D'acord, GENIAL, accents oberts, gràcies!!
Pau.
El 22 / agost / 2009 01:13, Roberto Torresblack.death.1...@gmail.com
va escriure:
2009/8/22 Pau Climent i Pérez pclimentpe...@gmail.com:
Vinga valents:
Imaginem que declarem una variable M:
$ M=hola món
$ echo $M
hello world
Ara tenim M
Hola,
22/08/09 @ 14:16 (+0200), thus spake Pau Climent i Pérez:
D'acord, GENIAL, accents oberts, gràcies!!
Si no tinc mal entès es recomana no utilitzar accents oberts
sinó $() que té el mateix efecte.
Ernest
--
To UNSUBSCRIBE, email to debian-user-catalan-requ...@lists.debian.org
with a
Vinga valents:
Imaginem que declarem una variable M:
$ M=hola món
$ echo $M
hello world
Ara tenim M declarada, i volem eliminar-li les aparicions de món per
afegir-hi, posem per exemple gent.
Com faríeu això?
Jo pensava fer servir sed:
$ echo $M | sed s/món/gent/
però ara com assigne el
2009/8/22 Pau Climent i Pérez pclimentpe...@gmail.com:
Vinga valents:
Imaginem que declarem una variable M:
$ M=hola món
$ echo $M
hello world
Ara tenim M declarada, i volem eliminar-li les aparicions de món per
afegir-hi, posem per exemple gent.
Com faríeu això?
Jo pensava fer
El ds 22 de 08 de 2009 a les 00:44 +0200, en/na Pau Climent i Pérez va
escriure:
Vinga valents:
Imaginem que declarem una variable M:
$ M=hola món
$ echo $M
hello world
Ara tenim M declarada, i volem eliminar-li les aparicions de món per
afegir-hi, posem per exemple gent.
Com
Jo personalment prefereixo fer servir
M=$(echo|sed s/món/gent/)
Els accents oberts es veuen poc
Ara bash permet simplificar lo anterior
proba M=${M/món/gent}
${parameter/pattern/string}
The pattern is expanded to produce a pattern just as in
pathname expan‐
sion.
If you don't mind specifying your reference file on the command line as a
shell script parameter, you can use $1 inside the shell script to pick its
name up and do things with it.
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with a subject of unsubscribe. Trouble? Contact [EMAIL
SENTHIL KUMAR wrote:
Hi,
i plan to run a program over some 20 files were i need to input
one files as REFERENCE and then the remaining as TEST. i have return
a script to take every file in the list to be taken a REF and the
remaining as TEST and it works well{all with all }. but i want
Hi,
i plan to run a program over some 20 files were i need to input
one files as REFERENCE and then the remaining as TEST. i have return
a script to take every file in the list to be taken a REF and the
remaining as TEST and it works well{all with all }. but i want to
specify which one to
Hallo,
Am Mit, 11 Okt 2006, David Haller schrieb:
chmod -R u+rwX,go+rwX .
Sollte natuerlich
chmod -R u+rwX,go+rX .
sein...
-dnh
--
Falls hier aber die Realität gemeint sein soll, so muss Ich sagen
Nichts ist so real wie die Wirklichkeit. Auch wenn manche das
nicht wahr haben
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Hash: SHA1
hallo, ich habe ein an sich triviales problem, aber bevor ich mich in
irgendwelchen script-monstern versteige möchte ich nach einem einfachen
weg fragen.
im prinzip muss ich nur files mit falschen permissions aussortieren, die
noch von
On Tuesday 10 October 2006 14:44, Andreas Grassl wrote:
im prinzip muss ich nur files mit falschen permissions aussortieren, die
noch von windows-partitionen stammen, d.h. sie sind alle 700, sollten
aber je nach typ 755 oder 644 sein.
mein ansatz
$ chmod 755 $(find -type d)
endet in
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Christoph Haas schrieb:
On Tuesday 10 October 2006 14:44, Andreas Grassl wrote:
$ chmod 755 $(find -type d)
endet in folgender ausgabe:
bash: /bin/chmod: Argument list too long
erste frage: wie kann ich das einfach umgehen?
ungetestet
find
Christoph Haas schrieb:
On Tuesday 10 October 2006 14:44, Andreas Grassl wrote:
im prinzip muss ich nur files mit falschen permissions aussortieren, die
noch von windows-partitionen stammen, d.h. sie sind alle 700, sollten
aber je nach typ 755 oder 644 sein.
mein ansatz
$ chmod 755 $(find
Gruesse!
* Andreas Grassl [EMAIL PROTECTED] schrieb am [10.10.06 14:44]:
im prinzip muss ich nur files mit falschen permissions aussortieren, die
noch von windows-partitionen stammen, d.h. sie sind alle 700, sollten
aber je nach typ 755 oder 644 sein.
mein ansatz
$ chmod 755 $(find
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Hash: SHA1
Jan Kohnert schrieb:
Christoph Haas schrieb:
On Tuesday 10 October 2006 14:44, Andreas Grassl wrote:
$ chmod 755 $(find -type d)
endet in folgender ausgabe:
bash: /bin/chmod: Argument list too long
erste frage: wie kann ich das einfach
Andreas Grassl wrote:
mein ansatz
$ chmod 755 $(find -type d)
endet in folgender ausgabe:
bash: /bin/chmod: Argument list too long
erste frage: wie kann ich das einfach umgehen?
zweite frage: wie kann ich leerzeichen in dateinamen verarbeiten
mit dieser methode?
Alternative
* Andreas Grassl (2006-10-10):
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hallo, ich habe ein an sich triviales problem, aber bevor ich mich in
irgendwelchen script-monstern versteige möchte ich nach einem einfachen
weg fragen.
im prinzip muss ich nur files mit falschen permissions
Hallo Andreas!
Andreas Grassl schrieb am Dienstag, den 10. Oktober 2006:
im prinzip muss ich nur files mit falschen permissions aussortieren, die
noch von windows-partitionen stammen, d.h. sie sind alle 700, sollten
aber je nach typ 755 oder 644 sein.
$ chmod 755 $(find -type d)
Falls
Hallo Andreas,
Andreas Grassl, 10.10.2006 (d.m.y):
hallo, ich habe ein an sich triviales problem, aber bevor ich mich in
irgendwelchen script-monstern versteige möchte ich nach einem einfachen
weg fragen.
im prinzip muss ich nur files mit falschen permissions aussortieren, die
noch von
Hallo,
Gerhard Brauer wrote:
* Andreas Grassl [EMAIL PROTECTED] schrieb am [10.10.06 14:44]:
$ chmod 755 $(find -type d)
endet in folgender ausgabe:
bash: /bin/chmod: Argument list too long
Du kannst Shell-Kommandos nur eine (einstellbar ?) begrenzte Anzahl von
Bytes als
Hallo,
Am Die, 10 Okt 2006, Bjoern Schliessmann schrieb:
Alternative Lösung:
for i in $(find -type d); do chmod 755 $i; done
$ touch 'abc' 'a b'
$ cd /tmp/test9/t
$ ls -b
a\ b abc
$ for i in $(find -type f); do ls $i; done
./abc
ls: ./a: No such file or directory
ls: b: No such file or
Hallo,
Am Die, 10 Okt 2006, Andreas Grassl schrieb:
Jan Kohnert schrieb:
Christoph Haas schrieb:
[..]
ungetestet
find . -type d -print0 | xargs -0 chmod 755
/ungetestet
Oder: find . -type d -print -exec chmod 755 {} \;
[..]
Interessanterweise klappt das exec auch ohne solche Umwege mit
Buenas, estoy haciendo un pequeño script par mantener backups diarias
de mis PC. Quiero que una vez realizada, me envie un mail diciendome
lo ocurrido, si todo fue bien,etc...
El problema es que no se como hacer para poder pasarle un body
directamente a mail...
du=du -h /Backup
df=df -h /Backup
Coordenadas temporales: Tue, Mar 28, 2006 at 05:28:44AM +0200
Sujeto: cocozz
Comunicaba sobre: Enviar un mail con shell scripting
Buenas, estoy haciendo un pequeño script par mantener backups diarias
de mis PC. Quiero que una vez realizada, me envie un mail diciendome
lo ocurrido, si todo fue
On Tue, Mar 28, 2006 at 05:36:13AM +0200, Emilio Santos wrote:
Sujeto: cocozz
Como le paso du y df directamente al comando mail para que me meta las
salida de los dos comandos como body ?
Una solución:
du -h /Backup /tmp/back
df -h /Backup /tmp/back
cat /tmp/back | mail -s
Bonjour,J'ai vu dans un vieux post qu'en raison d'un grand nombre de questions relatives au shell scripting, une liste avait était créée. Quelqu'un aurait-il plus d'infos ? Par avance, merci.
-- Le bon sens est la chose du monde la mieux partagée.
On Tue, Mar 07, 2006 at 09:51:52AM +0100, Toto Capuccino wrote:
Bonjour,
Bonjour
J'ai vu dans un vieux post qu'en raison d'un grand nombre de questions
relatives au shell scripting, une liste avait était créée. Quelqu'un
aurait-il plus d'infos ?
regarde ici : http://debianworld.org/?q=node
Hola.
Como de introducción al Shell scripting sobre Bash.
http://xinfo.sourceforge.net/documentos/bash-scripting/bash-script-2.0.html
Muy interesante ;-)
--
Un saludo de Willy Walker ;o)
Usuario Linux: 100651
Debian Sid 2.6.5
Uno pregunta, muchos responden, todos aprenden.
martin f krafft [EMAIL PROTECTED] said on Fri, 21 May 2004 01:39:55 +0200:
--ikeVEW9yuYc//A+q
Content-Type: text/plain; charset=iso-8859-15
Content-Disposition: inline
Content-Transfer-Encoding: quoted-printable
also sprach Martin McCormick [EMAIL PROTECTED] [2004.05.20.2126 +=
0200]:
Is it just more efficient in resources to use plain #! /bin/sh
rather than bash?
No, it just makes your script more portable to systems that might not
have bash.
Some systems that /do/ have bash installed have /bin/sh linked to it,
but some don't have bash by default or choice (Solaris,
A response to another poster peaked my curiosity.
Incidentally, there is no reason to make it a bash
script rather than vanilla sh, and you can simplify the script by using
exec:
I have been writing shell scripts for a bit over fourteen
years so I am not new to this, but I
also sprach Martin McCormick [EMAIL PROTECTED] [2004.05.20.2126 +0200]:
Is it just more efficient in resources to use plain #! /bin/sh
rather than bash?
surely not. /bin/sh is generally linked to bash (... by default,
that is).
--
Please do not CC me when replying to lists; I read them!
Hopefully somebody here has perhaps seen this oddity and can provide some
insight into the cause.
I have a very simple shell script as follows:
mail:~/scripts# more topcheck
#!/bin/bash
date /usr/local/apache/htdocs/topout.txt
echo \c ; top -n 1 /usr/local/apache/htdocs/topout.txt
I have also
On Wed, May 19, 2004 at 08:11:35AM -0500, Michael Martinell wrote:
} Hopefully somebody here has perhaps seen this oddity and can provide some
} insight into the cause.
}
} I have a very simple shell script as follows:
} mail:~/scripts# more topcheck
} #!/bin/bash
} date
Pete Clarke declaimed:
The core loop is clearly
for file in $1
do
filename=${file%.*}
echo Adding $file to $filename.zip...
$ZIP $ARGS $filename $file /dev/null
let nofiles += 1
done
Hint: try
How do you call your script?
Aurel
Quoting Pete Clarke [EMAIL PROTECTED]:
Hi there people,
I have a script that performs batch zipping of files. Trouble is that it
only does one file at a time (kind of going against the batch idea).
Could someone point out the silly mistake I am
you can save the script as batchzip.sh
then make it an executable..
chmod 755 batchzip.sh
and then run it
./batchzip.sh
HTH,
Pritpal Dhaliwal
[EMAIL PROTECTED] wrote:
How do you call your script?
Aurel
Quoting Pete Clarke [EMAIL PROTECTED]:
Hi there people,
I have a script that performs
Hi Pete,
I have a script that performs batch zipping of files. Trouble is that it
only does one file at a time (kind of going against the batch idea).
Could someone point out the silly mistake I am obviously making?
[...]
#!/bin/sh
#
# batch zip
# invoke with batchzip filespec
#
# this
and make packages of it?
Dunno much about php, but I guess so.
3. Has anybody else used php for shell scripting purposes?
Any sites with shell scripts? I've seen a lot site with scripts
for webcreation but none with shell scripts.
It's exactly the same, AFAIK. Seriously though, Perl is a much
it?
Is this the debian specific way when you want a package
that isn't available yet?
2. If i would want to use extra modules, i suppose i should
compile them too and make packages of it?
3. Has anybody else used php for shell scripting purposes?
Any sites with shell scripts? I've seen a lot site with scripts
* Cuenta la leyenda que Emilio J. Padrón ([EMAIL PROTECTED]) escribió:
¡Gracias pedazo CRACK!
Me ha venido de perlas tu solución, con lo que ya me ahorro los
problemas que estaba teniendo con lo de pasar valores entre shells.
Finalmente me ha quedado así (a lo mejor es un poco chapuza, pero no
On Thu, Nov 28, 2002 at 11:18:11AM -0300, German Gutierrez wrote:
Te lo corrijo y te tiro otro dato:
awk '
BEGIN{ MAX=nada; MIN=nada; CONT=0; MEDIA=0; ETIQUETA=TOTAL TIME}
$0 ~ ETIQUETA{
if ( (MAX == nada) || ( MAX $3 ) ) MAX = $3;
if ( (MIN == nada) || ( MIN $3 ) ) MIN = $3;
CONT
Hola lista,
¿cómo puede hacer para que una variable de entorno que hereda una
subshell se pueda modificar en esa subshell y el valor modificado quede
actualizado en la shell padre?
Es decir, algo como:
$export var=555
$echo $var
555
bash
echo $var
555
var=888
echo $var
888
exit
echo $var
888
El mié, 27 de nov de 2002, a las 12:14:04 +0100, Emilio J. Padrón comentaba ..
¿cómo puede hacer para que una variable de entorno que hereda una
subshell se pueda modificar en esa subshell y el valor modificado quede
actualizado en la shell padre?
Es decir, algo como:
$export var=555
$echo $var
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Hola
On Wednesday 27 November 2002 12:14, Emilio J. Padrón wrote:
Hola lista,
¿cómo puede hacer para que una variable de entorno que hereda una
subshell se pueda modificar en esa subshell y el valor modificado quede
actualizado en la shell padre?
On Wed, Nov 27, 2002 at 12:53:57PM +0100, Victor Calzado Mayo wrote:
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Hola
On Wednesday 27 November 2002 12:14, Emilio J. Padrón wrote:
Hola lista,
¿cómo puede hacer para que una variable de entorno que hereda una
subshell se pueda
* Cuenta la leyenda que Emilio J. Padrón ([EMAIL PROTECTED]) escribió:
On Wed, Nov 27, 2002 at 12:53:57PM +0100, Victor Calzado Mayo wrote:
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Hola
On Wednesday 27 November 2002 12:14, Emilio J. Padrón wrote:
Hola lista,
¿cómo puede
On Wed, Nov 27, 2002 at 11:21:35AM -0300, German Gutierrez wrote:
* Cuenta la leyenda que Emilio J. Padrón ([EMAIL PROTECTED]) escribió:
On Wed, Nov 27, 2002 at 12:53:57PM +0100, Victor Calzado Mayo wrote:
-BEGIN PGP SIGNED MESSAGE-
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Hola
On Wednesday 27
Hi,
* Matt Price [EMAIL PROTECTED] [02-10-07 23:44]:
I have a webstie which I manage both from home (mac) and work
(woody). I use sitecopy at work but my access is only via ftp, so
sitecopy can't identify newer files on the website. So after I
upsdate the site form home, I usually log in to
Michael,
thanks for the ref to ncftpput. I'l\l look into that and lftp, as
someone else suggested. But re: the here script: does this work if I
write it ina simple shell script? That is do I write a script thus:
cd ~!/website
ftp -i origin.chass.utoronto.caEOF
mget * */* */*/*
EOF
I seem to
On Mon, Oct 07, 2002 at 07:03:25PM -0400, Matt Price wrote:
But re: the here script:
They're actually usually called here documents or heredocs.
does this work if I write it ina simple shell script?
Yes.
That is do I write a script thus:
cd ~!/website
That ! looks odd ... did you mean
Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
cat just gives me the odd error of files not being found, however, I can't
see why the files wouldn't be found ... hrrm ...
thanks much for any info you might
, 2001 6:04 PM
Subject: Shell Scripting Question
Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
cat just gives me the odd error of files not being found, however, I can't
see why the files wouldn't be found
Hi Sunny!
Sunny Dubey wrote:
Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
because $i contains lines like:
drw-r--r--2 sunnysunny12345 Oct 23 14:09 hello.c
which is very unlikely to
--- Sunny Dubey [EMAIL PROTECTED] wrote: Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
the problem is with -l switch
cat just gives me the odd error of files not being
found, however, I can't
Apparently, on Mon, Nov 05, 2001 at 09:04:38PM -0500, Sunny Dubey wrote:
Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
cat just gives me the odd error of files not being found, however, I can't
see
Sunny Dubey wrote:
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
cat just gives me the odd error of files not being found, however, I can't
see why the files wouldn't be found ... hrrm ...
thanks much for
john wrote:
Sunny Dubey wrote:
Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
because $i contains lines like:
drw-r--r--2 sunnysunny12345 Oct 23 14:09 hello.c
which is
Craig Dickson wrote:
john wrote:
Sunny Dubey wrote:
Hey,
how come the followind doesn't seem to work ...
for i in `ls -1 /some/dir` ; do
cat /some/dir/$i /usr/fruits.txt
done
because $i contains lines like:
drw-r--r--2 sunnysunny
Use another loop:
for $fruit in `cat fruits`; do
while [ -z $ANS ]; do
echo -n Do you like $fruit
read ANS
done
done
hey, thanks for the code above :)
It works, I just needed to add a unset ANS above the while [ blah blah ] line
thanks
Sunny
hey,
Say i'm doing a loop in which I ask someone a question ...
for $fruit in `cat /usr/fruits.txt` ;
do
echo -n Do you like $fruit
read ANS
if [ -z $ANS ] ; then
# NEED HELP WITH CODE HERE
fi
done
On Tue, Oct 30, 2001 at 06:48:42PM -0500, Sunny Dubey wrote:
for $fruit in `cat /usr/fruits.txt` ;
do
echo -n Do you like $fruit
read ANS
if [ -z $ANS ] ; then
# NEED HELP WITH CODE HERE
fi
done
how can
On Tue, Oct 30, 2001 at 06:48:42PM -0500, Sunny Dubey wrote:
for $fruit in `cat /usr/fruits.txt` ;
do
echo -n Do you like $fruit
read ANS
if [ -z $ANS ] ; then
# NEED HELP WITH CODE HERE
fi
done
how can I
(sorry if this got send to the list twice)
Hi,
I have a file which as a list of varions itmes (example below)
# /usr/food/fruits.txt
banana medium yellow
apple small red
watermelon big green
plum small red
etc etc etc ...
when I create the following loop ...
for $fruit in `cat
on Sat, Oct 27, 2001 at 06:10:35PM -0400, Sunny Dubey ([EMAIL PROTECTED]) wrote:
(sorry if this got send to the list twice)
Hi,
I have a file which as a list of varions itmes (example below)
# /usr/food/fruits.txt
banana medium yellow
apple small red
watermelon big green
plum small
Hello,
I believe something like this should work:
for $fruit in 'cat /usr/food/fruits.txt'
do echo -n $fruit
done
echo -n omits the trailing newline as seen in 'man echo'
Enjoy,
Cameron Matheson
On Sat, Oct 27, 2001 at 06:10:35PM -0400, Sunny Dubey wrote:
I have a file which as a
Hi, (I am not programmer)
On Sat, Oct 27, 2001 at 06:10:35PM -0400, Sunny Dubey wrote:
I have a file which as a list of varions itmes (example below)
# /usr/food/fruits.txt
banana medium yellow
apple small red
watermelon big green
plum small red
etc etc etc ...
when I create the
On Sat, Oct 27, 2001 at 06:10:35PM -0400, Sunny Dubey wrote:
my question is, how do I get it to print the list with each line as the
variable $fruit, as opposed to $fruit being each word.
Set the IFS variable to exclude spaces. (Normally it contains a space, a
tab, and a newline.) $IFS
Could someone point me in a good direction to start learning some shell
scripting? I can do the extreme basic stuff, but I'd like to learn a lot
more than I already know. Any references for learning perl? shell scripting
in general?
TIA
--
dave wiard ([EMAIL PROTECTED])
The light that burns
Could someone point me in a good direction to start learning some shell
scripting? I can do the extreme basic stuff, but I'd like to learn a lot
more than I already know. Any references for learning perl? shell scripting
in general?
If you like bash, then get Learning the Bash Shell from
Could someone point me in a good direction to start learning some shell
scripting? I can do the extreme basic stuff, but I'd like to learn a lot
more than I already know. Any references for learning perl? shell scripting
in general?
Shell Scripting - for starters look at /etc/init.d and figure
* David Wiard [EMAIL PROTECTED] writes:
Could someone point me in a good direction to start learning some shell
scripting? I can do the extreme basic stuff, but I'd like to learn a lot
O'Reilly's Debian book has a chapter about bash. Here is the online version:
http://www.ora.com/catalog
Hello al I was wondering if someone can tell me of some website taht
talk aboput shell scripting in great detail thanks
Yeah, try www.perl.com
Oh, I just crack myself up...
I learned to do shell scripting using a copy of Unix in a Nutshell, but I
think there's a thin O'Reilly book on bash
Hello al I was wondering if someone can tell me of some website taht
talk aboput shell scripting in great detail thanks
On Thu, Apr 01, 1999 at 03:46:27PM -0600, Craig Hancock wrote:
Hello al I was wondering if someone can tell me of some website taht
talk aboput shell scripting in great detail thanks
The definitive reference for bash can be found (among other places)
at: http://www.gnu.org/manual/bash-2.02
i want to write a little shell script called addmailuser w/c executes
adduser --conf /etc/mailuser.conf where mailuser.conf contains info like
making their default shell /bin/false.
in dos, that addmailuser file will simply contain adduser --conf
/etc/mailuser.conf %1 where %1 is the first
I want to write a little shell script called addmailuser w/c
executes adduser --conf /etc/mailuser.conf where mailuser.conf
contains info like making their default shell /bin/false. in dos,
that addmailuser file will simply contain adduser --conf
/etc/mailuser.conf %1 where %1 is the first
in dos, that addmailuser file will simply contain adduser --conf
/etc/mailuser.conf %1 where %1 is the first parameter that it sees, that
is, the name of the user i want to add ... but how do u do this in linux ???
$1 is the first parm with bash. For simple scripts, see the files in
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