Date regex

[Samuz]

Someone tell me why this returns true when it should return false

 

$date = '1/2/2004';
if(!preg_match('/(\d{1,2})\/(\d{1,2})\/(\d{4})/', $date)) :
            echo 'false';
        else :
            echo 'true';
          
        endif;
    }

This works all fine and dandy and returns true.

 

But if I were to do something like

 

$date = '1/2/20033';

It'll return false.

 

How can I make it so that it's only limited to the (dd/mm/yyyy) format and nothing else?

[AyKay47]

Someone tell me why this returns true when it should return false

 

$date = '1/2/2004';
if(!preg_match('/(\d{1,2})\/(\d{1,2})\/(\d{4})/', $date)) :
            echo 'false';
        else :
            echo 'true';
          
        endif;
    }

This works all fine and dandy and returns true.

 

But if I were to do something like

 

$date = '1/2/20033';

It'll return false.

 

of course 1/2/20033 will return false, as it should be.

[Samuz]

Someone tell me why this returns true when it should return false

 

$date = '1/2/2004';
if(!preg_match('/(\d{1,2})\/(\d{1,2})\/(\d{4})/', $date)) :
            echo 'false';
        else :
            echo 'true';
          
        endif;
    }

This works all fine and dandy and returns true.

 

But if I were to do something like

 

$date = '1/2/20033';

It'll return false.

 

of course 1/2/20033 will return false, as it should be.

Sorry. I meant it returned true.

[AyKay47]

if you truly want it in the format of dd/mm/yyyy then it will always have two digits as the day and month (e.g 01 not 1) and 4 digits as the year. With that being said, this is what I have come up with.

 

$date = '01/02/2003';
if(!preg_match('~(?:0[1-9]|[12][0-9]|3[01])/(?:0[1-9]|1[012])/(?:19|20)\d{2}~', $date))
    echo "false";
else
echo "true";

 

now if you want it in the format of d/m/yyyy , the pattern will need adjusted.

[Samuz]

if you truly want it in the format of dd/mm/yyyy then it will always have two digits as the day and month (e.g 01 not 1) and 4 digits as the year. With that being said, this is what I have come up with.

 

$date = '01/02/2003';
if(!preg_match('~(?:0[1-9]|[12][0-9]|3[01])/(?:0[1-9]|1[012])/(?:19|20)\d{2}~', $date))
    echo "false";
else
echo "true";

 

now if you want it in the format of d/m/yyyy , the pattern will need adjusted.

Thanks for that. But i'm still getting the same problem with

 

$date = '01/02/20033';

Problem being it returning true

[AyKay47]

if you truly want it in the format of dd/mm/yyyy then it will always have two digits as the day and month (e.g 01 not 1) and 4 digits as the year. With that being said, this is what I have come up with.

 

$date = '01/02/2003';
if(!preg_match('~(?:0[1-9]|[12][0-9]|3[01])/(?:0[1-9]|1[012])/(?:19|20)\d{2}~', $date))
    echo "false";
else
echo "true";

 

now if you want it in the format of d/m/yyyy , the pattern will need adjusted.

Thanks for that. But i'm still getting the same problem with

 

$date = '1/2/20033';

 

because 1/2/2004 is not in the format of dd/mm/yyyy. the correct way to write it would be 01/02/2004

 

if the day and the month are allowed to be a single digit, which it appears you want, the pattern will look like this:

 

$date = '1/2/2003';
if(!preg_match('~(?:[1-9]|[12][0-9]|3[01])/(?:[1-9]|1[012])/(?:19|20)\d{2}~', $date))
    echo "false";
else
echo "true";

[requinix]

The expression needs ^ and $ anchors at the beginning and end, respectively.

[AyKay47]

The expression needs ^ and $ anchors at the beginning and end, respectively.

 

if the date is stand-alone, which judging solely from the example it is, then yes it will, thanks.

[Samuz]

Thanks all.

 

Solved.