Using filter_var() to verify date?

Question

I\'m obviously not using filter_var() correctly. I need to check that the user has entered a valid date, in the form \"dd/mm/yyyy\".

This simply returns whatever I p

Answer 1

$myregex = '~^\d{2}/\d{2}/\d{4}$~';

The regex matched because you just require that pattern anywhere in the string. What you want is only that pattern and nothing else. So add ^ and $.

Note that this still doesn't mean the value is a valid date. 99/99/9999 will pass that test. I'd use:

if (!DateTime::createFromFormat('d/m/Y', $string))

Answer 2

Why using a heavy RegEx, while there is a native DateTime PHP-class?

$value = 'badbadbad';

try {
    $datetime = new \DateTime($value);

    $value = $datetime->format('Y-m-d');
} catch(\Exception $e) {
    // Invalid date
}

Always returns the format you were expecting, if a valid datetime string was inserted.

Answer 3

Using regex to validate date is a bad idea .. Imagine 99/99/9999 can easily be seen as a valid date .. you should checkdate

bool checkdate ( int $month , int $day , int $year )

Simple Usage

$date = "01/02/0000";
$date = date_parse($date); // or date_parse_from_format("d/m/Y", $date);
if (checkdate($date['month'], $date['day'], $date['year'])) {
    // Valid Date
}

Answer 4

The better solution is posted by @baba, but the code posted doesn't work fine in certain conditions.

For example if you try to validate this date "2019-12-313";

$date = date_parse("2019-12-313");

the function returns this array:

Array
(
    [year] => 2019
    [month] => 12
    [day] => 31
    [hour] => 
    [minute] => 
    [second] => 
    [fraction] => 
    [warning_count] => 0
    [warnings] => Array
        (
        )

    [error_count] => 1
    [errors] => Array
        (
            [10] => Unexpected character
        )

    [is_localtime] => 
)

so, if you perform the command:

checkdate($date['month'], $date['day'], $date['year']);

checkdate will returns a true value because the value of $date['day'] is 31 instead of 313.

to avoid this case maybe you have to add a check to the value of $date['error_count']

the @baba example updated

$date = "2019-12-313";
$date = date_parse($date); // or date_parse_from_format("d/m/Y", $date);
if ($date['error_count'] == 0 && checkdate($date['month'], $date['day'], $date['year'])) {
    // Valid Date
}

Answer 5

$myregex = "~^(19|20)\d\d[- /.](0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])$~";

print filter_var("bad 01/02/2012 bad",FILTER_VALIDATE_REGEXP,array("options"=>array("regexp"=> $myregex)));

Answer 6

You received a lot of answers that indicated the RegEx would allow for an answer of 99/99/9999. That can be solved by adding a little more to the RegEx. If you know you'll never have a date outside of the years 1900-2099... you can use a RegEx like this:

/^(0[1-9]|[12][0-9]|3[01])[/](0[1-9]|1[012])[/](19|20)\d\d$/

That will validate a date of this format: dd/mm/YYYY

Between: 01/01/1900 - 31/12/2099

If you need more years, just add them near the end. |21|22|23|24|25 (would extend it to the year 2599.)