Re: increment # but keep # of digits same

There are two ways to accomplish what you want that immediately come to mind: use sprintf() to ensure the proper number of digits and use the magic increment. The first way is simply $newbuildnum = sprintf("%05d", $oldbuildnum+1) Adjust the 5 as needed. %05d is a format specifier that says output an integer that has a width of 5 characters and is 0 padded.

The second method is kind of sneaky. When you have a variable that holds a string, for instance, $a= "foo" and you do $a++, perl will turn that "foo" into "fop". Another $a++ will get you "foq" and so on. This behavior continues wrapping around at "z" (i.e., $a = "foz"; $a++; would result in $a = "fpa") When the string consists of numbers, perl does the same thing only wrapping at "9". Thus:

 
#!/usr/bin/perl

$a = "00001";
for (0..20) { print $a; $a++ }
__END__
00001
00002
00003
00004
00005
00006
00007
00008
00009
00010
00011
00012
00013
00014
00015
00016
00017
00018
00019
00020
00021
[download]

duff

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Re^2: increment # but keep # of digits same by Hero Zzyzzx (Curate) on Apr 26, 2005 at 14:51 UTC
That could simply be: `#!/usr/bin/perl foreach("00001".."00021"){ print "$_\n"; }` [download] Edit: Just showing that the increment operator is a bit magical, but now I see I've probably missed the point a bit. -Any sufficiently advanced technology is indistinguishable from doubletalk. My Biz	[reply] [d/l]