perl string match

Hi All,

Need a bit of help in pattern matching :)  How to accomplish the
code below of the string "test" is inside a variable? (e.g. $x="test")



if ($k =~ /^test/)
{
print "1\n";
}

Kelvin wrote:
> Hi All,
>
>   Need a bit of help in pattern matching :)  How to accomplish the
> code below of the string "test" is inside a variable? (e.g. $x="test")
>
>
>
> if ($k =~ /^test/)
>  {
>   print "1\n";
>  }
my $searchString = "test";
if ( $k =~ /^$searchString/) {


}

man perlop

also check out the /o switch

Mark

Mark Clements wrote:

> man perlop
>
> also check out the /o switch

Hmmm.  'perldoc perlop' does not state whether the lack of /o has as
much of a performance impact in perl-5.8 as it had in earlier versions.
-Joe

Xref: number1.nntp.dca.giganews.com comp.lang.perl.misc:551897

Joe Smith wrote:
> Mark Clements wrote:
> 
>
>
> Hmmm.  'perldoc perlop' does not state whether the lack of /o has as
> much of a performance impact in perl-5.8 as it had in earlier versions.
>     -Joe
a quick test with 5.8 on Solaris 9 shows very little difference.

use strict;
use warnings;

use Benchmark::Timer;

my $searchExpression=shift;
my $text=shift;
my $iterations=shift;

my $timer=Benchmark::Timer->new();
$timer->start("overall");
for(my $ii=0;$ii<$iterations;$ii++){
$timer->start("iteration");
$text=~/$searchExpression/;
$timer->stop("iteration");
$timer->start("iterationwitho");
$text=~/$searchExpression/o;
$timer->stop("iterationwitho");
}
$timer->stop("overall");

$timer->report();



redwood 24170 $ perl ./timere.pl ^asdf thisissometext 10000
1 trial of overall (1.014s total)
10000 trials of iteration (100.802ms total), 10us/trial
10000 trials of iterationwitho (98.526ms total), 9us/trial

redwood 24171 $ perl ./timere.pl ^asdf thisissometext 10000
1 trial of overall (1.011s total)
10000 trials of iteration (100.544ms total), 10us/trial
10000 trials of iterationwitho (96.909ms total), 9us/trial

redwood 24172 $ perl ./timere.pl ^asdf thisissometext 10000
1 trial of overall (938.571ms total)
10000 trials of iteration (93.197ms total), 9us/trial
10000 trials of iterationwitho (89.684ms total), 8us/trial

assuming my test is correct...

Mark

if ($k =~ /^test/)
{
$x=$_;
}
kcassidy@kakelma.mine.nu (Kelvin) wrote in message news:<26b862d9.0409262250.22e46f72@posti
ng.google.com>...
> Hi All,
>
>   Need a bit of help in pattern matching :)  How to accomplish the
> code below of the string "test" is inside a variable? (e.g. $x="test")
>
>
>
> if ($k =~ /^test/)
>  {
>   print "1\n";
>  }

"Oliver S?der" <osoeder@gmx.de> wrote in message
news:45c6c8e6.0409270408.67e5eec1@posting.google.com...
> kcassidy@kakelma.mine.nu (Kelvin) wrote in message
news:<26b862d9.0409262250.22e46f72@posting.google.com>... 
$x="test") 
>
> if ($k =~ /^test/)
>   {
>    $x=$_;
>   }

Please post your reply below what you are replying to.

Can you please explain exactly what it is you think this code is doing?

Paul Lalli

On Sun, 26 Sep 2004 23:50:04 -0700, Kelvin wrote:
>   Need a bit of help in pattern matching :)  How to accomplish the
> code below of the string "test" is inside a variable? (e.g. $x="test")
>
> if ($k =~ /^test/)
>  {
>   print "1\n";
>  }

If you just need to find out if a variable is inside another variable, you
shouldn't use regular expressions.  This will do (and it's faster);

my $k = 'This is a test';
my $x = 'test';

if ( index($k, $x) >= 0 ) {
# Match
}

Eventually, you can check if 'index(...)' equals 0 if you want to check if
the string begins with $x.


--
Tore Aursand <tore@aursand.no>
"Writing modules is easy. Naming modules is hard." (Anno Siegel, on
comp.lang.perl.misc)