preg_replace - don't include string if $4 is blank

Posted by bradenkeith on Stack Overflow See other posts from Stack Overflow or by bradenkeith
Published on 2010-04-27T15:51:40Z Indexed on 2010/04/27 15:53 UTC
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I have this expression:

        $regex_phone = '/^(?:1(?:[. -])?)?(?:\((?=\d{3}\)))?([2-9]\d{2})' 
                .'(?:(?<=\(\d{3})\))? ?(?:(?<=\d{3})[.-])?([2-9]\d{2})' 
                .'[. -]?(\d{4})(?: (?i:ext)\.? ?(\d{1,5}))?$/';
        if(!preg_match($regex_phone, $data['phone'])){ 
           $error[] = "Please enter a valid phone number."; 
        }else{ 
           $data['phone'] = preg_replace($regex_phone, '($1) $2-$3 ext.$4', $data['phone']);
        }

That will take a phone number such as: 803-888-8888 ext 2 as well as 803-888-8888

First number formats as: (803) 888-8888 ext.2 -- the desired effect

Second number formats as: (803) 888-8888 ext. -- blank extension

How can I set it so that if $4 is blank, that ext. won't show?

Thanks so much for any help you can offer. I hope this was clear.

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