I wrote my self this subroutine... which comes in very handy :-)
#
# strip
#
=head2 strip
=over
=item Description
Strips out leading and training whitespaces from references...
=item Input
* Reference to an array, hash or string
or
A string
=item Return
* If input was a reference then, None - The reference passed as input is modified...
Else, the stripped string
=item Method Type
* This method can be used as a class or object method
=back
=cut
sub strip {
my $self = shift;
my $ref = shift;
if (! ref($ref)) {
$ref =~ s/(^[\s\t]*)|([\s\t]*$)//g if (defined $ref);
return $ref;
}
elsif (ref($ref) =~ /^ARRAY$/i) {
foreach my $i (0 .. $#$ref) {
$ref->[$i] =~ s/(^[\s\t]*)|([\s\t]*$)//g;
}
}
elsif (ref($ref) =~ /^HASH$/i) {
while (my ($key, $value) = each %$ref) {
delete $ref->{$key};
$key =~ s/(^[\s\t]*)|([\s\t]*$)//g;
$value =~ s/(^[\s\t]*)|([\s\t]*$)//g;
$ref->{$key} = $value;
}
}
elsif (ref($ref) =~ /^SCALAR$/i) {
$$ref =~ s/(^[\s\t]*)|([\s\t]*$)//g;
}
else {
die "Unknown reference type";
}
}
On Wed, 2003-10-15 at 15:28, Sudarshan Raghavan wrote:
> [EMAIL PROTECTED] wrote:
>
> > Can someone hlpe me clean up this trim?
> >
> > Rule: remove all trailing blanks and newline/LF
>
> perldoc -q 'How do I strip blank space from the beginning/end of a
> string'
>
>
> >
> >
> > Do I need a chomp here somewhere?
>
> No, the \s+ will take care of that
>
>
> >
> >
> > sub trim
> > { my $z = $_[0];
> >
> > $z =~ s/^\s+//;
>
> You don't need the above statement if you only need to remove trailing
> blanks and newlines
>
>
> >
> > $z =~ s/\s+$//;
> >
> > return $z;
> > }
>
> You could maybe write the sub as
> sub trim {
> s/\s+$// foreach (@_);
> }
>
> This will also trim an entire list if needed
>
>
> >
> >
> > thanks,
> > -rkl
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