Simple Regular Expression Q

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October 26th, 2009, 05:06 AM

jch09

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Simple Regular Expression Q

Hello to all community,

I am newbie to Perl & Regular Expresiions too and I was looking
for an answer to the following problem...

I am trying to read a line of the form: xxyyzzz.pdf or xx.yy.zzz.pdf etc

Originally I used $line=~ s/^.*\///; to capture the filename but
this doesn't work in the second case, when dot(s) are inside the filename.

How I can do that with a regular expression ?

many thanks,

John Ch.

October 26th, 2009, 05:17 AM

ishnid

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The regexp you've posted removes everything up to (and including) the first '/' character in the string. There aren't any '/' characters in the filenames you've posted.

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October 26th, 2009, 05:26 AM

jch09

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Regexp newbie

ok let me reformylate my question:

Assuming that I have strings of the form xx.yy.zzz.pdf

Which is the regexp to get the filenames including dots (i.e capture the "xx.yy.zzz") ?

thanks ishnid for your immediate answer!

October 26th, 2009, 05:37 AM

ishnid

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You could do this:

Code:

my ( $no_extension ) = $filename =~ /^(.+)\./;

Or if they always end in .pdf you could just strip that from the end directly.

The main problem with capturing everything up to the last dot in your string is that you can get some filenames with double file extensions, such as "foo.tar.gz". Something to keep in mind.

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Axweildr agrees!

October 26th, 2009, 12:01 PM

shawnhcorey

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Quote:

Originally Posted by jch09

It would be better to use a module for it. Both File::Basename and File::Spec could and both are standard modules; they come with Perl.

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October 26th, 2009, 12:14 PM

ishnid

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Don't you need to expressly supply a suffix (i.e. '.pdf' in this case) for File::Basename?

October 26th, 2009, 02:13 PM

FishMonger

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Quote:

Originally Posted by ishnid

Don't you need to expressly supply a suffix (i.e. '.pdf' in this case) for File::Basename?

No, you can use a regex.

Taken from the perldoc.

Code:

If @suffixes are given each element is a pattern (either a string or a qr//) matched against the end of the $filename. The matching portion is removed and becomes the $suffix.

     # On Unix returns ("baz", "/foo/bar/", ".txt")
     fileparse("/foo/bar/baz.txt", qr/\.[^.]*/);

October 26th, 2009, 02:49 PM

shawnhcorey

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Quote:

Originally Posted by FishMonger

No, you can use a regex.

Taken from the perldoc.

Code:

If @suffixes are given each element is a pattern (either a string or a qr//) matched against the end of the $filename. The matching portion is removed and becomes the $suffix.

     # On Unix returns ("baz", "/foo/bar/", ".txt")
     fileparse("/foo/bar/baz.txt", qr/\.[^.]*/);

Wouldn't that be?

perl Code:

Original - perl Code
# On Unix returns ("baz", "/foo/bar/", ".txt") fileparse("/foo/bar/baz.txt", qr/\.[^.]*$/);

# On Unix returns ("baz", "/foo/bar/", ".txt")
fileparse("/foo/bar/baz.txt", qr/\.[^.]*$/);

October 31st, 2009, 04:49 AM

thangdd

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How to convert string to an expression in PERL

I have a quesion for my problem with using mysql and perl programming.

- data from mysql is logic expression. After query I get it's value on string format:

my $data = "$sex < 1 && $age >= 17 && $age <= 40 && ',1,2,3,' =~m/(,)$optid(,)/";

In this expression $sex,$age,$optid is variables

- How to make Perl understanding this function?
my $sex,$age,$optid;
if ($data) {
........
........
}

as

if ($sex < 1 && $age >= 17 && $age <= 40 && ',1,2,3,' =~m/(,)$optid(,)/) {
........
........
}

help me please!

#10

October 31st, 2009, 05:22 AM

Axweildr

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you would need to parse that logic expression, and break on &&

Code:

(@atoms)=split(/&&/, $data);
$if_string="if (";
while (@atoms) {
  $if_string.="($_) && ";
}
#remove the last two & and a space
$if_string=substr($if_string, 0, -3).") {";
print $if_string;

This what you're after, NOT TESTED BTW

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#11

October 31st, 2009, 05:23 AM

Axweildr

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something seems to have gone a tad arseways here, these two threads just collided???

#12

October 31st, 2009, 05:54 AM

thangdd

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How to convert string to an expression in PERL

Quote:

Originally Posted by Axweildr

you would need to parse that logic expression, and break on &&

Code:

(@atoms)=split(/&&/, $data);
$if_string="if (";
while (@atoms) {
  $if_string.="($_) && ";
}
#remove the last two & and a space
$if_string=substr($if_string, 0, -3).") {";
print $if_string;

This what you're after, NOT TESTED BTW

------------------------------------------------------------
Thank for your helped, but I want to make that perl understanding

my $data = "$sex < 1 && $age >= 17 && $age <= 40 && ',1,2,3,' =~m/(,)$optid(,)/";

my $sex,$age,$optid;
if ($data) {
........
........
}

Likes

if ($sex < 1 && $age >= 17 && $age <= 40 && ',1,2,3,' =~m/(,)$optid(,)/) {
........
........
}

Ex:

Code:

my $age=1;#wrong data for expression -> return fales
my $sex = 0;
my $optid = 100;#Wrong data for expression  -> return fales
my $data = "$sex < 1 && $age >= 17 && $age <= 40 && ',1,2,3,' =~m/(,)$optid(,)/";
if ($data) {
  print "True1";
} else {
  print "False1";
}

if ($sex < 1 && $age >= 17 && $age <= 40 && ',1,2,3,' =~m/(,)$optid(,)/) {
  print "True 2";
} else {
  print "false 2";
}

I wanted the same value printed as

false 1
false 2

But the first expression - $data is a string and is defined so the result is "True 1"
True 1
false 2

your final result is string too, not my target

#13

October 31st, 2009, 06:09 AM

Axweildr

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perl Code:

Original - perl Code
#!/usr/bin/perl $data="test2"; if ($data) { print "[$data]OK\n"; } else { print "[$data]NOK\n"; } $data="x"; if ($data) { print "[$data]OK\n"; } else { print "[$data]NOK\n"; } $data=""; if ($data) { print "[$data]OK\n"; } else { print "[$data]NOK\n"; }

#!/usr/bin/perl
$data="test2";
if ($data) {
    print "[$data]OK\n";
} else {
    print "[$data]NOK\n";
}
$data="x";
if ($data) {
    print "[$data]OK\n";
} else {
    print "[$data]NOK\n";
}
$data="";
if ($data) {
    print "[$data]OK\n";
} else {
    print "[$data]NOK\n";
}

The code above shows what's happening to your if statements, you need to eval your if statement in an if block as per below

perl Code:

Original - perl Code

my $data = '$sex < 1 && $age >= 17 && $age <= 40 && \',1,2,3,\' =~m/(,)$optid(,)/';
print $data;
print "=====================================\n";
(@atoms)=split(/&&/, $data);
$if_string="if (";
for (@atoms) {
  $if_string.="($_) && ";
}
#remove the last two & and a space
$if_string=substr($if_string, 0, -3).") {\n";
$if_string.=" print \"True2\";
 } else {
  print \"false 2\";
 }
 ";
print $if_string;
eval ($if_string);

Hope that helps
--Ax

#14

October 31st, 2009, 06:13 AM

FishMonger

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perldoc -f eval

This Q is cross posted to at least one other forum.
perlguru.com

#15

November 1st, 2009, 09:10 PM

thangdd

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How to convert string to an expression in PERL

Quote:

Originally Posted by Axweildr

perl Code:

Original - perl Code
#!/usr/bin/perl ...... print $if_string; eval ($if_string);

#!/usr/bin/perl
 ...... 
print $if_string;
eval ($if_string);

Hope that helps
--Ax

It works - thank you very much!

Simple Regular Expression Q

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