Suzanne — Sun, 14 May 2000 00:04:00 +0000

I think that you are right -- at least that's what the Perl Cookbook asks when it tries to tell me how to do with without a hash.

*sigh*

I would be doing it that way, but I'm not up to that lesson yet. Guess it's back to the books for a moment.

Thanks very much for taking the time to drill commonsense into my head!

Suzanne

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Suzanne
Tables DeMystified

anti — Sat, 13 May 2000 23:52:00 +0000

Hi,

why don't you use a hash ?
If you setup like this:
$array{ "wathever" } = "http://wherever/here.html";
.
.
.

Then you can just use:
print "Location: ".( $array{ $ENV{QUERY_STRING} } )." \n\n";

I'm not sure that this is what you want, but it works for me.

ciao
Anti

Suzanne — Sat, 13 May 2000 22:04:00 +0000

Yes, that does help, thanks!

$lookfor in this case would be $ENV{QUERY_STRING}?

What I need to do is where you have print... in the if statement, is have a new variable set there.

So if the name is found (which it will be if they just click on the link, because the name is generated), then the item's location will be returned?

$lookfor = $array[$number][0]

I think I am coming at this the wrong way.

Hmmm.

Thanks for the response, I am going to go off and mutter to myself for awhile.

Suzanne

Rob Pengelly — Sat, 13 May 2000 21:43:00 +0000

code:

@array = ("whatever|something|someimg.gif|","whatever|something|someimg2.gif|");
$lookfor = "someimg.gif";

foreach $line (@array)
{
	if($line =~ m/$lookfor/)
	{
		print "Found\n";
	}
	else
	{
		print "Not Found\n";
	}		
}
[/code]

I am not sure if that is exactly what you wanted, but it will look for the $lookfor in each element in an array.  With the first element in the array, it outputed "Found", and for the next one, it outputed "Not Found".

Does that help at all?

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[This message has been edited by robp (edited 13 May 2000).]