Regex to validate date in PHP using format as YYYY-MM-DD

Question

I am trying to make a date regex validator. The issue I\'m having is that I\'m using an input field with \"date\" type, which works like a charm in Chrome; it o

Answer 1

Do not use regex for this, you can get the same result by using DateTime::createFromFormat

// specify your date's original format, in this example m/d/Y (e.g. 08/31/2013)
$format = "m/d/Y";
$hireDate = DateTime::createFromFormat($format, $_POST['hiredate']);
if(!$hireDate) {
 // createFromFormat returns false if the format is invalid;
} else {
   //change it to any format you want with format() (e.g. 2013-08-31)
   echo $hireDate->format("Y-m-d");
}

you can read more here:

http://php.net/manual/en/datetime.createfromformat.php

However, it seems like the issue is totally unrelated to PHP.

PHP runs on the back end, and it seems like you have a front end problem.

I also doubt the problem is the input type you use. If one browser doesn't support the input type you specified, then it defaults to text. See it here:

http://jsfiddle.net/FKGCA/

My browser doesn't know what the <input type="whatever" /> is, so it defaults the input type to "text". If I wrap those 4 inputs in a <form action="myForm.php" method="POST"></form> tag, the browser sends the inputs to the server because the server doesn't care/know if the inputs were hidden, radio buttons, selects, texts, or password. The servers only receives raw-data.

More than likely, your issue is with your Javascript, and not with your PHP. Try to see if the browser that doesn't display your widget tells you that there's an error of some kind in your page.

Safari and Firefox have development/debugging tools, not so sure about IE.

Answer 2

Your regex didn't work because you had unescaped / delimiter.

The regex that would validate date in format YYYY-MM-DD as follows:

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

It will validate that the year starts with 19 or 20, that the month is not greater than 12 and doesn't equal 0 and that the day is not greater than 31 and doesn't equal 0.

Example Online

Using your initial example, you could test it like this:

$date_regex = '/^(19|20)\d\d[\-\/.](0[1-9]|1[012])[\-\/.](0[1-9]|[12][0-9]|3[01])$/';
$hiredate = '2013-14-04';

if (!preg_match($date_regex, $hiredate)) {
    echo '<br>Your hire date entry does not match the YYYY-MM-DD required format.<br>';
} else {
    echo '<br>Your date is set correctly<br>';      
}

Example Online

Answer 3

Check and validate YYYY-MM-DD date in one line statement

function isValidDate($date) {
    return preg_match("/^(\d{4})-(\d{1,2})-(\d{1,2})$/", $date, $m)
        ? checkdate(intval($m[2]), intval($m[3]), intval($m[1]))
        : false;
}

See the details in my answer here.

Don't use blindly DateTime::createFromFormat to validate dates. Let's take non-existent date 2018-02-30 and see:

$d = DateTime::createFromFormat("Y-m-d", "2018-02-30");
var_dump((bool) $d); // bool(true)

Yes, it returns true, not false as you may expected. More interesting:

$d = DateTime::createFromFormat("Y-m-d", "2018-99-99");
var_dump((bool) $d); // bool(true)

Also true... So, it validates just the number of digits. One more try:

$d = DateTime::createFromFormat("Y-m-d", "ABCD-99-99");
var_dump($d); // bool(false)

At last false.

What is going on here we can see from this snippet:

$d = DateTime::createFromFormat("Y-m-d", "2018-02-30");
var_dump($d);

// var_dump OUTPUT
object(DateTime)#1 (3) {
  ["date"]=>
  string(26) "2018-03-02 16:41:34.000000"
  ["timezone_type"]=>
  int(3)
  ["timezone"]=>
  string(3) "UTC"
}

As you can see when we pass non-existent 2018-02-30, the DateTime object contains 2018-03-02. I assume that it's because February 2018 has 28 days, i.e. the maximum date is 2018-02-28, and when we pass the day 30, createFromFormat just adds 30 days to the 2018-02-01 and create new date without any preceding date validation.