regular expression newbee. plz help!!

Paul_help_

Hi.

I need some help with the following code:

$myPets = “favoritePet=Lucky, Rover=dog, Lucky=cat”;

preg_match( ‘/favoritePet\=(\w+).*\1\=(\w+)/’, $myPets, $matches );

echo “My favorite pet is a “ . $matches[2] . “ called “ . $matches[1] . “.”;

This code displays the following output:

 “My favorite pet is a cat called Lucky.”

The thing i can't understand is how the regular expression jumps past the Rover=dog part of the sting.

could someone explain how it does this?

NogDog

http://www.php.net/manual/en/regexp.reference.back-references.php

Paul_help_

could u tell me in more detail. yourself? as ive read this link and i seem to be missing something.

Weedpacket

\1 matches whatever the first parenthesised part matched. \2 matches the second, and so on.

johanafm

While the entire pattern would match nothing in this case. Perhaps this is where the confusion arises?

Paul_help_

could someone just tell me what part of this regular expression tells it to jump past Rover=dog (i understand everything else):

‘/favoritePet\=(\w+).*\1\=(\w+)/’

bradgrafelman

".*" -- any character zero or more times.

Paul_help_

right. i understand what that little snippet means. but how does it jump past Rover=dog,??

Im guessing it does it in conjunction with something else?

bradgrafelman

Yes, it "skips" Rover=dog because "Rover" is not what was matched in the first subgroup (which is what the backreference "\1" refers to).

NogDog

/favoritePet\=([COLOR="#0000FF"]\w+[/COLOR])[COLOR="#FF0000"].*[/COLOR][COLOR="#00CC00"]\1[/COLOR]\=([COLOR="#FF8C00"]\w+[/COLOR])/

holds the favorite pet
skips over stuff until...
...it finds the favorite pet's name... (back reference)
...at which point it gets that pet's type

Paul_help_

ok. i think i get it.. so the word 'Lucky' within the sting is crucial to jumping past Rover=dog?

bradgrafelman

It's not so much that it is "jumping past" the "Rover=dog" -- that part still gets matched by the "." because the next part of the pattern is looking for "Lucky=(any word characters one or more times)" due to the backreference. Since the "." isn't surrounded by parentheses, it isn't placed in a subgroup - unlike the "any word characters one or more times" that follow "Lucky=".

Paul_help_

through the use of (\w+)........ becoz lucky is stored inside (\w+).

bradgrafelman

Paul help!;11033253 wrote:
becoz lucky is stored inside (\w+).

No; "cat" is matched by the (\w+).

EDIT: Think I see the confusion here... I thought you were referring to the second instance of (\w+). "Lucky" is indeed matched by the first subgroup... which also happens to be (\w+).

Paul_help_

but i thought the back reference \1 refer to Lucky??

bradgrafelman

Right, so in this:

\1\=(\w+)

\1 matches "Lucky"
\= matches a literal equal sign (which is apparently escaped for no reason at all)
(\w+) matches "cat"

NogDog

Paul help!;11033261 wrote:
but i thought the back reference \1 refer to Lucky??

Yes, in this example, "Lucky" was extracted from the first "(\w+)" sub-pattern, and then looked for again where the "\1" is, at which point the type assigned to that name in the second "(\w+)" sub-pattern will be saved and then available as $matches[2] in the result array.

Paul_help_

thanks guys. i'll take these notes and reexamine stuff.