Taking the chicken way out but I can't think of the right tricky
search strings to uncover a tried and true way to output a 3 digit
padded numeric series. In this case its for file names. And needs to
roll over to 4 digit in the event there are enough files.
There is no problem of clobbering since files are being renamed as
they are moved to a new clean directory. I just can't recall how to
make my incremented counter start at 000 and go:
001.ext
002.ext
003.ext
etc.
Its just the numeric part I need a jump start on.

Sat, 11 Mar 2006, Harry Putnam wrote:

Its just the numeric part I need a jump start on.

Have you looked at sprintf yet? `perldoc -f sprintf`

That's probably the easiest way.

Alternatively, you could do some kind of silly subroutine that padded
two zeroes if $n 10, one zero if $n 100, etc. It wouldn't be that
hard to do, but I think sprintf will be much easier.

Harry Putnam wrote:
Taking the chicken way out but I can't think of the right tricky
search strings to uncover a tried and true way to output a 3 digit
padded numeric series. In this case its for file names. And needs to
roll over to 4 digit in the event there are enough files.

There is no problem of clobbering since files are being renamed as
they are moved to a new clean directory. I just can't recall how to
make my incremented counter start at 000 and go:
001.ext
002.ext
003.ext
etc.

Its just the numeric part I need a jump start on.

$ perl -le'
for my $number ( 0 4, 997 1003 ) {
my $filename = sprintf q[%03d.ext], $number;
print $filename;
}
'
000.ext
001.ext
002.ext
003.ext
004.ext
997.ext
998.ext
999.ext
1000.ext
1001.ext
1002.ext
1003.ext

John

3/11/06, Chris Devers <cdevers (AT) pobox (DOT) comwrote:
Sat, 11 Mar 2006, Harry Putnam wrote:

Its just the numeric part I need a jump start on.

Have you looked at sprintf yet? `perldoc -f sprintf`

That's probably the easiest way.

Alternatively, you could do some kind of silly subroutine that padded
two zeroes if $n 10, one zero if $n 100, etc. It wouldn't be that
hard to do, but I think sprintf will be much easier.
--

Harry Putnam wrote:
Taking the chicken way out but I can't think of the right tricky
search strings to uncover a tried and true way to output a 3 digit
padded numeric series. In this case its for file names. And needs to
roll over to 4 digit in the event there are enough files.

There is no problem of clobbering since files are being renamed as
they are moved to a new clean directory. I just can't recall how to
make my incremented counter start at 000 and go:
001.ext
002.ext
003.ext
etc.

Its just the numeric part I need a jump start on.

Believe it or not, write the number as a string.

#!/usr/bin/perl

use strict;
use warnings;

for ( '000' '100' ){
print "$_\n";
}

__END__

Robin Sheat wrote:
Sunday 12 March 2006 18:22, Shawn Corey wrote:
>Believe it or not, write the number as a string.
>for ( '000' '100' ){
>print "$_\n";
>}
Note that if you have a string such as "000", you can treat it like a
number:
my $a = '000';
$a++;
print "$a\n";

This will print '001'.

To have it go to 4 digits, well, that's as easy as you might expect:
my $a = '999';
$a++;
print "$a\n";

This will print '1000'.

This trick also works on letters:

#!/usr/bin/perl

use strict;
use warnings;

for ( 'a0' 'z9' ){
print "$_\n";
}

__END__

See `perldoc perlop` and search for "Auto-increment and Auto-decrement"

"Chas " <chas.owens (AT) gmail (DOT) comwrites:
sprintf is definitly the correct answer, but just to prove TIMTWTDI

#!/usr/bin/perl

use strict;

print zeropad(5,100), "\n";

sub zeropad {
my ($count, $n) = @_;
return substr '0' x $count . $n, -$count;
}

I probably wasn't clear enough in my request
This ouputs:

./test.pl
00100

I'm after:
001 thru 999 and then 1000

Robin Sheat <robin (AT) kallisti (DOT) net.nzwrites:
Note that if you have a string such as "000", you can treat it like a
number:
my $a = '000';
$a++;
print "$a\n";
Haaa, there it the obvious way I didn't see in Shawns post.
thanks.

"John W. Krahn" <krahnj (AT) telus (DOT) netwrites:

>Its just the numeric part I need a jump start on.
>
$ perl -le'
for my $number ( 0 4, 997 1003 ) {
my $filename = sprintf q[%03d.ext], $number;
print $filename;
}
'

Ahh. Nice, thanks. I didn't recognize the
`q[]' usage but it appears to operate the same as

"%03d,ext", $number;

Shawn Corey <shawnhcorey (AT) magma (DOT) cawrites:

[]

Believe it or not, write the number as a string.

#!/usr/bin/perl

use strict;
use warnings;

for ( '000' '100' ){
print "$_\n";
}

__END__

Thanks

This has the right output but its not very obvious how one would use
this to increment a counter and files are being renamed.

Harry Putnam wrote:
Ahh. Nice, thanks. I didn't recognize the
`q[]' usage but it appears to operate the same as

"%03d,ext", $number;

No, it would be the same as '%03d.ext' the other hand, qq[%03d.ext]
would be the same as "%03d.ext"

See `perldoc perlop` and search for "Quote and Quote-like "

Harry Putnam wrote:
This has the right output but its not very obvious how one would use
this to increment a counter and files are being renamed.

As you would any other variable that contains a string.

my $file = "$_.ext";

or

my $file = $_ . '.ext';

or

my $file = sprintf( '%s.ext', $_ );

or even

my $file = sprintf( q/%03d.ext/, $_ );

though this sort of defeats the purpose of auto-incrementing a string.

Robin Sheat <robin (AT) kallisti (DOT) net.nzwrites:

Note that if you have a string such as "000", you can treat it like a
number:
my $a = '000';
$a++;
print "$a\n";

This will print '001'.

That looks like the tidiest way to do it in this case.
Just start the count with my $a and were done.

Thanks