[gentoo-user] About sed

[gentoo-user] About sed

[Rafael Barreto]

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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,

CLOCK="anything" # set clock of the system

and

CLOCK="anything"

are different.

Other thing... if i put:

sed '/^CLOCK=*/p' /etc/conf.d/clock

the return will be anything that contains CLOCK. Why?

I suppose that I didn't undestand the use of regular expression
metacharacters. So, please, anyone to explain me that?

Thanks a lot and sorry by my english...

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Re: [gentoo-user] About sed

[Willie Wong]

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
-- 
Q: Why won't Heisenberg's operators live in the suburbs?
A: They don't commute.
Sortir en Pantoufles: up 4 days,  5:24
-- 
gentoo-user@gentoo.org mailing list

Re: [gentoo-user] About sed

[Rafael Barreto]

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:) Thanks very much.

Now that I read your answear I searched in a book the significant of *,+ and
others and I suppose I understood.

Thanks

2005/11/7, Willie Wong <wwong@princeton.edu>:
>
> 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
> --
> Q: Why won't Heisenberg's operators live in the suburbs?
> A: They don't commute.
> Sortir en Pantoufles: up 4 days, 5:24
> --
> gentoo-user@gentoo.org mailing list
>
>

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Re: [gentoo-user] About sed

[gentuxx]

-----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-----

-- 
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Re: [gentoo-user] About sed

[Rafael Barreto]

[-- Attachment #1: Type: text/plain, Size: 3723 bytes --]

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@gmail.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@gentoo.org mailing list
>
>

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Re: [gentoo-user] About sed

[Rafael Barreto]

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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@gmail.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@gmail.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@gentoo.org mailing list
> >
> >
>

[-- Attachment #2: Type: text/html, Size: 5295 bytes --]

Re: [gentoo-user] About sed

[gentuxx]

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

Rafael Barreto wrote:

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

Sorry for things being out of order, but you top-posted.  The answer
to your first question is below.

The /^CLOCK=/ is NOT necessary, in this instance, and may actually
cause problems if you're not 100% sure of your regular expression.  It
is there, simply because I was expanding your original pattern.  What
this does, though, is it tells sed to match the line with the regular
expression before making any substitutions and/or replacements.  Say
for example, you had 2 lines in the same file with different values
but similar structure, and you only wanted to change one of them.
Let's go with the example you used.  Let's assume that our example
file looks something like this:

# This is a comment
CLOCK="foo1"
# This is another comment
TIMEZONE="GMT"
# This is yet another comment
CLOCK="${CLOCK}foo2"

So, let's say that you wanted to change the second instance.  The
/^CLOCK=/ tells sed to ignore all lines that do NOT begin with
'CLOCK='.  So the comments ( # This is a comment ) and the 'TIMEZONE'
directive will be ignored.  Actions will only be taken on the lines:

CLOCK="foo1"
CLOCK="${CLOCK}foo2"

...because they match the expression we gave sed as an "address" (
/^CLOCK=/ ).  We would furthur specify how to change that line by our
search and replace command (as described below):

s/^\(CLOCK=\"\)\${.*}\(\".*\)$/\1foo3\2/                                 
 


It is important to note here that the left and right brace characters
( { } ) are not escaped.  If they were, they would have been handled
as metacharacters, which produces a drastically different result.

If you really want to get to know sed (and awk), I HIGHLY recommend
getting the 2 O'reilley books:  Sed & Awk, and Mastering Regular
Expressions.  Both of these two books have taught me almost everything
I know on the matter, and I refer to them frequently.

HTH


> 2005/11/7, Rafael Barreto <rafaelmbarreto@gmail.com
<mailto:rafaelmbarreto@gmail.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@gmail.com <mailto:gentuxx@gmail.com>>:
>

I will try to answer both of your questions with this one email.

The '\1' takes the first element, or element group, captured by the
parentheses metacharacters ( \( \) ) and uses it as a variable.  The
variable '\0', as you found out, represents the entire contents of the
"pattern space" or basically, the entire line since you're matching
beginning ( ^ ) to end ( $ ).  The variable '\0' is also synonymous
with the ampersand metacharacter ( & ).  So, if you want to change the
value of the "CLOCK" variable in /etc/conf.d/clock, while leaving any
comments intact, you could do something like this:

sed 's/^\(CLOCK=\"\).*\(\".*\)$/\1foo2\2/' /etc/conf.d/clock

This matches and captures the contents of the first set of
parentheses, and places it into a variable accessible by '\1'.  Then
it matches 0 or more of any character type, until it matches a literal
double-quote ( \" ).  Then it matches and captures the contents of the
second set of parentheses, and places it into a variable accessible by
'\2'.  So, to change the value, you would issue a substitution command
( s/// ) which matches the first set and replaces it with the second
set.  The variables '\1' and '\2' are interpolated, so the values
captured are printed.  I.e. the result should be:

was:
CLOCK="foo1"      # some comments here

now:
CLOCK="foo2"      # some comments here



> 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


- --
gentoo-user@gentoo.org <mailto:gentoo-user@gentoo.org> mailing list





- --
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)

iD8DBQFDbvZBLYGSSmmWCZMRAu2TAJ9QluvFRh2Hvu2T1p6DxOkhsD+gUgCgvYjC
qQ12BqJsAQHxMEfBkLTodAw=
=LvhF
-----END PGP SIGNATURE-----

-- 
gentoo-user@gentoo.org mailing list

Re: [gentoo-user] About sed

[Willie Wong]

On Sun, Nov 06, 2005 at 09:03:01PM -0800, gentuxx wrote:
> sed -n '/^CLOCK=/s/^\(CLOCK=".*"\).*$/\1/p /etc/conf.d/clock
> 

Ah, yes, I misunderstood the OP. I thought he didn't want the lines
with trailing comments at all. 

But is it necessary to give the address for an s// replacement? As I
understand it that sed is a stream editor and will try the replacement
on every "line" it encounters. The -n flag would guarantee only the
line changed would be printed anyway. 

I guess what I am saying is that 
  sed -n 's/^\(CLOCK=".*"\).*$/\1/p' /etc/conf.d/clock
would do just as fine, no?

W
-- 
The particle physicists use order parameter fields, too. Their 
order parameter fields also hide lots of details about what their
quarks and gluons are composed of. The main difference is that
they don't know of what their fields are composed. It ought to
be reassuring to them that we don't always find our greater
knowledge very helpful.
   ~James P. Sethna "Order Parameters, Broken Symmetry, and Topology"
Sortir en Pantoufles: up 4 days,  8:10
-- 
gentoo-user@gentoo.org mailing list

Re: [gentoo-user] About sed

[Willie Wong]

On Mon, Nov 07, 2005 at 03:42:05AM -0200, Rafael Barreto wrote:
> Other thing... Why was necessary to ^CLOCK= before
> s/^\(CLOCK=".*"\).*$/\1/p? And which the necessity of the ( ) between the
> regular expression?

as I just posted in another post, the /^CLOCK/ should not be strictly
necessary. 

See below for an explanation of the ()
> 
> 
> 2005/11/7, Rafael Barreto <rafaelmbarreto@gmail.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.
> >

Do you know what fool2 would be? If so:

   s/^\(CLOCK="\)[^["]]*\(".*\)$/\1fool2\2/p

should do what you want. (I think sed greedy matches by default? Hum,
I honestly don't remember, better safe than sorry.)

See below for the use of the \( .. \) and \1 \2 variables. 

The [^["]] specifies a class of characters to match, in this case, it
means anything except for the quotation mark. (The caret ^ negates the
class ["] which means only the quotation mark.) So what this sed
expression is really saying is that we want to keep the beginning of
the line up to the first quotation mark the same, and keep the end of
the line starting from the second quotation mark the same, and change
everything in between to 'fool2'.

> > 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...
> >

\0 is the entire matching expression/line.
\n is tokenizing. By using parentheses in the regexp, you can reuse
the texts in the parentheses in the substitution. 

For example, consider the expression

   s/\(.at\).*\(.ox\)/\2 not \1/p

If we send it 

   The bat scared the fox.

it would print 

   fox not bat

because \2 gets replaced by the second matching token, which in this
case, is fox, because it matches ".ox". Similarly for bat and the
first token. 

If we send the same expression

   The cat died in the box.

it would print

   box not cat

I can't explain the concept very well, and I hope the examples are
good enough. 

Seriously, one of the best way to learn sed is to read the manual from 

  `info sed'

W
-- 
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No Les No more.
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Re: [gentoo-user] [OT] About sed

[Willie Wong]

On Sun, Nov 06, 2005 at 10:37:53PM -0800, gentuxx wrote:
> If you really want to get to know sed (and awk), I HIGHLY recommend
> getting the 2 O'reilley books:  Sed & Awk, and Mastering Regular
> Expressions.  Both of these two books have taught me almost everything
> I know on the matter, and I refer to them frequently.
> 

Ah yes. I definitely agree. My copy disappeared when I moved out of
the dorm last. That fact has caused me much pain and inconvenience in
the past few months.

Useful as 'info sed' maybe, the book makes looking for help that much 
easier. 

W
-- 
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Re: [gentoo-user] About sed

[gentuxx]

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

Willie Wong wrote:

>On Sun, Nov 06, 2005 at 09:03:01PM -0800, gentuxx wrote:
>
>>sed -n '/^CLOCK=/s/^\(CLOCK=".*"\).*$/\1/p /etc/conf.d/clock
>>
>
>Ah, yes, I misunderstood the OP. I thought he didn't want the lines
>with trailing comments at all.
>

He didn't really make that clear in his first post.  ;-)

>But is it necessary to give the address for an s// replacement? As I
>understand it that sed is a stream editor and will try the replacement
>on every "line" it encounters. The -n flag would guarantee only the
>line changed would be printed anyway.
>

No, the address is NOT necessary for this case.  I stated that in my
reply to his post.  However, sed will only replace the first instance
found, unless the 'g' flag is specified ( s///g ).  Then it'll search
and replace globally.  The '-n' option suppresses unchanged output, so
if the output of the sed command was being redirected to a file, only
the changed lines would be printed.  So, in this case, you would NOT
want to use the '-n' option, because you would want the entire file
output *with* the changes.

>I guess what I am saying is that
> sed -n 's/^\(CLOCK=".*"\).*$/\1/p' /etc/conf.d/clock
>would do just as fine, no?
>
>W

(See above).  To actually make changes to the file, once you had
tested your regular expressions to STDOUT, you would most likely want
to redirect it to a file to replace the original file.  If you include
the '-n' option, all of the unchanged lines would not be printed or
redirected - so you would have your changes, but nothing else.

So for your final "run", you would want something without the '-n'
option.  And without the '-n' option, the print command ( s///p ) is
moot.  (See my most recent reply to the OP for a final solution.)

- --
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
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Version: GnuPG v1.4.1 (GNU/Linux)

iD8DBQFDbwOjLYGSSmmWCZMRAmG9AJ46OE4xLI3pCKnb3RTQSCIGWNKSBgCfWOOl
V9zFRF6QNqmVSK0LHjt4e9c=
=tnMC
-----END PGP SIGNATURE-----

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Re: [gentoo-user] About sed

[Alexander Skwar]

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.

No, that's not what it does. It matches "CLOCK" at "beginning
of line" (^) followed by = followed by any number of " followed
by another ", which is the last character of the line.

You forgot the . before the *.

> Other thing... if i put:
> 
> sed '/^CLOCK=*/p' /etc/conf.d/clock
> 
> the return will be anything that contains CLOCK. Why?

No, it will return all lines, that start with CLOCK, followed
by any numbers of =. You again missed the . before the *.

> Thanks a lot and sorry by my english...

No problem at all, but please do not send HTML junk to
the list. That's not so much excusable (well, that's of
course an exaggeration by me...).

Alexander Skwar
-- 
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Re: [gentoo-user] About sed

[Walter Dnes]

On Mon, Nov 07, 2005 at 03:38:00AM -0200, Rafael Barreto wrote
> Is this right? Well, what I really want is replace just CLOCK="fool1"
> by CLOCK="fool2" keeping the comments in line.

  That is not what sed is designed to do.  "sed" is "Streaming EDitor".
You specify an input file, and the changed file goes to STDOUT.  If you
want to change the original file, you need to use "ed".  For details,
"man ed".

-- 
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My musings on technology and security at http://tech_sec.blog.ca
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Re: [gentoo-user] About sed

[Neil Bothwick]

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On Tue, 8 Nov 2005 23:37:52 -0500, Walter Dnes wrote:

>   That is not what sed is designed to do.  "sed" is "Streaming EDitor".
> You specify an input file, and the changed file goes to STDOUT.  If you
> want to change the original file, you need to use "ed".

Or use sed's -i or --in-place argument.


-- 
Neil Bothwick

Never drink coffee that's been anywhere near a fish.

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