Re: [SLUG] Perl Question
Rick Welykochy wrote:
Peter Abbott wrote:
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box, = $1;
$name = $2;
$place = $4;
$time = $5;
$margin = $6;
$name =~ s/\.//;
}
Please write maintainable code that is easy to understand.
I'll put my money where my mouth is ...
$match = qr/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*). ... etc/;
if (m/$match/)
{
($box,$name,$place,$time,$margin) = ($1,$2,$4,$5,$6);
$name =~ s/\.//;
}
i.e. my rule of matching: grab the $placement variables in the statement,
to be sure, to be sure :)
The qr/.../ creates a pattern match, precompiled. And it can be used
multiple times in your code. Easier to maintain.
It's the one thing missing from python: ease of handling regexs. Yes, yes,
there is the 're' package, but that is an O-O implementation. ot part of
the language.
It is for that reason alone that my GLPs (grungy little programs) are still
whipped up in perl instead of python.
cheers
rickw
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Re: [SLUG] Perl Question
On Wed, Apr 02, 2008 at 03:09:44PM +1100, Peter Abbott wrote:
Can anyone with greater knowledge than myself explain this regex
behaviour.
[ ... ]
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$name =~ s/\.//;
$place = $4;
$time = $5;
$margin = $6;
}
Undoubtedly something simple that a self taught dummy has failed to
grasp.
Doing s/// for $name there resets the $1, $2, .. counters
Matt
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[SLUG] Perl Question
Can anyone with greater knowledge than myself explain this regex
behaviour.
This code segment works as intended:
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$place = $4;
$time = $5;
$margin = $6;
$name =~ s/\.//;
}
However if the $name substitution is inserted above $place like so it
fails to allocate values to $place, $time and $margin.
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$name =~ s/\.//;
$place = $4;
$time = $5;
$margin = $6;
}
Undoubtedly something simple that a self taught dummy has failed to
grasp.
Regards,
Peter.
Happy proud to be,
100% Microsoft free.
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Re: [SLUG] Perl Question
On Wed, Apr 2, 2008 at 3:09 PM, Peter Abbott [EMAIL PROTECTED] wrote:
Can anyone with greater knowledge than myself explain this regex
behaviour.
This code segment works as intended:
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$place = $4;
$time = $5;
$margin = $6;
$name =~ s/\.//;
}
However if the $name substitution is inserted above $place like so it
fails to allocate values to $place, $time and $margin.
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$name =~ s/\.//;
$place = $4;
$time = $5;
$margin = $6;
}
Undoubtedly something simple that a self taught dummy has failed to
grasp.
I suspect that the added line resets the back-references (the
$1..$6). Try moving it under the $margin assignment (i.e. until
after you are done with the back-references).
--Amos
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Re: [SLUG] Perl Question
On Wed, Apr 2, 2008 at 1:09 AM, Peter Abbott [EMAIL PROTECTED]
wrote:
[..snip..]
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$name =~ s/\.//;
$place = $4;
$time = $5;
$margin = $6;
}
Undoubtedly something simple that a self taught dummy has failed to
grasp.
Hi Peter,
I'm not too sure about this, but I believe when you do $name =~, you're
effectively starting a new regular expression which means it clears $1, $2,
$3, $4, etc (unless you use brackets inside the new regular expression).
Maybe someone else can explain it better than me? :-)
HTH!
- Gonzalo
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[SLUG] Perl Question
Thanks to all who replied. I am the wiser now. - Regards, Peter. -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Question
Peter Abbott wrote:
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$place = $4;
$time = $5;
$margin = $6;
$name =~ s/\.//;
}
correcto
However if the $name substitution is inserted above $place like so it
fails to allocate values to $place, $time and $margin.
if ( $_ =~ m/^Box(\d) ([A-Z]+(\. | ?)[A-Z]* ?[A-Z]*).*?(\d).*?(\d\d\.\d
\d).*?(\d*\.\d\d)L/) {
$box = $1;
$name = $2;
$name =~ s/\.//;
$place = $4;
$time = $5;
$margin = $6;
}
incorrecto and hard to maintain
You've also introduced a very subtle bug that I could not, at first
glance, see. The $name substitution resets $1 ... $6.
Please write maintainable code that is easy to understand.
I wouldn't want to be given the second code example buried amongst
1000+ lines of perl and then asked to fix the obscure bug.
cheers
rickw
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[SLUG] Perl question
Quick question How can I do this print This is the value .$x-SomeFunc(); as print This is the value $x-SomeFunc(); seems like the substition stops at the - on the second line Alex signature.asc Description: Digital signature -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl question
Alexander Samad wrote:
Quick question
How can I do this
print This is the value .$x-SomeFunc();
as
print This is the value $x-SomeFunc();
You can go:
print This is the value @{[$x-SomeFunc()]};
But are you sure you want to do that? ;)
Mick.
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[SLUG] Perl question
Hi All,
OK this is a perl Q not Linux but it's Sunday night. :-)
Am having probs opening several files in a perl script using indirect
filehandles.
#!/usr/bin/perl -w
# This script reads the ASF's Apache log files and extracts accesses by
specific users. These accesses are regularly emailed to the users by a
cron
script.
use strict;
use FileHandle;
my $fileUsers="users"; # I use IN for the file handle for this.
my $fh;
my $fhUser; # File handle for files.
# Here is where I open a user list and read it in. Im used to this way.
# I am used to using direct filehandles like this ie IN.
open (IN, "$fileUsers") || die "Cannot open file $fileUsers for
reading\!";
while(IN) {
($users_login, $users_name, $users_email) = split(/:/, $_);
push(@users, $users_login);
}
close (IN) || die "Cannot close file $fileUsers\!";
All the above is OK.
Now here I have probs. For each user I want to open a file with the
filename being the same as the name of the user.
I want to have this file open for the duratiion of parsing an Apache
log file and write to each one so I need a filehandle for each user
file.
If I try this below:
foreach $user (@users)
{
$fhUser = "handle".$user;
open ($shUser, " $user") || die "Cannot open file $user for
reading\!";
print $fhUser "Hi $user\n";
}
I get the error upon running of:
"Can't use string ("handlemike") as a symbol ref while "strict refs" in
use at ./process line 90."
So I tried this from the Programming Perl book, chap 7 using the
FileHandle method:
foreach $user (@users)
{
$fh = new FileHandle " $user";
$fhUser = $fh.$user;
if (defined $fhUser) {
print $fhUser "Hi $user\n";
print $fhUser $subject;
print $fhUser "Date File from IP Address\n";
}
else {
print "Cannot open file $user for writing\!";
}
}
and I get
"Can't use string ("FileHandle=GLOB(0x1202e1600)mike") as a symbol ref
while "strict refs" in use at ./process line 82."
Then all I want to do is the processing
while ()
{
Here is where I do processing and writing stuff to those open
files
print $fhUser "$words[3] $words[4] $words[6] from $words[0]\n";
etc..
}
and then close all those open files...
foreach $user (@users)
{
print "Total logins: $users{$user}\n";
$fhUser-close;
}
It's just that I have not used indirect filehandles before and the
Learning Perl book only has examples of direct filehandles.
Mike
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Re: [SLUG] Perl question
Michael Lake was once rumoured to have said:
Hi All,
OK this is a perl Q not Linux but it's Sunday night. :-)
Am having probs opening several files in a perl script using indirect
filehandles.
[Snip]
foreach $user (@users)
{
$fh = new FileHandle " $user";
$fhUser = $fh.$user;
if (defined $fhUser) {
print $fhUser "Hi $user\n";
print $fhUser $subject;
print $fhUser "Date File from IP Address\n";
}
else {
print "Cannot open file $user for writing\!";
}
}
Drop the $fhUser bit, - this section is *all* wrong. you use:
print $fh
to print to the file, you don't mangle the object rej.
But given what it looks like you're trying to do, you probably want to
keep many filehandles open at once.. [one for each user], you probably
want a construct like:
my %userfh;
foreach my $user (@users) {
$userfh[$user] = new FileHandle " $user"; # note, this is WRONG.
if (defined $userfh[$user]) {
}
}
you can then close all of these filehandles using:
foreach my $user (keys %userfh) {
$userfh[$user]-close;
}
Then all I want to do is the processing
while ()
{
Here is where I do processing and writing stuff to those open
files
print $fhUser "$words[3] $words[4] $words[6] from $words[0]\n";
etc..
}
Looks fine... 'cept that $fhUser bit should be $userfh[$user] if I'm
understanding what you're trying to do.
and then close all those open files...
foreach $user (@users)
{
print "Total logins: $users{$user}\n";
$fhUser-close;
}
More crack. $foo{blah} construcuts are valid for hashes, not arrays.
It's just that I have not used indirect filehandles before and the
Learning Perl book only has examples of direct filehandles.
You've got more than just that wrong.
I seriously suggest you brush up on Perl Operators, datatypes and
start using perldoc more.
C.
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Re: [SLUG] Perl question
Crossfire was once rumoured to have said:
But given what it looks like you're trying to do, you probably want to
keep many filehandles open at once.. [one for each user], you probably
want a construct like:
my %userfh;
foreach my $user (@users) {
$userfh[$user] = new FileHandle " $user"; # note, this is WRONG.
if (defined $userfh[$user]) {
}
}
GRAH! I'm on crack! this is what I get for answering questions before
I leave for work.
just replace all my ravings of $userfh[$user] with $userfh{$user}!
[x] is an array subscript.
{x} is a hash subscript.
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