Comments on: Regular expression challenge

By: Gareth Heyes

Gareth Heyes — Fri, 19 Oct 2007 09:44:56 +0000

Well done Thijs!

Thanks for point that error out, I’ll look into that bug.

By: Thijs Lensselink

Thijs Lensselink — Fri, 19 Oct 2007 09:36:38 +0000

Bit late as usual. But here goes:

reg exp: (from|to|via)\s(.+?(?=( ?from | ?to | ?via |$)))
Replacement: ${1}: “${2}”

By the way i got an error on submitting
Warning: join() [function.join]: Bad arguments. in /path/to/plugins/spambam/spambam.php on line 43
We don’t allow comment spam here. Javascript is required to submit a comment.

By: Gareth Heyes

Gareth Heyes — Fri, 19 Oct 2007 09:36:12 +0000

Here’s mine:-
$output = preg_replace(”/\s?\b(from|to|via)\b \b([\w-]+([,][\s][\w]+([\s][\w]+)?)?)/i”, ‘ $1: “$2″ ‘, $string);

By: Gareth Heyes

Gareth Heyes — Fri, 19 Oct 2007 00:20:15 +0000

Well done everyone who has entered so far, I shall post my solution tomorrow and let you know which one I thought was best. Some of them look at lot better than mine

I can’t offer any prizes unless I could get a sponsor only my praise

By: Peter

Peter — Fri, 19 Oct 2007 00:14:03 +0000

foreach($strings as $string) {
$output = preg_replace(”/(from|to|via) (\S*+(?: (?!from\b|to\b|via\b)\S*+)*+)/i”, ‘\1: “\2″‘, $string);
echo “{ ” . $output . ” }
”;
}

By: The-Wildcat

The-Wildcat — Thu, 18 Oct 2007 23:28:04 +0000

foreach($strings as $string) {
$output = preg_replace(”/(from|to|via)\s{1}([a-z\-]{0,})(?(?=\,)([a-z\s,]{0,16}\S|to))/i”,
‘ $1: “$2$3″ ‘, $string);
echo “{” . $output . “}
”;
}

Very bad, but it works xD
And now, i go to bed

By: Adnan Ali

Adnan Ali — Thu, 18 Oct 2007 22:54:50 +0000

Sorry for the spam, but I figured that could be simplified to:

“/(to|from|via) (.*?)(?= (?1) |$)/i”
‘\1: “\2″′

/embarrased

By: Adnan Ali

Adnan Ali — Thu, 18 Oct 2007 22:38:36 +0000

Ah, clever, comments are moderated so you see them first.

Here it is:
match:
“/((to|from|via) )(.*?)( ?)(?=(?1)|$)/i”

replace:
‘\2: “\3″\4′

By: Adnan Ali

Adnan Ali — Thu, 18 Oct 2007 22:33:31 +0000

I got it! Should I post the solution here?

By: Josh Johnston

Josh Johnston — Thu, 18 Oct 2007 21:55:07 +0000

I had to add the U modifier at the end to make this one work so there is probably something better.
Enjoy

$output = preg_replace(’/(from|via|to)\s(.+)(?=(?: from| via| to|$))/iU’, ‘$1: “$2″‘, $string);
?>