f.withResolverStyle ( ResolverStyle.LENIENT );

The results:

  • ResolverStyle.LENIENT
    ld: 2000-03-02
  • ResolverStyle.SMART
    ld: 2000-02-29
  • ResolverStyle.STRICT
    ERROR: java.time.format.DateTimeParseException: Text '31/02/2000' could not be parsed: Invalid date 'FEBRUARY 31'

We can see that in ResolverStyle.LENIENT mode, the invalid date is moved forward an equivalent number of days. In ResolverStyle.SMART mode (the default), a logical decision is made to keep the date within the month and going with the last possible day of the month, Feb 29 in a leap year, as there is no 31st day in that month. The ResolverStyle.STRICT mode throws an exception complaining that there is no such date.

All three of these are reasonable depending on your business problem and policies. Sounds like in your case you want the strict mode to reject the invalid date rather than adjust it.

About java.time

The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.

The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.

To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.

You may exchange java.time objects directly with your database. Use a JDBC driver compliant with JDBC 4.2 or later. No need for strings, no need for java.sql.* classes.

Where to obtain the java.time classes?

The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.

java.time

With the Date and Time API (java.time classes) built into Java 8 and later, you can use the LocalDate class.

public static boolean isDateValid(int year, int month, int day) {
    boolean dateIsValid = true;
    try {
        LocalDate.of(year, month, day);
    } catch (DateTimeException e) {
        dateIsValid = false;
    }
    return dateIsValid;
}

An alternative strict solution using the standard library is to perform the following:

1) Create a strict SimpleDateFormat using your pattern

2) Attempt to parse the user entered value using the format object

3) If successful, reformat the Date resulting from (2) using the same date format (from (1))

4) Compare the reformatted date against the original, user-entered value. If they're equal then the value entered strictly matches your pattern.

This way, you don't need to create complex regular expressions - in my case I needed to support all of SimpleDateFormat's pattern syntax, rather than be limited to certain types like just days, months and years.

Building on answer of @Pangea to fix the problem pointed out by @ceklock, I added a method to verify that the dateString doesn't contain any invalid character.

Here is how I do:

private boolean isDateCorrect(String dateString) {
    try {
        Date date = mDateFormatter.parse(dateString);
        Calendar calendar = Calendar.getInstance();
        calendar.setTime(date);
        return matchesOurDatePattern(dateString);    //added my method
    }
    catch (ParseException e) {
        return false;
    }
}

/**
 * This will check if the provided string matches our date format
 * @param dateString
 * @return true if the passed string matches format 2014-1-15 (YYYY-MM-dd)
 */
private boolean matchesDatePattern(String dateString) {
    return dateString.matches("^\\d+\\-\\d+\\-\\d+");
}

I suggest you to use org.apache.commons.validator.GenericValidator class from apache.

GenericValidator.isDate(String value, String datePattern, boolean strict);

Note: strict - Whether or not to have an exact match of the datePattern.

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;
}

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.

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;
    }
}

Two comments on the use of SimpleDateFormat.

it should be declared as a static instance if declared as static access should be synchronized as it is not thread safe

IME that is better that instantiating an instance for each parse of a date.

Above methods of date parsing are nice , i just added new check in existing methods that double check the converted date with original date using formater, so it works for almost each case as i verified. e.g. 02/29/2013 is invalid date. Given function parse the date according to current acceptable date formats. It returns true if date is not parsed successfully.

 public final boolean validateDateFormat(final String date) {
        String[] formatStrings = {"MM/dd/yyyy"};
        boolean isInvalidFormat = false;
        Date dateObj;
        for (String formatString : formatStrings) {
            try {
                SimpleDateFormat sdf = (SimpleDateFormat) DateFormat.getDateInstance();
                sdf.applyPattern(formatString);
                sdf.setLenient(false);
                dateObj = sdf.parse(date);
                System.out.println(dateObj);
                if (date.equals(sdf.format(dateObj))) {
                    isInvalidFormat = false;
                    break;
                }
            } catch (ParseException e) {
                isInvalidFormat = true;
            }
        }
        return isInvalidFormat;
    }

Here's what I did for Node environment using no external libraries:

Date.prototype.yyyymmdd = function() {
   var yyyy = this.getFullYear().toString();
   var mm = (this.getMonth()+1).toString(); // getMonth() is zero-based
   var dd  = this.getDate().toString();
   return zeroPad([yyyy, mm, dd].join('-'));  
};

function zeroPad(date_string) {
   var dt = date_string.split('-');
   return dt[0] + '-' + (dt[1][1]?dt[1]:"0"+dt[1][0]) + '-' + (dt[2][1]?dt[2]:"0"+dt[2][0]);
}

function isDateCorrect(in_string) {
   if (!matchesDatePattern) return false;
   in_string = zeroPad(in_string);
   try {
      var idate = new Date(in_string);
      var out_string = idate.yyyymmdd();
      return in_string == out_string;
   } catch(err) {
      return false;
   }

   function matchesDatePattern(date_string) {
      var dateFormat = /[0-9]+-[0-9]+-[0-9]+/;
      return dateFormat.test(date_string); 
   }
}

And here is how to use it:

isDateCorrect('2014-02-23')
true

// 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;
}

Here is I would check the date format:

 public static boolean checkFormat(String dateTimeString) {
    return dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}") || dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}\\s\\d{2}:\\d{2}:\\d{2}")
            || dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}T\\d{2}:\\d{2}:\\d{2}") || dateTimeString
            .matches("^\\d{4}-\\d{2}-\\d{2}T\\d{2}:\\d{2}:\\d{2}Z") ||
            dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}\\s\\d{2}:\\d{2}:\\d{2}Z");
}
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