Re: problems with scalar(@array)

2002-12-18 Thread Pam Derks
" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Tuesday, December 17, 2002 4:54 PM Subject: Re: problems with scalar(@array) > my final goal is to loop through each filename and count how many times each filename appears, example: > archive/summer.html 2 > arts.

Re: problems with scalar(@array)

2002-12-17 Thread Rob Dixon
close IN; } #end get_data HTH, Rob - Original Message - From: "Pam Derks" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Tuesday, December 17, 2002 4:54 PM Subject: Re: problems with scalar(@array) > my final goal is to loop through each filename

RE: problems with scalar(@array)

2002-12-17 Thread Paul Kraus
] > Subject: Re: problems with scalar(@array) > > > > my final goal is to loop through each filename and count how many > > times each filename appears, example: archive/summer.html 2 > arts.html > > 2 arttherapy.html 3. > > > > > > for right now,

Re: problems with scalar(@array)

2002-12-17 Thread tallison
> my final goal is to loop through each filename and count how many times > each filename appears, example: archive/summer.html 2 > arts.html 2 > arttherapy.html 3. > > > for right now, I'm just trying to loop through the array @url. > Next step would be to compare whats in $line with what's in $ur

Re: problems with scalar(@array)

2002-12-17 Thread Chris Ball
>> On Tue, 17 Dec 2002 11:59:46, <[EMAIL PROTECTED]> said: > Also, you might change 'scalar(@url)' to '$#url' for simplicity. More than just simplicity: void:chris~ % perl -le '@a=(5,10,15,20); print scalar @a,"\n",$#a' 4 3 $# signifies the index of the last element in an arr

Re: problems with scalar(@array)

2002-12-17 Thread Pam Derks
my final goal is to loop through each filename and count how many times each filename appears, example: archive/summer.html 2 arts.html 2 arttherapy.html 3. for right now, I'm just trying to loop through the array @url. Next step would be to compare whats in $line with what's in $url[$i] and if

Re: problems with scalar(@array)

2002-12-17 Thread tallison
> Howdy, > > why isn't $i incrementing? > >24for ($i=0; $i<$num; $i++){ >25 print ("$i = $url[$i]\n"); >26} >27 It's working perfectly! :) 21$num = scalar(@url); 22print ("num = $num\n"); output: num = 1 0 = archive

Re: problems with scalar(@array)

2002-12-17 Thread Rob Dixon
Hi Pam $i is incrementing. Your loop starts with $i at zero, as shown by your trace. The print statement executes, $i is incremented to 1, your test $i < $num is made and found to be false ($i == $num) so the loop executes. I'm not sure exactly what you're wanting. The line split(/ /, $line) will

RE: problems with scalar(@array)

2002-12-17 Thread Kipp, James
you have told i not to increment past 0 but making $num your count. as your print out shows, $num never gets past 1. I am not sure what you want $num to be? do you want it to be a count of chars in the scalar string $url ? if so, try length() > -Original Message- > From: Pam Derks [mailto