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From: Michael Peppler <mpeppler at peppler dot org>
Subject: RE: Remove Leading Whitespaces
Date: Sep 27 1999 11:24PM
Xin_Li@WellspringRes.com writes:
> ^ means bitwise logical xor. How can I make sense of ^ in s/^\s*//?
No.
^ means bitwise logica xor when used like this:
$baz = $foo ^ $bar;
However, inside of a m// or s/// operation there are characters that
serve as operators that are quite different from the normal stuff.
>From the command line, run
perldoc perlre
to read the perl manual on regular expressions, and also go to your
local bookstore and browse the perl section - there are a few very
good books that your should look into (I like Effective Perl
Programming, but I'm not sure how good that book is for a beginner...)
Michael
> > -----Original Message-----
> > From: Ravipudi, Murali [SMTP:Murali.Ravipudi@gs.com]
> > Sent: Monday, September 27, 1999 2:44 PM
> > To: SybPerl Discussion List
> > Subject: RE: Remove Leading Whitespaces
> >
> >
> >
> > > From: Xin_Li@WellspringRes.com [SMTP:Xin_Li@WellspringRes.com]
> > > Sent: Monday, September 27, 1999 2:41 PM
> > > To: SybPerl Discussion List
> > > Subject: Remove Leading Whitespaces
> > >
> > > Thanks. But if I want to remove the leading spaces or whitespaces, how
> > > should I do?
> > >
> > [Ravipudi, Murali]
> >
> > A s/^\s*// will do
> >
> >
> >
> > > > -----Original Message-----
> > > > From: Michael Peppler [SMTP:mpeppler@peppler.org]
> > > > Sent: Monday, September 27, 1999 2:30 PM
> > > > To: SybPerl Discussion List
> > > > Subject: RE: where does it come from? help please.
> > > >
> > > > Xin_Li@WellspringRes.com writes:
> > > > > What's the meaning of $ in /\s*$/
> > > >
> > > > End of the string. The reg-exp means "all the whitespace at the end of
> > > > the string"
> > > >
> > > > Michael
> > > >
> > > > >
> > > > > > -----Original Message-----
> > > > > > From: Simon Haven [SMTP:Simon.Haven@wdr.com]
> > > > > > Sent: Monday, September 27, 1999 1:05 PM
> > > > > > To: SybPerl Discussion List
> > > > > > Subject: Re: where does it come from? help please.
> > > > > >
> > > > > >
> > > > > > >
> > > > > > > But in this case, if some one has space(s) at the end in
> > his/her
> > > > > > password,
> > > > > > > it fails...
> > > > > >
> > > > > > No it won't fail, it's doing exactly what rtrim() would do.
> > > > > > /\s*$/ will find zero or more occurrences of whitespace before
> > the
> > > > end of
> > > > > > line.
> > > > > > However, if I read my definition of "\s" right in the Camel book,
> > > it
> > > > WILL
> > > > > > remove the newline - which may not be a problem, but might be
> > worth
> > > > > > knowing.
> > > > > >
> > > > > > Simon
> > > > > >
> > > > > > >
> > > > > > > Murali
> > > > > > >
> > > > > > > > -----Original Message-----
> > > > > > > > From: Michael Peppler [SMTP:mpeppler@peppler.org]
> > > > > >
> > > > > >
> > > > > > > >
> > > > > > > > > NB, make sure you're using chop, not chomp, which only
> > > removes
> > > > a
> > > > > > > > newline.
> > > > > > > >
> > > > > > > > Use s/// instead of chop (eg $var =~ s/\s*$//;)
> > > > > > > >
> > > > > > > > Michael
> > > > > >
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> > > >
> > > > --
> > > > Michael Peppler -||- Data Migrations Inc.
> > > > mpeppler@peppler.org -||- http://www.mbay.net/~mpeppler
> > > > Int. Sybase User Group -||- http://www.isug.com
> > > > Sybase on Linux mailing list: ase-linux-list@isug.com
>
--
Michael Peppler -||- Data Migrations Inc.
mpeppler@peppler.org -||- http://www.mbay.net/~mpeppler
Int. Sybase User Group -||- http://www.isug.com
Sybase on Linux mailing list: ase-linux-list@isug.com
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