Re: [gentoo-user] About sed

[Rafael Barreto]

Other thing... Why was necessary to ^CLOCK= before
s/^\(CLOCK=".*"\).*$/\1/p? And which the necessity of the ( ) between the
regular expression?

Thanks again


2005/11/7, Rafael Barreto <rafaelmbarreto@×××××.com>:
>
> For that I understood, this command will return the line of CLOCK= in
> /etc/conf.f/clock without any comments. Is this right? Well, what I really
> want is replace just CLOCK="fool1" by CLOCK="fool2" keeping the comments in
> line.
>
> By the way, \1 do really what? If i put \0 the result is the entire line.
> So, could you explain me this a little more? Thanks...
>
> 2005/11/7, gentuxx < gentuxx@×××××.com>:
> >
> > -----BEGIN PGP SIGNED MESSAGE-----
> > Hash: SHA1
> >
> > Willie Wong wrote:
> >
> > >On Mon, Nov 07, 2005 at 01:44:42AM -0200, Rafael Barreto wrote:
> > >
> > >>Hi,
> > >>
> > >>I'm learning about the use of the sed command and I have some
> > questions. I'm
> > >>trying to read in /etc/conf.d/clock the CLOCK variable with:
> > >>
> > >>sed '/^CLOCK="*"$/p' /etc/conf.d/clock
> > >>
> > >>This command, in principe, must print in screen the line that contains
> > >>CLOCK= in the begin, contains anything between double quotes and ends.
> >
> > Well,
> > >>this doesn't return anything. If I enter the above command without $,
> > all is
> > >>ok. But, if I would like to return just that line contains
> > CLOCK="anything"
> > >>and nothing more? For example,
> > >
> > >
> > >No it doesn't. What you want is the regexp ^CLOCK=".*"$ if you want
> > >anything (including nothing) between the double quotes, or
> > >^CLOCK=".+"$ if you want something (excluding nothing) between the
> > >double quotes.
> > >
> > >The reason that removing the trailing $ worked is that it matched the
> > >CLOCK=" part, the * character specifies 0 or more iterates of the
> > >previous character, which is "
> > >
> > >HTH
> > >
> > >W
> >
> > Also, as you pointed out, lines with trailing comments would not be
> > returned based on the expression (even as modified):
> >
> > sed '/^CLOCK=".*"$/p /etc/conf.d/clock
> >
> > This is because the expression, as is, does not allow for anything
> > after the last double quote ("). The following expression should
> > match the line you want, and print out ONLY the 'CLOCK="foo"':
> >
> > sed -n '/^CLOCK=/s/^\(CLOCK=".*"\).*$/\1/p /etc/conf.d/clock
> >
> > How this works is as follows (since you're trying to learn sed):
> >
> > 1) the '-n' suppresses all output except that which was changed by
> > your expression/commands.
> > 2) the first expression ( /^CLOCK=/ ) gives sed the "address" at which
> > to make the changes.
> > 3) the second expression ( s/^\(CLOCK=".*"\).*$/\1/p )tells sed what
> > to do when it reaches that address. This is better broken down into
> > smaller steps:
> > a) the first half of the substitution expression (
> > s/^\(CLOCK=".*"\).*$/ ) tells sed to match the capital letters C
> > - -L-O-C-K which start a line ( ^ ),
> > b) followed by an equals sign (=), a double-quote ("),
> > c) followed by 0 or more of any character type - except newlines
> > - - ( .* ),
> > d) followed by another double-quote (").
> > e) Then, because of the parentheses metacharacters ( \( \) ),
> > store the match in the holding space (memory).
> > f) Then match 0 or more of any character type ( .* ), ending the
> > line ( $ ).
> > g) the second half ( /\1/ ) substitutes the characters "captured"
> > in the parentheses metacharacters, for the whole line
> > h) and prints ( /p ) the result
> >
> > So, while Willie's suggestion is correct, this should give you a more
> > complete solution.
> >
> > HTH
> >
> > - --
> > gentux
> > echo "hfouvyAdpy/ofu" | perl -pe 's/(.)/chr(ord($1)-1)/ge'
> >
> > gentux's gpg fingerprint ==> 34CE 2E97 40C7 EF6E EC40 9795 2D81 924A
> > 6996 0993
> > -----BEGIN PGP SIGNATURE-----
> > Version: GnuPG v1.4.1 (GNU/Linux)
> >
> > iD8DBQFDbuAELYGSSmmWCZMRAoxdAKDZTA89tDCO+I67qhZwba6oJ28TrgCdHIkT
> > Lctx2b5xRczC3bXl+emMrOs=
> > =780W
> > -----END PGP SIGNATURE-----
> >
> > --
> > gentoo-user@g.o mailing list
> >
> >
>