正则表达式验证不解析字符串，如果PHP

Question

Here is the deal...

I am suppose to parse a Canadian Postal Code (CapsLetterNumberCapsLetterNumberCapsLetterNumber: exemple A1B2C3 or G2V3V4) IF there is one.

I have this code (PHP):

//Create new SESSION variable to store a warning
$_SESSION['msg'] = "";
//IF empty do nothing, IF NOT empty parse, IF NOT match regex put message in msg
if(!preg_match('^([A-Z][0-9][A-Z][0-9][A-Z][0-9])?$^', $_POST['txtPostalCode']) && $_POST['txtPostalCode'] != "")
{
    $_SESSION['msg'] .= "Warning invalide Postal Code";
}

then the code goes on to display $_SESSION['msg'] .

The problem is that whatever I enter in $_POST['txtPostalCode'] it NEVER get parse by the REGEX.

Answer 1

You made the entire capturing group optional:

^([A-Z][0-9][A-Z][0-9][A-Z][0-9])?$^
                                 ^

It's also not a good idea to use regex metadata characters as your delimiter. Try this regex, which matches an uppercase letter and a number three times:

/^((?:[A-Z][0-9]){3})$/

You don't need to make the capturing group optional because you handle the logic for when the user doesn't submit a code with the && $_POST['txtPostalCode'] != "" part of the if statement.

Finally, since you're not even using the matches from this regex, you don't need the capturing group:

/^(?:[A-Z][0-9]){3}$/

Answer 2

Your regex will match invalid postal codes.

A quick Google search for "canadian postal code regex" bought up

^[ABCEGHJKLMNPRSTVXY]{1}\d{1}[A-Z]{1} *\d{1}[A-Z]{1}\d{1}$

You may also want to put your $_POST['txtPostalCode'] != "" condition first since there's no point in executing a regex if the value is empty to begin with.

Edit: As pointed out by the comments, the quantifiers are redundant:

^[ABCEGHJKLMNPRSTVXY]\d[A-Z] *\d[A-Z]\d$