Re: putting ";" as a replacement in the substitution.
--- Michael Alipio <[EMAIL PROTECTED]> wrote:
> It would be pointless to put () in my regexp when testing with if, unless
> I'm grouping something or I want to do something with $1.
Correct.
> if /(^\w+)\s+/
>
> But if I am assigning something, like:
>
> my $captured =~ /^(\w+)\s+/
>
> I should put it inside parenthesis.
That doesn't make sense. Since you have 'my $captured', it's a lexical
variable just being declared and it doesn't yet have a value. You could try
this:
my $captured;
if ( $some_var =~ /^(\w+)\s+/ ) {
$captured = $1;
}
if ( $captured ) { ... }
> I also noticed that $capture here will always contain the first catched
> match ($1).
No, it doesn't in your example. The only way to make that work would be to
use 'list context' and default to matching the '$_' variable:
my ($capture) = /^(\w+)\s+/;
Note that because we're using parentheses on the left side, that forces list
content and because we're using the assignment operator, '=', instead of the
binding operator, '=~', the regular expression matches against '$_' instead
of the left hand side. This is probably a bit more of an advanced usage,
though. You might use it like this:
while () {
next if /^#/; # skip comments
my ($capture) = /^(\w+)\s+/; # grab first 'word'
...
}
> The, (?:) as suggested by someone is also good when I want to avoid
> something being stored in $n... I have read about it and a lot
> more(particularly the extended regexp features) in perlre but not quite
> sure what they mean.
Yes, that's correct. Sometimes you want to group something but not capture
it to a variable:
if ( $some_var =~ /(?:foo|bar)/ ) { # matched by not captured
...
}
Hope that helps.
Cheers,
Ovid
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Re: Pattern Matching
On 1/19/07, John W. Krahn <[EMAIL PROTECTED]> wrote:
Dharshana Eswaran wrote:
> Hi All,
Hello,
> I have a string as shown below:
>
> $string =
>
"{[0]=0x53,[1]=0x65,[2]=0x63,[3]=0x75,[4]=0x72,[5]=0x69,[6]=0x74,[7]=0x79,[8]=0x43,[9]=0x6F,[10]=0x64,[11]=0x65,[12]=0x00}"
>
>
> This is stored as a string in a variable. I need to pull out only the
> numbers and store them in a array. Like:
>
> @array = (53, 65, 63, 75, 72, 69, 74, 79, 43, 6F, 64, 65, 00);
>
> I am unable to get a pattern to try pattern matching and spliting
it.
>
> How do i do this?
$ perl -le'
my $string =
"{[0]=0x53,[1]=0x65,[2]=0x63,[3]=0x75,[4]=0x72,[5]=0x69,[6]=0x74,[7]=0x79,[8]=0x43,[9]=0x6F,[10]=0x64,[11]=0x65,[12]=0x00}";
my @array = $string =~ /[[:xdigit:]]{2}/g;
print "@array";
'
53 65 63 75 72 69 74 79 43 6F 10 64 11 65 12 00
John
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Thank you John.
Thanks and Regards,
Dharshana
Re: Pattern Matching
On 1/19/07, Rob Dixon <[EMAIL PROTECTED]> wrote:
Igor Sutton wrote:
> I have an update:
>
>> my @data = $string =~ m/0x(\d{2})/g;
>
> my @data = $string =~ m/0x(\S{2}),?/g;
>
> Now I think it is right :)
my @data = $string =~ m/=0x(..)/g;
:)
Rob
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Thanks for the alternative option.
Thanks and Regards,
Dharshana
Re: Pattern Matching
On 1/19/07, Igor Sutton <[EMAIL PROTECTED]> wrote:
I have an update:
> my @data = $string =~ m/0x(\d{2})/g;
my @data = $string =~ m/0x(\S{2}),?/g;
Now I think it is right :)
--
Igor Sutton Lopes <[EMAIL PROTECTED]>
I used the above expression and it worked for me. Thanks you so much.
Thanks and Regards,
Dharshana
Trouble appending into gzipped file using Compress::Zlib (Re: compressing files into tar.gz (which module do you prefer)
Hi,
- Original Message
From: Rob Dixon <[EMAIL PROTECTED]>
To: Michael Alipio <[EMAIL PROTECTED]>
Cc: beginners@perl.org
Sent: Monday, January 22, 2007 10:46:40 AM
Subject: Re: compressing files into tar.gz (which module do you prefer)
Michael Alipio wrote:
> Hi,
>
> After parsing a log and writing it into a file, I now, have to compress it
> into tar.gz. Right now, I'm doing a search at CPAN and there where too many
> modules out there with "compress" or "archive" search keyword.
>
> What do you suggest?
> > Archive::Tar
Right now, I'm playing with Compress::Zlib.
I have a logfile wich contains logs from different devices.
Now I have a sub which extracts and separates each device's logs.
sub extractlog {
my $clientname=shift;
my $log=shift;
my $year=shift;
my $date=shift;
## create the necessary path in "/client_name/year" notation
mkpath ($clientname.'/'.$year,1);
my $datedlog = $clientname.'/'.$year.'/'.$date.'.log';
## the actual log file destination (e.g; client1/2007/2007-01-12.log)
open FH, '>>', $datedlog or die $!;
print FH $log;
close FH;
}
After the all the extraction have been finished, when I looked into a
particular client's log:
#ls -alsh
22624 -rw-r--r-- 1 root wheel22M Jan 22 11:03 2007-01-17.log
# wc 2007-01-17.log
44018 268264 23148149 2007-01-17.log
It's around 22 Mb and 44018 lines.
Now, if I change my sub to this:
sub extractlog {
my $clientname=shift;
my $log=shift;
my $year=shift;
my $date=shift;
## create the necessary path in "/client_name/year" notation
mkpath ($clientname.'/'.$year,1);
my $datedlog = $clientname.'/'.$year.'/'.$date.'.log'.'.gz';
## the actual log file destination (e.g; client1/2007/2007-01-12.log.gz)
my $gzlogfile = gzopen($datedlog, "ab9");
## append that log entry into the $datedlog.
gzlogfile->gzwrite($log);
$gzlogfile->gzclose();
}
And surprisingly, when I looked into the same client:
#ls -alsh
14352 -rw-r--r-- 1 root wheel14M Jan 22 11:13 2007-01-17.log.gz
It has grown to 14Mb. The compression is almost only 50% of the original 22Mb
size.
what's more surprising is that, when I tried to gunzip it to uncompress it:
#ls -lahs
7904 -rw-r--r-- 1 root wheel 7.7M Jan 22 11:13 2007-01-17.log
It's only 7.7Mb.. What has happened here?
#wc 2007-01-17.log
15936 96734 8066422 2007-01-17.log
Seems like I have lost 28,082 log entries for that particular client :-(
Any idea what went wrong?
I was following this example:
open(LOG,"<$log_file1")or print "cld not open $log_file1 for reading \n";
$HTML_LOG = gzopen("$log_file2", "wb") or print "cld not open
$log_file2 for writing \n";
while ()
{
$HTML_LOG->gzwrite($_)
}
close(LOG);
$HTML_LOG->gzclose();
But it differs from my case as I am not continuously reading the logfile and
'gzwriting' it.
In my case, I'm only reading the logfile, then for every line, decide which
client it belongs, then call throw it to sub extractlog which in turn writes it
into gzipped file..
> > Rob
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Re: compressing files into tar.gz (which module do you prefer)
Michael Alipio wrote: Hi, After parsing a log and writing it into a file, I now, have to compress it into tar.gz. Right now, I'm doing a search at CPAN and there where too many modules out there with "compress" or "archive" search keyword. What do you suggest? Archive::Tar Rob -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Global variable - No of hits counter
On 01/21/2007 04:55 PM, tom tom wrote:
Hi,
In my mod_perl (Authentication module). I have global
variable defined as follows
use vars qw( $SESSION_CLEANUP_COUNTER);
$SESSION_CLEANUP_COUNTER=0;
my intention is to count no of times it is getting
executed by clients (no of hits). I am incrementing it
within Authenticate method as follows.
$SESSION_CLEANUP_COUNTER ++,
But everytime it prints 1.
How can I get this going,
I am not an expert in mod_perl.
Mod_perl is good with making program loading more efficient, but my
experience has been that it's not very good a storing
constantly-changing data such as the counter you created.
Apache starts several processes to handle requests, and mod_perl is
active in each process, but each process has its own data, and you can't
guarantee that the same process that handles *this* request will handle
the *next* request.
For example, take these four processes:
USERPID PROGRAM
root5105 apache2
www-data5106 apache2
www-data5107 apache2
www-data5108 apache2
Any requests will be served by either pid 5106 or 5107 or 5108. If 5106
gets the first HTTP request, the instance of mod_perl in 5106 will set
$SESSION_CLEANUP_COUNTER in 5106, but 5107 and 5108 will be unaffected.
If the next request goes to 5107, $SESSION_CLEANUP_COUNTER will start
off with an "undef" value. If you keep quickly hitting "refresh" on the
page, you should eventually see some incrementing--but it will be uneven
because each apache2 process will have a different
$SESSION_CLEANUP_COUNTER, and the assigning of HTTP requests to apache
sub-processes seems to be semi-random.
Mod_perl is great for making static data and module procedures more
efficient; however, I think that data that *must* be updated in a linear
fashion yet shared between several apache processes should be stored
externally--in a database or a file (on ramdisk?) or in IPC.
Again, this is my €0.0002 from my limited experience with mod_perl and a
lot of deduction.
Well perhaps there was a little induction too: try this program to prove
to yourself that each mod_perl process has a different copy of the $sc
variable:
use strict;
use warnings;
our $sc;
$sc++;
print "Content-Type: text/html\n";
print "\n";
print qq{
Session Cleanup Test
This is a test: $sc .
PID: $$
};
__END__
Keep hitting refresh. So long as the PID stays the same, $sc increments
normally, but when apache assigns a new sub-process to deal with
requests, $sc restarts at the beginning (1).
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compressing files into tar.gz (which module do you prefer)
Hi, After parsing a log and writing it into a file, I now, have to compress it into tar.gz. Right now, I'm doing a search at CPAN and there where too many modules out there with "compress" or "archive" search keyword. What do you suggest? Thanks Be a PS3 game guru. Get your game face on with the latest PS3 news and previews at Yahoo! Games. http://videogames.yahoo.com/platform?platform=120121
Re: character classes vs regexp alternatives (using "( )" or "[ ]"
Rob Dixon schreef:
> Dr.Ruud:
>> Michael Alipio:
>>> $log = "date=2007-01-12 blah blah";
>>> [...]
>>> ($date) = $log =~
>>> /date=(\S+?)[\s+|,]/;
>>
>> /date=([0-9]{4}-[0-9]{2}-[0-9]{2})\b/;
>>
>> /date=([0-9]{4}(?:-[0-9]{2}){2})\b/;
>>
>> /date=([0-9-]{10})\b/;
>>
>> /date=([0-9-]+)\b/;
>>
>> (untested)
>
> Why would you not test the code you posted?
No time.
> $log =~ /date=([\d-]+)/;
>
> will do what is required here.
I prefer [0-9] to \d, and to have that \b in there. Why didn't you ask
why? :)
Your "required" is not mine, I prefer to test the full lay out of a date
(see my 1st and 2nd alternative), unless specifically asked not to.
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"Gewoon is een tijger."
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Re: putting ";" as a replacement in the substitution.
- Original Message
From: Ovid <[EMAIL PROTECTED]>
To: beginners@perl.org
Sent: Sunday, January 21, 2007 10:26:20 PM
Subject: Re: putting ";" as a replacement in the substitution.
--- Michael Alipio <[EMAIL PROTECTED]> wrote:
> No, not correct. The regular expression is what's being matched. Period.
> The capturing parentheses merely capture some of all of the regular
> expression into a 'dollar digit' variable ($1, $2, and so on).
> So for this:
>$var =~ s/foo(bar)/$1/;
> The 'foo(bar)' is what is being matched and the 'bar' is captured to the $1
> variable. For this:
>if ( $var =~ /foo(bar)/ ) { ... }
> The 'foo(bar)' is *still* what is being matched and the 'bar' is *still* what
> is being captured to the $1 variable.
> The first version is when you want to alter the string you're matching. The
> second version is good when you want to take action based upon a match and
> possibly extract data out of the string.
I see.. Now it's a lot more clearer.
It would be pointless to put () in my regexp when testing with if, unless I'm
grouping something or I want to do something with $1.
if /(^\w+)\s+/
But if I am assigning something, like:
my $captured =~ /^(\w+)\s+/
I should put it inside parenthesis.
I also noticed that $capture here will always contain the first catched match
($1).
The, (?:) as suggested by someone is also good when I want to avoid something
being stored in $n... I have read about it and a lot more(particularly the
extended regexp features) in perlre but not quite sure what they mean. The
topic on backtracking when using quantifier is also a good read.
Anyway, thanks for your help!
Have a nice day!
> Cheers,
> Ovid
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Global variable - No of hits counter
Hi, In my mod_perl (Authentication module). I have global variable defined as follows use vars qw( $SESSION_CLEANUP_COUNTER); $SESSION_CLEANUP_COUNTER=0; my intention is to count no of times it is getting executed by clients (no of hits). I am incrementing it within Authenticate method as follows. $SESSION_CLEANUP_COUNTER ++, But everytime it prints 1. How can I get this going, Looking for earth-friendly autos? Browse Top Cars by "Green Rating" at Yahoo! Autos' Green Center. http://autos.yahoo.com/green_center/ -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: character classes vs regexp alternatives (using "( )" or "[ ]"
On 1/20/07, Michael Alipio <[EMAIL PROTECTED]> wrote: Hi, I'm a bit confused here: I have a regexp: ($date) = $log =~ /date=(\S+?)[\s+|,]/; so if I have: "date=2007-01-12 blah blah" or "date=2007-01-12,blah,blah" I was able to retrieve "2007-01-12" However, just recently after reading my notes on perl, I read that I should use parenthesis on regexp alternatives. so this: [\s+|,] should be written as (\s+|) But if I use that parenthesis my regexp: ($date) = $log =~ /date=(\S+?)[\s+|,]/; As I remember my regexp notes, If I want to match something, I will have to put it inside parenthesis. And if I change those "[ ]" into "( )" I'm afraid that I might also match those \s+ or "," Can you shed some light on this? Thanks. There have been some regex solutions posted already, but in the general case, you're right that normal parenthesis both group and capture. If you want non-capturing parentheses, you can use (?:). So... you could use ($date) = $log =~ /date=(\S+?)(?:\s+|,)/; See perldoc perlre for details. - Jen -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Could someone help me with this source code?
On 1/20/07, Caduceus <[EMAIL PROTECTED]> wrote: I'm trying to run this perl script called "salter" on activestates komodo. I hope to use it with Mozilla Thunderbird. I've read Learning Perl, another perl book, went to perl.com, perl.org, pm.org, and cpan.com but nothing seems to help. I will show you the script. Any help will be appreciated. What does it do that it shouldn't, or what does it fail to do that it should? Can you find a line of code that isn't doing what you think it should? Cheers! --Tom Phoenix -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: putting ";" as a replacement in the substitution.
--- Michael Alipio <[EMAIL PROTECTED]> wrote:
> I see... so in substitutions, all patterns in the left side are those that
> have to be substituted, regardless of which is enclosed in parenthesis.
Well, with a lot of hand-waving, then yes, that's basically correct.
(There are exceptions, but those are generally for advanced features you're
unlikely to encounter right now and that would be a distraction to the
current topic).
> However in a plain regexp look ups, only those inside the parenthesis are
> being matched...
No, not correct. The regular expression is what's being matched. Period.
The capturing parentheses merely capture some of all of the regular
expression into a 'dollar digit' variable ($1, $2, and so on).
So for this:
$var =~ s/foo(bar)/$1/;
The 'foo(bar)' is what is being matched and the 'bar' is captured to the $1
variable. For this:
if ( $var =~ /foo(bar)/ ) { ... }
The 'foo(bar)' is *still* what is being matched and the 'bar' is *still* what
is being captured to the $1 variable.
The first version is when you want to alter the string you're matching. The
second version is good when you want to take action based upon a match and
possibly extract data out of the string.
Cheers,
Ovid
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Re: character classes vs regexp alternatives (using "( )" or "[ ]"
Dr.Ruud wrote:
Michael Alipio schreef:
$log = "date=2007-01-12 blah blah";
[...]
($date) = $log =~
/date=(\S+?)[\s+|,]/;
/date=([0-9]{4}-[0-9]{2}-[0-9]{2})\b/;
/date=([0-9]{4}(?:-[0-9]{2}){2})\b/;
/date=([0-9-]{10})\b/;
/date=([0-9-]+)\b/;
(untested)
Why would you not test the code you posted?
$log =~ /date=([\d-]+)/;
will do what is required here.
Rob
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Re: Date and time
M. Lewis wrote:
>
> Given the following code, if I were to want $day, $month, $hour, $minute
> & $sec to have a leading zero (ie 01 for Jan rather than 1), is my only
> option to use printf? Or is there a better way.
>
> What I'm searching for here is the *correct* method to get $day, $month,
> etc for uses like naming backup files (databackup-2007-01-21.tar.gz).
>
> Thanks,
> Mike
>
>
> #!/usr/bin/perl
>
> use strict;
> use warnings;
>
> my($sec, $min, $hour, $day, $month, $year)=(localtime)[0 .. 5];
>
> print "day=$day\n";
> print "month=".($month+1)."\n";
> print "year=".($year+1900)."\n\n";
> print "hour=$hour\n";
> print "minute=$min\n";
> print "second=$sec\n\n";
Hi Mike
I'm not sure exactly what you're looking for, as your code doesn't address the
question of naming backup files.
I would write a short subroutine to 'fix' the output from localtime, and use
sprintf to build the filename. I hope this is what's wanted.
Rob
use strict;
use warnings;
sub date {
my @date = localtime;
$date[5] += 1900;
$date[4]++;
@date;
}
my $backup = sprintf 'databackup-%d-%02d-%02d.tar.gz', (date)[5,4,3];
print $backup, "\n";
**OUTPUT**
databackup-2007-01-21.tar.gz
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Re: putting ";" as a replacement in the substitution.
Michael Alipio schreef: > #I have this string: > > my $string = 'vd=root,status='; > > #Now, I want to transform it into: > > 'vd=root;status=' > > #That is replace the comma(,) between root and status with semicolon > (;); > > > $string =~ s/vd=\w+(,)/;/; > print $string,"\n"; > > #And it prints: > > ;status= > > Can you tell me why it has ate up vd= as well? > And how to get around with it.. Because the "vd=\w+" is part of the search-string, it will be replaced. The () around the comma don't do what you expect. You can't make the "vd=\w+" a zero-width positive look-behind assertion yet (see `perldoc perlre`) because the length of \w+ is not fixed. So you need to capture that part: $string =~ s/\b(vd=\w+),/$1;/; -- Affijn, Ruud "Gewoon is een tijger." -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: putting ";" as a replacement in the substitution.
Michael Alipio am Sonntag, 21. Januar 2007 13:07: > D. Bolliger <[EMAIL PROTECTED]> > > Because everything matched - that is: vd=\w+(,) - is replaced with the > > semicolon. > > > > You seem to misunderstand the meaning of the capturing parenthesis '()' > > on the left part of the substitution: They do not indicate the part of > > the string that is to be replaced; replaced is what the left side of the > > substitution matches. > > I see... so in substitutions, all patterns in the left side are those that > have to be substituted, regardless of which is enclosed in parenthesis. Yes (and the whole left side constitutes *the* search pattern). > However in a plain regexp look ups, only those inside the parenthesis are > being matched... [snipped] No, only those inside the parenthesis are being *catched* into the $1..$n variables. For the catching taking place, the *whole* regex must match, and what's catched is/are part(s) of what matched. Dani -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Date and time
"M. Lewis" schreef:
> if I were to want $day, $month, $hour,
> $minute & $sec to have a leading zero (ie 01 for Jan rather than 1),
> is my only option to use printf? Or is there a better way.
>
> What I'm searching for here is the *correct* method to get $day,
> $month, etc for uses like naming backup files
> (databackup-2007-01-21.tar.gz).
(s)printf is fine:
my $i = 0;
my $fname = sprintf "databackup-%04d-%02d-%02d.tar.gz",
map {$_ + (1900,1,0)[$i++]} (localtime)[5,4,3];
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"Gewoon is een tijger."
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Re: putting ";" as a replacement in the substitution.
- Original Message From: D. Bolliger <[EMAIL PROTECTED]> To: beginners@perl.org Sent: Sunday, January 21, 2007 7:43:19 PM Subject: Re: putting ";" as a replacement in the substitution. > Because everything matched - that is: vd=\w+(,) - is replaced with the > semicolon. > You seem to misunderstand the meaning of the capturing parenthesis '()' on > the > left part of the substitution: They do not indicate the part of the string > that is to be replaced; replaced is what the left side of the substitution > matches. I see... so in substitutions, all patterns in the left side are those that have to be substituted, regardless of which is enclosed in parenthesis. However in a plain regexp look ups, only those inside the parenthesis are being matched... I tried explaining my experiment here. http://www.mail-archive.com/beginners%40perl.org/msg82761.html > And how to get around with it.. One way is: $string =~ s/(vd=\w+),/$1;/; It did work, as someone has already suggested. Thanks. Now, it's a lot clearer. > There are several man pages, where also the capturing parenthesis and the $1.. > $n variables are explained: > perldoc perlre > perldoc perlretut > perldoc perlrequick > Hope this helps! Dani -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ Looking for earth-friendly autos? Browse Top Cars by "Green Rating" at Yahoo! Autos' Green Center. http://autos.yahoo.com/green_center/
Re: character classes vs regexp alternatives (using "( )" or "[ ]"
Michael Alipio schreef:
> $log = "date=2007-01-12 blah blah";
> [...]
> ($date) = $log =~
> /date=(\S+?)[\s+|,]/;
/date=([0-9]{4}-[0-9]{2}-[0-9]{2})\b/;
/date=([0-9]{4}(?:-[0-9]{2}){2})\b/;
/date=([0-9-]{10})\b/;
/date=([0-9-]+)\b/;
(untested)
--
Affijn, Ruud
"Gewoon is een tijger."
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Re: maximum file size for while() loop?
David Moreno Garza am Sonntag, 21. Januar 2007 07:50:
> On Sat, 2007-01-20 at 09:31 +1100, Ken Foskey wrote:
> > > What's exactly the difference between:
> > > ++$lines and $lines++; ?
> >
> > Nothing in this context.
>
> What about other contexts?
Hi David
#!/usr/bin/perl
use strict;
use warnings;
{ # preincrement
my (%h, $i);
$h{++$i}='hi';
print keys %h, ", $i\n";
}
{ # postincrement
my (%h, $i);
$h{$i++}='hi';
print keys %h, ", $i\n";
}
__END__
1, 1
0, 1
The difference is the order of "read current value" (used as hash key value)
and "increment current value" (done by ++ operator).
There's no difference between "standalone" ++$lines and $lines++ because only
increment takes place, and the result value is not used in the same
expression.
See also perldoc perlop, "Auto-increment and Auto-decrement".
Hope this helps!
Dani
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Re: putting ";" as a replacement in the substitution.
Michael Alipio am Sonntag, 21. Januar 2007 04:08: > Hi, Hi Michael > my $string = 'vd=root,status='; > > #Now, I want to transform it into: > 'vd=root;status=' > #That is replace the comma(,) between root and status with semicolon (;); > > $string =~ s/vd=\w+(,)/;/; > print $string,"\n"; > > #And it prints: > > ;status= > > Can you tell me why it has ate up vd= as well? Because everything matched - that is: vd=\w+(,) - is replaced with the semicolon. You seem to misunderstand the meaning of the capturing parenthesis '()' on the left part of the substitution: They do not indicate the part of the string that is to be replaced; replaced is what the left side of the substitution matches. > And how to get around with it.. One way is: $string =~ s/(vd=\w+),/$1;/; There are several man pages, where also the capturing parenthesis and the $1.. $n variables are explained: perldoc perlre perldoc perlretut perldoc perlrequick Hope this helps! Dani -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Regexp in lookup differs in regexp in substitution (Re: putting ";" as a replacement in the substitution.)
- Original Message From: Michael Alipio <[EMAIL PROTECTED]> To: [EMAIL PROTECTED] Cc: begginers perl.org Sent: Sunday, January 21, 2007 7:11:47 PM Subject: Regexp in loopup differs in regexp in substitution (Re: putting ";" as a replacement in the substitution.) - Original Message From: Ovid <[EMAIL PROTECTED]> To: Michael Alipio <[EMAIL PROTECTED]> Sent: Sunday, January 21, 2007 4:49:41 PM Subject: Re: putting ";" as a replacement in the substitution. --- Michael Alipio <[EMAIL PROTECTED]> wrote: > Hi, > > #I have this string: > > my $string = 'vd=root,status='; > > #Now, I want to transform it into: > > 'vd=root;status=' > > #That is replace the comma(,) between root and status with semicolon > (;); > > $string =~ s/vd=\w+(,)/;/; > print $string,"\n"; > > #And it prints: > > ;status= > > Can you tell me why it has ate up vd= as well? > And how to get around with it.. > Try: > $string =~ s/(vd=\w+),/$1;/; > The parentheses are for capturing a match and the 'dollar digit' > variables are for accessing the results of whatever the parentheses > captured. However, when using a substitution (s///), the left hand > side is the regex to match text and the right hand side is what you > replace the entire *match* with. > So your substitution "vd=\w+(,)" was matching "vd=root,", capturing the > comma, ignoring said capture (since you weren't using $1), and then > replacing the entire match with ';'. It did work but this is quite confusing. As far as I can understand, parenthesis in regexp are used to indicate whatever you are looking for. and if I am going to do something like: s/(vd=\w+),/$1;/; $1 should contain only "vd=\w+" and not including the ",". So as I understand it, the expression above will result to replacing "vd=\w+" with "vd=root;," my $string = 'devid=234FG,vd=root,status=ok,logid=235'; print "There is an alphanumeric word at the beginning that is followed by a =\n" if $string =~ /^(\w+)=/; print "My first match contains $1\n"; print "My entire match contains $&\n"; See, in my first match, given the regexp(\w+) which is surrounded by ( ), it captures "devid" And goes without saying the entire match consist of "devid" and "=". Now, in substitutions: my $string = 'devid=234FG,vd=root,status=ok,logid=235'; $string =~ s/(vd=\w+),/$1;/; print "My first match is $1\n"; print "My entire match is $&\n"; How come my $1 which contains only "vd=root" when replaced with "vd=root;", the comma in the regexp pattern was also included in $1?? > See 'perldoc perlre' for more information. > Cheers, > Ovid -- Buy the book -- http://www.oreilly.com/catalog/perlhks/ Perl and CGI -- http://users.easystreet.com/ovid/cgi_course/ The fish are biting. Get more visitors on your site using Yahoo! Search Marketing. http://searchmarketing.yahoo.com/arp/sponsoredsearch_v2.php Do you Yahoo!? Everyone is raving about the all-new Yahoo! Mail beta. http://new.mail.yahoo.com
Regexp in loopup differs in regexp in substitution (Re: putting ";" as a replacement in the substitution.)
- Original Message From: Ovid <[EMAIL PROTECTED]> To: Michael Alipio <[EMAIL PROTECTED]> Sent: Sunday, January 21, 2007 4:49:41 PM Subject: Re: putting ";" as a replacement in the substitution. --- Michael Alipio <[EMAIL PROTECTED]> wrote: > Hi, > > #I have this string: > > my $string = 'vd=root,status='; > > #Now, I want to transform it into: > > 'vd=root;status=' > > #That is replace the comma(,) between root and status with semicolon > (;); > > $string =~ s/vd=\w+(,)/;/; > print $string,"\n"; > > #And it prints: > > ;status= > > Can you tell me why it has ate up vd= as well? > And how to get around with it.. > Try: > $string =~ s/(vd=\w+),/$1;/; > The parentheses are for capturing a match and the 'dollar digit' > variables are for accessing the results of whatever the parentheses > captured. However, when using a substitution (s///), the left hand > side is the regex to match text and the right hand side is what you > replace the entire *match* with. > So your substitution "vd=\w+(,)" was matching "vd=root,", capturing the > comma, ignoring said capture (since you weren't using $1), and then > replacing the entire match with ';'. It did work but this is quite confusing. As far as I can understand, parenthesis in regexp are used to indicate whatever you are looking for. and if I am going to do something like: s/(vd=\w+),/$1;/; $1 should contain only "vd=\w+" and not including the ",". So as I understand it, the expression above will result to replacing "vd=\w+" with "vd=root;," my $string = 'devid=234FG,vd=root,status=ok,logid=235'; print "There is an alphanumeric word at the beginning that is followed by a =\n" if $string =~ /^(\w+)=/; print "My first match contains $1\n"; print "My entire match contains $&\n"; See, in my first match, given the regexp(\w+) which is surrounded by ( ), it captures "devid" And goes without saying the entire match consist of "devid" and "=". Now, in substitutions: my $string = 'devid=234FG,vd=root,status=ok,logid=235'; $string =~ s/(vd=\w+),/$1;/; print "My first match is $1\n"; print "My entire match is $&\n"; How come my $1 which contains only "vd=root" when replaced with "vd=root;", the comma in the regexp pattern was also included in $1?? > See 'perldoc perlre' for more information. > Cheers, > Ovid -- Buy the book -- http://www.oreilly.com/catalog/perlhks/ Perl and CGI -- http://users.easystreet.com/ovid/cgi_course/ The fish are biting. Get more visitors on your site using Yahoo! Search Marketing. http://searchmarketing.yahoo.com/arp/sponsoredsearch_v2.php
Re: Date and time
M. Lewis wrote: > > Given the following code, if I were to want $day, $month, $hour, > $minute & $sec to have a leading zero (ie 01 for Jan rather than 1), > is my only option to use printf? Or is there a better way. > > What I'm searching for here is the *correct* method to get $day, > $month, etc for uses like naming backup files > (databackup-2007-01-21.tar.gz). > > Thanks, > Mike > > > #!/usr/bin/perl > > use strict; > use warnings; > > my($sec, $min, $hour, $day, $month, $year)=(localtime)[0 .. 5]; > > print "day=$day\n"; > print "month=".($month+1)."\n"; > print "year=".($year+1900)."\n\n"; > print "hour=$hour\n"; > print "minute=$min\n"; > print "second=$sec\n\n"; > > use POSIX 'strftime'; my $dd = strftime "%Y-%m-%d",localtime; my $backup = "databackup-.$dd."tar.gz""; goksie -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Could someone help me with this source code?
Hi:
I'm trying to run this perl script called "salter" on activestates komodo.
I hope to use it with Mozilla Thunderbird. I've read Learning Perl, another
perl book, went to perl.com, perl.org, pm.org, and cpan.com but nothing
seems to help. I will show you the script. Any help will be appreciated.
TIA Steve
--
#!/usr/bin/perl -w
# Salter single-threaded email address salter
# (c) 2003, 2004 Julian Haight, http://www.julianhaight.com/
# All Rights Reserved under GPL:http://www.gnu.org/licenses/gpl.txt
# Current version available here: http://www.julianhaight.com/salter
# Version history
# 7/19/04 V1.2
# added stripsender feature
# fixed missing newline between header & body
# 3/26/04 V1.1
# cleaned up smtp sending code, added envonly mode, added version
# 3/12/04
# give each recipient their own, permanent random virtual sender
# move config to user-dir, not /etc.
# 9/29/03 - changed to use only lowercase-alpha, avoid spam filters
# Also, added final response after quit (worked without for pine, but not
moz)
use strict; use Socket; use FileHandle; use Digest::MD5;
my($CONFIG) = ($ENV{HOME} . '/.salter');
my($MAPFN) = "$CONFIG/map.txt";
my($EOL) = "\015\012";
my($debug) = 0;
my($SMTPTO) = 10; # 10 second timeout
my($VERSION) = 'V1.2';
my($SAMP) = '
# here is a sample config file:
listenport 2525
listenip 127.0.0.1
sendport 25
sendip your_isps_mailserver.example.com
maxclient 5
# 1 for unsafe but fast!, 0 for slow & steady (not yet available)
buffermode 1
# 1 remaps only envelope, not header, good if you want to filter bad bounces
envonly 0
# 1 strips sender field (for pine or whatever)
stripsender 1
# From this address To random @ this domain!
# -
remap [EMAIL PROTECTED] salty.you.example.com
remap [EMAIL PROTECTED] foo.example.com
# to set your identity per-recipient (email or part)
# - use workplace address for work recipients
hardwireworkplace.example.com [EMAIL PROTECTED]
# - use mailing list subscription address when posting to list.
hardwire[EMAIL PROTECTED][EMAIL PROTECTED]
# end sample config!
';
my(%config, %remap, %map, %hardwire);
unless (-e $CONFIG) { mkdir($CONFIG); }
readConfig(); # read the config file into %config
readMap();
listenLoop(); # work 'til you die!
exit 0;
# listen for one connection at a time, and call the proxy for each one.
# die if there are errors
sub listenLoop {
my($cliaddr, $cliip, $cliport);
socket(SOCK, PF_INET, SOCK_STREAM, getprotobyname('tcp')) ||
die "Socket: $!";
setsockopt(SOCK, SOL_SOCKET, SO_REUSEADDR, pack('l', 1)) ||
die "Setsockopt: $!";
bind(SOCK, sockaddr_in($config{'listenport'},
inet_aton($config{'listenip'}))) ||
die "bind: $!";
listen(SOCK, $config{'maxclient'}) ||
die "listen: $!";
while ($cliaddr = accept(CLI, SOCK)) {
# print STDERR "got connection\n";
($cliport, $cliip) = (sockaddr_in($cliaddr));
CLI->autoflush(1);
if ($_ = proxyIt(\*CLI)) {
print STDERR "<< 550 Proxy error: $_\n";
print CLI "550 Proxy error: $_\n";
}
close CLI;
}
}
sub proxyIt {
my($CLI) = @_;
my($cmds, $head, $body, $cmd);
$cmds = '';
unless ($config{buffered}) {
print $CLI "500 No safe delivery mode yet, sorry!$EOL";
close($CLI);
die "No safe mode yet, sorry!";
}
# read smtp
print $CLI "220 localhost SMTP pretender: salter $VERSION $EOL";
while ($cmd = <$CLI>) {
$cmds .= $cmd || '';
if (lc($cmd) eq "data$EOL") { last; }
if (lc(substr($cmd, 0, 4)) eq 'ehlo') {
print $CLI "451 EHLO is soo complicated$EOL";
} else {
print $CLI "250 Buffering$EOL";
}
}
print $CLI "354 Ready for data$EOL";
# read head
while ($cmd = <$CLI>) {
if ($cmd eq $EOL) { last; }
if ((!$config{stripsender}) || ($cmd !~ m/^sender:/i)) {
$head .= $cmd;
}
}
# read body
while ($cmd = <$CLI>) {
if ($cmd eq ".$EOL") { last; }
$body .= $cmd;
}
while ($CLI && print $CLI "250 Buffering$EOL") {
$cmd = <$CLI>;
$cmds .= $cmd;
if (lc($cmd) eq "quit$EOL") { last; }
}
print $CLI "221 Bye bye, hopefully it'll work!$EOL";
close $CLI;
deliverAll($cmds, $head, $body);
}
sub deliverAll {
my($cmds, $head, $body) = @_;
my($recipmap, $message, $line, $remap, $recip, $sender,
$sremap, $cmd, $val, $S,
@recips, $from);
#print STDERR "Deliverall:\n$cmds\n==\n$head\n--\n$body\n++\n";
while ($cmds =~ m/([^:\n]*): ?\\n]*[^\s\>])?\>?/g) {
$cmd = lc($1); $val = $2;
# print "cmd: $cmd = $val\n";
if ($cmd eq 'mail from') {
$sender = $val;
} elsif ($cmd eq 'rcpt to') {
$recip = $val;
$remap = getRecipMapping($recip);
# print STDERR "remap $recip to $remap\n";
push(@{$recipmap->{$remap}}, $recip);
}
}
#print STDERR "Done w/commands\n";
while ($_ = smtpOpen(*S
Re: maximum file size for while() loop?
On Sat, 2007-01-20 at 09:31 +1100, Ken Foskey wrote: > > What's exactly the difference between: > > > > ++$lines; > > > > and > > > > $lines++; ? > > > Nothing in this context. What about other contexts? David. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Date and time
On Sun, 21 Jan 2007 02:17:08 -0500 "M. Lewis" <[EMAIL PROTECTED]> wrote: > > Given the following code, if I were to want $day, $month, $hour, $minute > & $sec to have a leading zero (ie 01 for Jan rather than 1), is my only > option to use printf? Or is there a better way. Look at sprintf perldoc -f sprintf Owen -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Date and time
M. Lewis wrote: > > Given the following code, if I were to want $day, $month, $hour, $minute > & $sec to have a leading zero (ie 01 for Jan rather than 1), is my only > option to use printf? Or is there a better way. > > What I'm searching for here is the *correct* method to get $day, $month, > etc for uses like naming backup files (databackup-2007-01-21.tar.gz). > > > #!/usr/bin/perl > > use strict; > use warnings; > > my($sec, $min, $hour, $day, $month, $year)=(localtime)[0 .. 5]; > > print "day=$day\n"; > print "month=".($month+1)."\n"; > print "year=".($year+1900)."\n\n"; > print "hour=$hour\n"; > print "minute=$min\n"; > print "second=$sec\n\n"; use POSIX 'strftime'; print strftime "day=%d\nmonth=%m\nyear=%Y\n\nhour=%H\nminute=%M\nsecond=%S\n\n", localtime; John -- Perl isn't a toolbox, but a small machine shop where you can special-order certain sorts of tools at low cost and in short order. -- Larry Wall -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
