php date validation

Question

Im trying to to set up a php date validation (MM/DD/YYYY) but I'm having issues. Here is a sample of what I got:

$date_regex = '%\A(0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])[- /.](19|20)\d\d\z%'; 

$test_date = '03/22/2010'; 
if (preg_match($date_regex, $test_date,$_POST['birthday']) ==true) {
    $errors[] = 'user name most have no spaces';`

consider how the date is supplied in the first instance.A client side date picker will greatly reduce the change of getting wrongly formatted dates. — user557846
– user557846, Commented Aug 19, 2012 at 23:43
Just try to create a DateTime object with the supplied string. If that fails then the date was invalid — GordonM
– GordonM, Commented Jun 29, 2017 at 8:37
@user557846 You shouldn't depend on client side to ensure correct data. It's so easily bypassed it should not be considered reliable. It should only be considered an aid to the user, in that it helps them make fewer invalid form submissions. — GordonM
– GordonM, Commented Jun 29, 2017 at 8:39

Nicolás Ozimica · Accepted Answer · 2012-08-20 00:00:43Z

93

You could use checkdate. For example, something like this:

$test_date = '03/22/2010';
$test_arr  = explode('/', $test_date);
if (checkdate($test_arr[0], $test_arr[1], $test_arr[2])) {
    // valid date ...
}

A more paranoid approach, that doesn't blindly believe the input:

$test_date = '03/22/2010';
$test_arr  = explode('/', $test_date);
if (count($test_arr) == 3) {
    if (checkdate($test_arr[0], $test_arr[1], $test_arr[2])) {
        // valid date ...
    } else {
        // problem with dates ...
    }
} else {
    // problem with input ...
}

edited Aug 20, 2012 at 0:00

answered Aug 19, 2012 at 23:35

Nicolás Ozimica

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8 Comments

Cody Caughlan Over a year ago

This is the best solution as it also checks for invalid dates, e.g. Feb 29. A pure regex approach based on formatting would allow invalid dates.

Madara's Ghost Over a year ago

+1 for a regexless solution. Not all problems need a regex solution.

salathe Over a year ago

Let's come into the 21st century. DateTime::createFromFormat() and DateTime::getLastErrors(). Thank me later.

Sk446 Over a year ago

@salathe has the best approach. Nicolás should really update his answer as it's providing an old approach of doing this.

Nicolás Ozimica Over a year ago

@rolinger Why should checkdate fail with such years??? First: are they invalid years? Second: the documentation for checkdate explicitly states that: "The year is between 1 and 32767 inclusive."

|

salathe · Accepted Answer · 2012-08-19 23:59:22Z

56

You can use some methods of the DateTime class, which might be handy; namely, DateTime::createFromFormat() in conjunction with DateTime::getLastErrors().

$test_date = '03/22/2010';

$date = DateTime::createFromFormat('m/d/Y', $test_date);
$date_errors = DateTime::getLastErrors();
if ($date_errors['warning_count'] + $date_errors['error_count'] > 0) {
    $errors[] = 'Some useful error message goes here.';
}

This even allows us to see what actually caused the date parsing warnings/errors (look at the warnings and errors arrays in $date_errors).

answered Aug 19, 2012 at 23:59

salathe

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6 Comments

s3m3n Over a year ago

It's worth to mention that you can check it straight with DateTime::createFromFormat() only, by checking if it's false, so if(DateTime::createFromFormat('m/d/Y', $test_date) === false) exit('bad date format');

salathe Over a year ago

Only very crudely @s3m3n. For example, without checking for warnings, "44/33/2211" is a valid m/d/Y date (it is 2nd Sep 2214 btw).

Nitsan Baleli Over a year ago

DateTime::createFromFormat(PHP 5 >= 5.3.0)

o0'. Over a year ago

This is not always the best approach. If you receive the input already in separated values, but you don't know how exactly they are formatted (M vs. MM, YYYY vs. YY), I think checkdate would be more flexible and accept anything that resembles a valid date. Additionaly, with DateTime, if I write 31 as the month it will be considered valid, and simply add 2 years and set it to July (since 31%12 = 7). I really don't think this would be expected behaviour by many people…

Brandon Over a year ago

Important to note that @s3m3n's comment may no longer be correct. In PHP8, attempting this with a date like 3/31/2023 will NOT return false while the date is actually invalid. Using DateTime::getLastErrors(); from the answer does work as it should in this example.

|

Community · Accepted Answer · 2017-05-23 12:34:37Z

32

Though checkdate is good, this seems much concise function to validate and also you can give formats. [Source]

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

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

The extra ->format() is needed for cases where the date is invalid but createFromFormat still manages to create a DateTime object. For example:

// Gives "2016-11-10 ..." because Thursday falls on Nov 10
DateTime::createFromFormat('D M j Y', 'Thu Nov 9 2016');

// false, Nov 9 is a Wednesday
validateDate('Thu Nov 9 2016', 'D M j Y');

edited May 23, 2017 at 12:34

CommunityBot

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answered Apr 8, 2014 at 16:20

Joel

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5 Comments

Gras Double Over a year ago

Why/when is the extra verification $d->format($format) == $date needed?

jjwdesign Over a year ago

@Gras Double - Validation. He's returning true/false, but ensuring that the DateTime object was created and that the formatted DateTime equals the function input $date.

Gras Double Over a year ago

Sure, but a return (bool) $d; would do it. My question was, as the createFromFormat succeeded, in what scenario could the $d->format() be different from the input?

Gras Double Over a year ago

I have found this: validateDate('Mon, 21-Jan-2041 15:24:52 GMT', DateTime::COOKIE);. Returns false because the ->format() produces "Monday" instead of "Mon". Though, it would be better to return true, as the input is valid. I'm updating the answer.

Gras Double Over a year ago

Found out a purpose for the ->format() verification. Answer updated again.

d.raev · Accepted Answer · 2017-03-02 15:59:04Z

10

Instead of the bulky DateTime object .. just use the core date() function

function isValidDate($date, $format= 'Y-m-d'){
    return $date == date($format, strtotime($date));
}

edited Mar 2, 2017 at 15:59

answered Aug 10, 2015 at 16:28

d.raev

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2 Comments

Andrew Rump Over a year ago

Why not just test if strtotime() === false?

user10905710 Over a year ago

Testing if strtotime() === false checks the validity of the date, but not that it's in the desired format.

user2028418 · Accepted Answer · 2013-01-31 09:02:07Z

Use it:

function validate_Date($mydate,$format = 'DD-MM-YYYY') {

    if ($format == 'YYYY-MM-DD') list($year, $month, $day) = explode('-', $mydate);
    if ($format == 'YYYY/MM/DD') list($year, $month, $day) = explode('/', $mydate);
    if ($format == 'YYYY.MM.DD') list($year, $month, $day) = explode('.', $mydate);

    if ($format == 'DD-MM-YYYY') list($day, $month, $year) = explode('-', $mydate);
    if ($format == 'DD/MM/YYYY') list($day, $month, $year) = explode('/', $mydate);
    if ($format == 'DD.MM.YYYY') list($day, $month, $year) = explode('.', $mydate);

    if ($format == 'MM-DD-YYYY') list($month, $day, $year) = explode('-', $mydate);
    if ($format == 'MM/DD/YYYY') list($month, $day, $year) = explode('/', $mydate);
    if ($format == 'MM.DD.YYYY') list($month, $day, $year) = explode('.', $mydate);       

    if (is_numeric($year) && is_numeric($month) && is_numeric($day))
        return checkdate($month,$day,$year);
    return false;           
}

Jared Drake · Accepted Answer · 2012-08-19 23:46:05Z

3

REGEX should be a last resort. PHP has a few functions that will validate for you. In your case, checkdate is the best option. http://php.net/manual/en/function.checkdate.php

answered Aug 19, 2012 at 23:46

Jared Drake

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Comments

FirmView · Accepted Answer · 2012-08-19 23:52:32Z

3

Nicolas solution is best. If you want in regex,

try this,

this will validate for, 01/01/1900 through 12/31/2099 Matches invalid dates such as February 31st Accepts dashes, spaces, forward slashes and dots as date separators

(0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])[- /.](19|20)[0-9]{2}

edited Aug 19, 2012 at 23:52

answered Aug 19, 2012 at 23:40

FirmView

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1 Comment

BillyTom Over a year ago

But why would you want regex in the first place?

Saranga Mapalagama · Accepted Answer · 2020-08-20 02:27:31Z

3

This function working well,

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

answered Aug 20, 2020 at 2:27

Saranga Mapalagama

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Comments

Rich R · Accepted Answer · 2014-12-17 18:03:25Z

1

I know this is an older post, but I've developed the following function for validating a date:

function IsDateTime($aDateTime) {
    try {
        $fTime = new DateTime($aDateTime);
        $fTime->format('m/d/Y H:i:s');
        return true;
    }
    catch (Exception $e) {
        return false;
    }
}

answered Dec 17, 2014 at 18:03

Rich R

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2 Comments

Gras Double Over a year ago

@ksimka Actually yes, see answer above.

Rich R Over a year ago

The call to format will throw an exception if the date is invalid.

David Nguyen · Accepted Answer · 2017-08-12 01:43:15Z

1

Try This

/^(19[0-9]{2}|2[0-9]{3})\-(0[1-9]|1[0-2])\-(0[1-9]|1[0-9]|2[0-9]|3[0-1])((T|\s)(0[0-9]{1}|1[0-9]{1}|2[0-3]{1})\:(0[0-9]{1}|1[0-9]{1}|2[0-9]{1}|3[0-9]{1}|4[0-9]{1}|5[0-9]{1})\:(0[0-9]{1}|1[0-9]{1}|2[0-9]{1}|3[0-9]{1}|4[0-9]{1}|5[0-9]{1})((\+|\.)[\d+]{4,8})?)?$/

this regular expression valid for :

2017-01-01T00:00:00+0000
2017-01-01 00:00:00+00:00
2017-01-01T00:00:00+00:00
2017-01-01 00:00:00+0000
2017-01-01

Remember that this will be cover all case of date and date time with (-) character

edited Aug 12, 2017 at 1:43

answered Jun 29, 2017 at 8:25

David Nguyen

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1 Comment

GordonM Over a year ago

This question has already had plenty of far superior answers, what's more it's years old. Your answer adds nothing. Please don't indulge in necromancing.

Berke Cagkan Toptas · Accepted Answer · 2015-05-06 20:07:22Z

0

Not sure if this answer the question or going to help....

$dt = '6/26/1970' ; // or // '6.26.1970' ;

$dt = preg_replace("([.]+)", "/", $dt);

$test_arr  = explode('/', $dt);

if (checkdate($test_arr[0], $test_arr[1], $test_arr[2]) && preg_match("/[0-9]{1,2}\/[0-9]{1,2}\/[0-9]{4}/", $dt))

     { echo(date('Y-m-d', strtotime("$dt")) . "<br>"); }

   else

     { echo "no good...format must be in mm/dd/yyyy"; }

edited May 6, 2015 at 20:07

Berke Cagkan Toptas

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answered May 6, 2015 at 19:36

Dennis

1

Comments

Plusce · Accepted Answer · 2017-03-11 19:47:39Z

We can use simple "date" input type, like below:

Birth date: <input type="date" name="userBirthDate" /><br />

Then we can link DateTime interface with built-in function 'explode':

public function validateDate()
    {
        $validateFlag = true;
        $convertBirthDate = DateTime::createFromFormat('Y-m-d', $this->birthDate);
        $birthDateErrors = DateTime::getLastErrors();

        if ($birthDateErrors['warning_count'] + $birthDateErrors['error_count'] > 0)
        {
            $_SESSION['wrongDateFormat'] = "The date format is wrong.";
        }

        else
        {
            $testBirthDate = explode('-', $this->birthDate);
            if ($testBirthDate[0] < 1900)
            {
                $validateFlag = false;
                $_SESSION['wrongDateYear'] = "We suspect that you did not born before XX century.";
            }
        }

        return $validateFlag;
    }

I tested it on Google Chrome and IE, everything works correctly. Furthemore, Chrome display simple additional interface. If you don't write anything in input or write it in bad format (correctly is following: '1919-12-23'), you will get the first statement. If you write everything in good format, but you type wrong date (I assumed that nobody could born before XX century), your controller will send the second statement.

Polash Khan · Accepted Answer · 2019-03-11 18:02:11Z

I think it will help somebody.

function isValidDate($thedate) {
    $data = [
        'separators' => array("/", "-", "."),
        'date_array' => '',
        'day_index' => '',
        'year' => '',
        'month' => '',
        'day' => '',
        'status' => false
    ];

    // loop through to break down the date
    foreach ($data['separators'] as $separator) {
        $data['date_array'] = explode($separator, $thedate);
        if (count($data['date_array']) == 3) {
            $data['status'] = true;
            break;
        }
    }

    // err, if more than 4 character or not int
    if ($data['status']) {
        foreach ($data['date_array'] as $value) {
            if (strlen($value) > 4 || !is_numeric($value)) {
                $data['status'] = false;
                break;
            }
        }
    }

    // get the year
    if ($data['status']) {
        if (strlen($data['date_array'][0]) == 4) {
            $data['year'] = $data['date_array'][0];
            $data['day_index'] = 2;
        }elseif (strlen($data['date_array'][2]) == 4) {
            $data['year'] = $data['date_array'][2];
            $data['day_index'] = 0;
        }else {
            $data['status'] = false;
        }
    }

    // get the month
    if ($data['status']) {
        if (strlen($data['date_array'][1]) == 2) {
            $data['month'] = $data['date_array'][1];
        }else {
            $data['status'] = false;
        }
    }

    // get the day
    if ($data['status']) {
        if (strlen($data['date_array'][$data['day_index']]) == 2) {
            $data['day'] = $data['date_array'][$data['day_index']];
        }else {
            $data['status'] = false;
        }
    }

    // finally validate date
    if ($data['status']) {
        return checkdate($data['month'] , $data['day'], $data['year']);
    }

    return false;
}

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