in reply to Re: Removing spaces, again.
in thread Removing spaces, again.

my $bit_o_line = substr ($line, 3, 7);
$bit_o_line =~ m/^\s*(.*)\s*$/;
$BKSIC_MODEL[$i] = $1;

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This is pretty much what he did do, and it doesn't work. It might help to change the regex slightly: 
my $bit_o_line = substr ($line, 3, 7);
$bit_o_line =~ m/^\s*(\S+)\s*$/;
$BKSIC_MODEL[$i] = $1;

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But it's still cumbersome compared to some of the earlier suggestions. If you must do it this way, why not follow the method in the FAQ: 
my $BKSIC_MODEL[$i] = substr ($line, 3, 7);
$BKSIC_MODEL[$i] =~ s/^\s+//;
$BKSIC_MODEL[$i] =~ s/\s+$//;

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buckaduck

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Re: Re: Re: Removing spaces, again.
by McD (Chaplain) on Mar 16, 2001 at 04:13 UTC
    buckaduck writes:

    This is pretty much what he did do, and it doesn't work.

    I think it half works. He's getting caught by the greedy *, which explain the trailing space on his match, but I don't see how he got a leading space. Anytime I scratch my head like that on my own code, I fall back to a technique like the one I suggested.

    All your suggestions are great - and as usual, TMTOWTDI - but I was hoping in my response to point akm2 towards ways he could help himself more, now and in the future. Maybe if he came up with something new, he'd post it back here.

    Peace,
    -McD

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      I've also been scratching my head over that leading space. The only thing I can think of is that he really wrote something like: 
      ($BKSIC_MODEL[$i] = substr ($line, 3, 7)) =~ m/^\s*(.*)\s*$/;

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      (which would simply capture the substr result and throw away the result of the pattern match)

      ... instead of the line he posted here: 

      ($BKSIC_MODEL[$i]) = substr ($line, 3, 7) =~ m/^\s*(.*)\s*$/;

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      (which should not capture the leading space)

      buckaduck

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