PHP DateTime CreateFromFormat Issue

Question

function validateDate($date, $format = 'm-Y') {
    $d = DateTime::createFromFormat($format, $date);
    return $d && $d->format($format) == $date;
}


validateDate('09-2017', 'm-Y');

^{function was copied from this answer or php.net}

I'm very puzzled to why this returns false while it returns true for the previous months. Any ideas?

Answer 1

It's because you're not supplying a day, so it's using the current day by default. Current day is 31, but September only has 30 days, so it skips to October 1st.

Look at this example:

function validateDate($date, $format = 'm-Y') {
    $d = DateTime::createFromFormat($format, $date);
    echo $d->format("d-".$format); // added the day for debugging
    return $d && $d->format($format) == $date;
}
var_dump(validateDate('08-2017', 'm-Y')); // 31-08-2017, true
var_dump(validateDate('09-2017', 'm-Y')); // 01-10-2017, there's no 31-09-2017, false

^{function was copied from this answer or php.net}

This is a little rudimentary, but you can detect if there's no d in the format and manually set it to 1 to avoid this:

<?php
function validateDate($date, $format = 'm-Y') {
    if (strpos($format, "d") === false) {
        $format = "d ".$format;
        $date = "01 ".$date;
    }
    $d = DateTime::createFromFormat($format, $date);
    return $d && $d->format($format) === $date;
}
var_dump(validateDate('08-2017', 'm-Y')); // 31-08-2017, true
var_dump(validateDate('09-2017', 'm-Y')); // 01-09-2017, true

Answer 2

If you are not supplying a day. current one is used; just like explained in previous answer. But this can be quickly solved with one format sign: ! , which resets all fields (year, month, day, hour, minute, second, fraction and timezone information) to the Unix Epoch; see format table .

Fix:

function validateDate($date, $format = 'm-Y') {
    $d = DateTime::createFromFormat('!'.$format, $date);
    return $d && $d->format($format) == $date;
}

^{function was copied from this answer or php.net}