php date validation

Php Problem Overview

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';`

Php Solutions

Solution 1 - Php

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 ...
}

Solution 2 - Php

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).

Solution 3 - Php

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');

Solution 4 - Php

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));
}

Solution 5 - Php

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;           
}

Solution 6 - Php

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}

Solution 7 - Php

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

Solution 8 - Php

This function working well,

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

Solution 9 - Php

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;
    }
}

Solution 10 - Php

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

Solution 11 - Php

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"; }

Solution 12 - Php

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.

Solution 13 - Php

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;
}

Content Type	Original Author	Original Content on Stackoverflow
Question	Pablo Lopez	View Question on Stackoverflow
Solution 1 - Php	Nicolás Ozimica	View Answer on Stackoverflow
Solution 2 - Php	salathe	View Answer on Stackoverflow
Solution 3 - Php	Joel	View Answer on Stackoverflow
Solution 4 - Php	d.raev	View Answer on Stackoverflow
Solution 5 - Php	user2028418	View Answer on Stackoverflow
Solution 6 - Php	FirmView	View Answer on Stackoverflow
Solution 7 - Php	Jared Drake	View Answer on Stackoverflow
Solution 8 - Php	Saranga Mapalagama	View Answer on Stackoverflow
Solution 9 - Php	Rich R	View Answer on Stackoverflow
Solution 10 - Php	David Nguyen	View Answer on Stackoverflow
Solution 11 - Php	Dennis	View Answer on Stackoverflow
Solution 12 - Php	Plusce	View Answer on Stackoverflow
Solution 13 - Php	Polash Khan	View Answer on Stackoverflow