Re: Increment a mixed alphanumeric (Updated)

[BrowserUk]

So, what should FOO123Z increment to? Showing a sequence of increments would be good.

Does this come close to your intent?:

#! perl -slw
use strict;

my $id = "F00123Z";

for ( 1 .. 11 ) {
    print $id;
    substr( $id, 1, -1 )++;
}

__END__
C:\test>junk72
F00123Z
F00124Z
F00125Z
F00126Z
F00127Z
F00128Z
F00129Z
F00130Z
F00131Z
F00132Z
F00133Z

Update: Or maybe this comes closer?:

#! perl -slw
use strict;

my $id = "F00123Z";

for ( 1 .. 29 ) {
    print $id;
    if( substr( $id, -1 ) eq 'Z' ) {
        substr( $id, -1 ) = 'A';
        substr( $id, 1, -1 )++;
    }
    else {
        substr( $id, -1 ) ++;
    }
}

__END__
C:\test>junk72
F00123Z
F00124A
F00124B
F00124C
F00124D
F00124E
F00124F
F00124G
F00124H
F00124I
F00124J
F00124K
F00124L
F00124M
F00124N
F00124O
F00124P
F00124Q
F00124R
F00124S
F00124T
F00124U
F00124V
F00124W
F00124X
F00124Y
F00124Z
F00125A
F00125B

[miketosh]

Right, here is a regexp for valid formats:

/^[A-Z]+[0-9]*[A-Z]*/;

N1234Z becomes N12350
N12359 becomes N1235A
N9999Z becomes NA0000


Ordering of the addition is less important than testing all possible variations. Technically NA0000 is invalid since it has a number, but the number part equals zero. I can handle that logic outside of this code construct.

[BrowserUk]

By extension:

#! perl -slw
use strict;

=comment
N1234Z becomes N12350
N12359 becomes N1235A
N9999Z becomes NA0000
=cut

my $id = "F00123Z";

for ( 1 .. 80 ) {
    print $id;
    last if substr( $id, -5 ) eq '9999Z';

    if( substr( $id, -1 ) eq 'Z' ) {
        substr( $id, -1 ) = '0';
        substr( $id, 1, -1 )++;
    }
    elsif( substr( $id, -1 ) eq '9' ) {
        substr( $id, -1 ) = 'A';
    }
    else {
        substr( $id, -1 ) ++;
    }
}