Finding string within a string with preg_match_all

[python72]

I am looking for a date within larger string, lets say the date is December 4, 2010.

To find it I use pattern and function below:

 

$Pattern='/[(January|February|March|April|May|June|July|August|September|October|November|December)] \d, \d\d\d\d/i';
preg_match_all($Pattern, $String, $Matches, PREG_OFFSET_CAPTURE, $NumberPosition);

 

The function finds the dates within the string but to my supprise the result I get in $Matches is: r 4, 2010

What I would like to get is: December 4, 2010 but don't know how it should be fixed. I thought that with the pattern I am using but obviously that is not the case.

[requinix]

[...]  means a character set. [abc] will match "a", "b", or "c"; [(a|b|c)] will match "a", "b", "c", "(", "|", and ")".

In other words, the (|)s mean nothing besides what they literally are.

 

Since you don't want a character set, remove the []s.

 

Also, what about a string

December 14, 2010

The day has two digits.

[python72]

I guess I have to follow the previous search with check for NULL in $Matches, I believe that there would be nothing stored in $Matches if string was not found so I think the if statement should look like this:

 

if ($Matches == Null){
$Pattern='/(January|February|March|April|May|June|July|August|September|October|November|December) \d\d, \d\d\d\d/i';
$Pattern='/\d, \d\d\d\d/';
preg_match_all($Pattern, $String, $Matches, PREG_OFFSET_CAPTURE, $NumberPosition);
}

 

Would this be right?

[requinix]

Would this be right?

What does the documentation say?

 

The expression now counts for two-digit days, but not for one digit. That second \d (in the first $Pattern) has to be optional.

[python72]

Did not realize that I can have optional attributes in my pattern but after reading some stuff I guess my pattern would look like this:

$Pattern='/(January|February|March|April|May|June|July|August|September|October|November|December) (\d)?\d, \d\d\d\d/i';

Please correct me if I am wrong.

[AbraCadaver]

I would probably do:

 

$Pattern='/(January|February|March|April|May|June|July|August|September|October|November|December) [\d]{1,2}, [\d]{4}/i';

[python72]

Thanks a lot guys, always learning something new:)