if ( $number =~ m/^\d+$/ and $number != 1 ) {
    # do stuff
}
else {
    die "There can be only 1!"
}
[download]

m/\A
  (?:
     [1]\d+  # if it starts with a 1,
             # it has to be followed by other digits
     |
     [023456789]\d* # otherwise one digit can be fine
  )
  \z
/x
[download]

Note that this matches only non-negative integers, and 01 is matched as well (no idea if that's OK or not, spec is unclear).

Note that you need some kind of anchoring (perhaps to a word boundary), but you don't necessarily need to anchor it to start and beginning of the string.

I need to match a number other than 1.

So you started with matching 1 ? How did you do that?

Read perlre. While you're at it, read How (Not) To Ask A Question, then post the code you tried.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

Am I missing the point here or isnt it just

if($number != 1) {

}
[download]

You miss the "match number" part.

qwurx [shmem] ~ > perl -wle '$foo = "bar"; print "not 1" if $foo != 1'
Argument "bar" isn't numeric in numeric ne (!=) at -e line 1.
not 1
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--shmem

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

if ($num =~ /^[^\D1]\d+$/) {
   print "does not begin with a 1\n";
}
[download]

The key to understanding this is that you are looking for anything that is not a non-digit or a 1. When you negate the negations in the above phrase, you'll see that it means you're matching 0, 2-9.

uh, update: that doesn't quite work, hang on a sec...

if ($num =~ /^(?:[^\D1]|1\d)\d*$/) {
   print "not a 1\n";
}
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2nd update: now it works. It is instructive to see how the pattern had to be modified to exclude 1, but allow 10, 19, 1111, ...

That doesn't match any single digit number, nor will it match 10, which is not 1 either.

I need to match a number other than 1. The input can be even four or five digit. But it should allow when it is not number 1

After trying to interpret your question (what is meant by "number 1" anyway), I'll add another version of the "detection of 1 game":

 ...
 my $num1 = qr{^ (?> ([\.\d]+) ) (??{$1 == '1' ? '(?!)' : '' }) }x;
 my $txt1 = qr{^ (?> ([\.\d]+) ) (??{$1 eq '1' ? '(?!)' : '' }) }x;

 my @stuff = qw' 0101  1 123 1.1  01  222 21  1.000 ';

 print
    map $_
       . "\t as num: " . (/$num1/ ? 'ok' : '--')
       . "\t as txt: " . (/$txt1/ ? 'ok' : '--')
       . "\n",
         @stuff;
 ...
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Regards

mwa

You don't have to generate any code. The (?(COND)TRUE|FALSE) assertion fits the glove here. $^N can be used instead of $1 here, making it copy-paste safe.

I'd also like to draw some attention to Regexp::Common::number. These changes would make

my $num1 = qr{^ (?> ([\.\d]+) ) (??{$1 == '1' ? '(?!)' : '' }) }x;
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become

use Regexp::Common qw/ number /;

my $num1 = qr{^ (?> ($RE{num}{dec}) ) (?(?{ $^N != '1' })|(?!)) }x;
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I reversed the logic in the COND block so that it reads better to me. "Match a number, the number isn't 1, done".

lodin

I tried that and got this warning:

Argument "\x{d6d}\x{bef}" isn't numeric in numeric eq (==) at (re_eval
+ 2) line 1.
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:)

if ($theNumberUnderTest == 42) {
    print "Matched a number other than 1\n";
}
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Yeah, I'd have to say this answers the "question" by testing for the ultimate answer, to life, the universe, and everything; but what was the question?

if ( $number =~ /^\d*$/ and $number != 1 )
{
  # not 1 stuff here
}
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if( m{^\d{2,}$} ) {
   # do stufff
}
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my $sequence = '12345';
if( index($sequence, '1') >= 0 ) {
    # found the 1 digit. Interrupt here.
}
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Which solution did you choose, kprasanna_79? What was the original problem that stumped such an experienced Monk as yourself?

if ( $number > 1 ) {
    print "$number greater than 1.\n";
} elsif ( $number < 1 ) {
    print "$number less than 1.\n";
} else {
    print "$number is 1.\n";
}
[download]

"Sometimes a cigar is just a cigar" Freud

if ($number - 1){
    #do sometning
}
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 if ($str =~ /^(?!1$)\d+$/) {
    do_stuff($str);
 }
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