incrementing strings

deathbygoats has asked for the wisdom of the Perl Monks concerning the following question:

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Re: incrementing strings by Roy Johnson (Monsignor) on Mar 14, 2005 at 22:00 UTC
perldoc perlop If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and matches the pattern `/^[a-zA-Z][0-9]\z/` So yes, it's the dot. Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re: incrementing strings by ikegami (Patriarch) on Mar 14, 2005 at 22:02 UTC
Quote perlop: The auto-increment operator has a little extra builtin magic to it. If you increment a variable that is numeric, or that has ever been used in a numeric context, you get a normal increment. If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and *matches the pattern `/^[a-zA-Z][0-9]\z/`*, the increment is done as a string, preserving each character within its range, with carry: The "." is indeed the problem.	[reply] [d/l]
Re: incrementing strings by hv (Prior) on Mar 15, 2005 at 00:52 UTC
I think the easiest fix is to change the match for a substitution to do the increment: `if ($alias =~ s/(\d+)$/$1 + 1/e) { print $alias; } elsif ...` [download] Hugo	[reply] [d/l]
Re: incrementing strings by Roy Johnson (Monsignor) on Mar 15, 2005 at 00:45 UTC
Shamefully, I didn't answer how you can fix that. Here ya go: `if ($result->entries \|\| grep(/\b$alias\b/i, @aliases)) { if (my ($fixed,$plusable) = $alias =~ /(.?)([a-z]\d+)$/i +) { ++$plusable; $alias = $fixed.$plusable; print $alias; } elsif` [download] Update. Ooh! Ooh! `if ($result->entries \|\| grep(/\b$alias\b/i, @aliases)) { if ($alias =~ s/([a-z]\d+)$/my $x=$1; ++$x/ei) { print $alias; } elsif` [download] Capturing the letters is optional. If you don't, johnq.smith9 becomes johnq.smith10. Otherwise, johnq.smith9 becomes johnq.smiti0. Caution:* Contents may have been coded under pressure.	[reply] [d/l] [select]
Re: incrementing strings by moot (Chaplain) on Mar 14, 2005 at 22:01 UTC
Strings don't increment this way, if what you are trying to do is take `$alias = "johnq.smith1"; $alias++;` [download] and turn $alias into `"johnq.smith2"` [download] . What you'll need to do is save the digits into a temporary variable, increment them, then append to the digitless string: `$alias = "johnq.smith1"; if ($alias =~ s/(\d+$)//) { $alias .= $1++; # $alias is now "johnq.smith2" }` [download] There are, of course, other ways to achieve this, but not via directly incrementing a string. Update: Ignore my response. Of course you can increment strings this way. Is it friday yet?	[reply] [d/l] [select]
Re^2: incrementing strings by ysth (Canon) on Mar 15, 2005 at 07:02 UTC
s/// is the way to go, even without the period, since I don't think you want a sequence "madonna8", "madonna9", "madonnb0", "madonnb1", as string incrementing would do.	[reply]
Re: incrementing strings by ambrus (Abbot) on Mar 15, 2005 at 16:46 UTC
In such cases, one is tempted to write this: `$string = "johnq.smith1"; $string =~ /(\d+$)/ or die "no number"; $1++ +; print $string, $/; # WRONG` [download] That's however invalid, so let's do this: `$string = "johnq.smith1"; $string =~ /(\d+$)/ or die "no number"; matc +h($string, 1)++; print $string, $/; sub match : lvalue { my $n = $_[1] \|\| 0; substr $_[0], $-[$n], $+[$n] +- $-[$n]; }` [download] that doesn't have much advantage over a simple substitution of course. One could even create a tied array for the modifiable versions of $1, $2, ... `{ package Matched; sub TIEARRAY { bless \my$x, $_[0]; } sub FETCH { my + $n = $_[1]; substr $_, $-[$n], $+[$n] - $-[$n]; } sub STORE { my $n += $_[1]; substr($_, $-[$n], $+[$n] - $-[$n]) = $_[2]; } tie @~, Match +ed::; } local $\ = $/; $_ = "johnq.smith1"; /(\d+$)/ or die "no number"; $~[1] +++; print;` [download]	[reply] [d/l] [select]
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