Re: ctrl-p not working as expected at bash command prompt on debian docker container

[Steve Dondley]

On 2021-04-26 02:43 PM, Steve Dondley wrote:

I downloaded and ran this docker image: https://hub.docker.com/_/debian

It works, but typically when I hit the ctrl-p key at the bash prompt,
it acts like the up arrow key and shows the previous command.

However, I have to hit ctrl-p twice to show the previous command and
twice each time to show the next previous command.

Ctrl-n works as expected with each single stroke going to the next 
command.


The debian machine doesn't have a lot of packages installed so I'm
guessing there is some basic config package I need to install.

Does anyone have any idea?


OK, issue fixed with: https://stackoverflow.com/a/41820278/1641112

Basically, modify the ~/.docker/config.json with somethis line:

"detachKeys": "ctrl-@",

ctrl-p not working as expected at bash command prompt on debian docker container

[Steve Dondley]

I downloaded and ran this docker image: https://hub.docker.com/_/debian

It works, but typically when I hit the ctrl-p key at the bash prompt, it 
acts like the up arrow key and shows the previous command.


However, I have to hit ctrl-p twice to show the previous command and 
twice each time to show the next previous command.


Ctrl-n works as expected with each single stroke going to the next 
command.


The debian machine doesn't have a lot of packages installed so I'm 
guessing there is some basic config package I need to install.


Does anyone have any idea?

Re: Bash command completion

[David]

On 7 July 2016 at 08:06, Lisi Reisz  wrote:
>
> So have you followed the suggestion to test whether it is in fact bash that
> you are in fact using?
>
> lisi@Tux-II:~$ echo $SHELL
> /bin/bash
> lisi@Tux-II:~$

In case anyone is unaware, it might be generally helpful to clarify what
this test actually does.

Here's a demo on my system:
# my login shell is bash:
[david@kablamm]$ echo $SHELL
/bin/bash
# now I run dash
[david@kablamm]$ dash
# now repeat the above test, inside dash:
$ echo $SHELL
/bin/bash
$
# and now I type ^D to return to the login bash
[david@kablamm]$

Which demonstrates that $SHELL does not report the running shell.
It reports the user's login shell. Which might be different to whatever
shell the echo command happens to be typed into.

$SHELL is not controlled by the running shell. It is supplied to it in
its runtime environment.

$SHELL is set by 'login' process and exported to child processes.
It is set to the value specified in /etc/passwd. This is mentioned
in 'man 1 login'.

Many shells are not capable of identifying themselves directly.
bash can do it like this:
[david@kablamm]$ echo $BASH_VERSION
Neither sh nor dash have this capability.

Re: Bash command completion

[Glenn English]

> On Jul 6, 2016, at 11:33 PM, Peter Ludikovsky  wrote:
> 
> No, chsh changes the login shell for the user within /etc/passwd. It
> won't affect any currently active shells.
> 
> What happens when you do an
>  /bin/bash --login
> That should start a login shell. If you still only get the tab
> character, check if you've got the line
>  set -o vi
> in /etc/profile, /etc/bash*, ~/.profile, or ~/.bash* anywhere.

Fired up the naughty computer, changed the /bin/sh link back to dash, did a 
chsh to /bin/bash in /etc/passwd, rebooted. Worked like a champ. The question 
now is who/what put /bin/sh in /etc/passwd...

Thank you all for your comments and help. This was making me crazy. I've 
learned IP and *nix on my own, and I get to know the corners by trial and 
error. This one was a pretty trivial, but beyond what I could figure out. 
Thanks all, I've learned a new corner...

Sure am glad I didn't have to deal with MS tech support :-)

-- 
Glenn English

Re: Bash command completion

[Peter Ludikovsky]

No, chsh changes the login shell for the user within /etc/passwd. It
won't affect any currently active shells.

What happens when you do an
  /bin/bash --login
That should start a login shell. If you still only get the tab
character, check if you've got the line
  set -o vi
in /etc/profile, /etc/bash*, ~/.profile, or ~/.bash* anywhere.

Am 07.07.2016 um 07:14 schrieb Glenn English:
> 
>> On Jul 6, 2016, at 10:38 PM, Peter Ludikovsky  wrote:
>>
>> After an chsh, you have to log out & in again.
> 
> I thought of that -- I logged out and back in, no joy. I rebooted, same thing.
> 
> I wasn't too surprised. I assumed that rebooting the machine would just put 
> stuff back the way it was. And that the problem was with the scripts in the 
> user directories.
> 
> That wasn't it, and I was working on the wrong things. Lisi's insight led me 
> to this afternoon's solution. And deloptes came up with another very 
> interesting thought (which I haven't investigated yet).
> 
> Does chsh change things for good -- in passwd or in a shell var or a link? I 
> thought it just started a shell, as if it was just starting a new program.
> 



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Re: Bash command completion

[Glenn English]

> On Jul 6, 2016, at 10:38 PM, Peter Ludikovsky  wrote:
> 
> After an chsh, you have to log out & in again.

I thought of that -- I logged out and back in, no joy. I rebooted, same thing.

I wasn't too surprised. I assumed that rebooting the machine would just put 
stuff back the way it was. And that the problem was with the scripts in the 
user directories.

That wasn't it, and I was working on the wrong things. Lisi's insight led me to 
this afternoon's solution. And deloptes came up with another very interesting 
thought (which I haven't investigated yet).

Does chsh change things for good -- in passwd or in a shell var or a link? I 
thought it just started a shell, as if it was just starting a new program.

-- 
Glenn English

Re: Bash command completion

[Peter Ludikovsky]

After an chsh, you have to log out & in again.

Am 07.07.2016 um 00:17 schrieb Glenn English:
> 
>> On Jul 6, 2016, at 4:06 PM, Lisi Reisz  wrote:
>>
>> So have you followed the suggestion to test whether it is in fact bash that 
>> you are in fact using?
> 
> Yes. And I wasn't -- it was dash.
> 
> So I:
> 
> 'chsh -s /bin/bash'
> 'ls Do\t'
> 
> and got a tab.
> 
>> lisi@Tux-II:~$ echo $SHELL
>> /bin/bash
>> lisi@Tux-II:~$
> 
> I'll try your codelet. Before and after chsh...
> 



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Re: Bash command completion [Fixed]

[Glenn English]

> On Jul 6, 2016, at 5:59 PM, deloptes  wrote:
> 
> What is the default for the user in /etc/passwd ?

Good question. Another very likely error. And I'd answer it if the massively 
obsolete box wasn't powered down and in the give-away bin :-)

I'll look into it tomorrow.

-- 
Glenn English

Re: Bash command completion [Fixed]

[deloptes]

Glenn English wrote:

> 
>> On Jul 6, 2016, at 4:06 PM, Lisi Reisz  wrote:
> 
> There were far too many 'sh's in scripts in /etc, so I changes /bin/sh
> from pointing at dash to pointing at bash.

What is the default for the user in /etc/passwd ?

> 
> That fixed it.
> 
> Lisi, as usual, found the problem :-)
> 

Re: Bash command completion [Fixed]

[Glenn English]

> On Jul 6, 2016, at 4:06 PM, Lisi Reisz  wrote:

There were far too many 'sh's in scripts in /etc, so I changes /bin/sh from 
pointing at dash to pointing at bash.

That fixed it.

Lisi, as usual, found the problem :-)

-- 
Glenn English

Re: Bash command completion

[Glenn English]

> On Jul 6, 2016, at 4:06 PM, Lisi Reisz  wrote:
> 
> lisi@Tux-II:~$ echo $SHELL
> /bin/bash
> lisi@Tux-II:~$

Ahah! As root, echo $SHELL says /bin/bash. As a user, it says /bin/sh. And sh 
is dash. That explains a *lot*. Maybe.

I'll see if I can find the dastardly script that does that. 

Thanks Lisi.

-- 
Glenn English

Re: Bash command completion

[Glenn English]

> On Jul 6, 2016, at 4:06 PM, Lisi Reisz  wrote:
> 
> So have you followed the suggestion to test whether it is in fact bash that 
> you are in fact using?

Yes. And I wasn't -- it was dash.

So I:

'chsh -s /bin/bash'
'ls Do\t'

and got a tab.

> lisi@Tux-II:~$ echo $SHELL
> /bin/bash
> lisi@Tux-II:~$

I'll try your codelet. Before and after chsh...

-- 
Glenn English

Re: Bash command completion

[Lisi Reisz]

On Wednesday 06 July 2016 22:52:58 Glenn English wrote:
> > On Jul 6, 2016, at 2:29 PM, Charlie Kravetz 
> > wrote:
> >
> > -BEGIN PGP SIGNED MESSAGE-
> > Hash: SHA256
> >
> > There should be a set of commands towards the bottom
> > of /etc/bash.bashrc to enable completion. The commands are:
> >
> > # enable bash completion in interactive shells
> > #if ! shopt -oq posix; then
> > #  if [ -f /usr/share/bash-completion/bash_completion ]; then
> > #. /usr/share/bash-completion/bash_completion
> > #  elif [ -f /etc/bash_completion ]; then
> > #. /etc/bash_completion
> > #  fi
> > #fi
>
> There were.
>
> > You usually have to uncomment these lines.
>
> I did and logged out and back in, and rebooted, but no change; it's still
> bent.

So have you followed the suggestion to test whether it is in fact bash that 
you are in fact using?

lisi@Tux-II:~$ echo $SHELL
/bin/bash
lisi@Tux-II:~$

Lisi

Re: Bash command completion

[Glenn English]

> On Jul 6, 2016, at 3:16 PM, Sven Arvidsson  wrote:
> 
> Are you sure that your user uses bash for the login shell? There was a
> transition from bash to dash some releases ago.

Nope. According to 'man sh', it's dash. I understood that dash is a fixed bash.

But why would it work for root and not for a user? And why is it OK on all the 
other wheezy boxes around here?

And how do I replace it? I'll poke around a bit more...

-- 
Glenn English

Re: Bash command completion

[Glenn English]

> On Jul 6, 2016, at 2:29 PM, Charlie Kravetz  
> wrote:
> 
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA256
> 
> There should be a set of commands towards the bottom
> of /etc/bash.bashrc to enable completion. The commands are:
> 
> # enable bash completion in interactive shells
> #if ! shopt -oq posix; then
> #  if [ -f /usr/share/bash-completion/bash_completion ]; then
> #. /usr/share/bash-completion/bash_completion
> #  elif [ -f /etc/bash_completion ]; then
> #. /etc/bash_completion
> #  fi
> #fi

There were.

> You usually have to uncomment these lines. 

I did and logged out and back in, and rebooted, but no change; it's still bent.

-- 
Glenn English

Re: Bash command completion

[Sven Arvidsson]

On Wed, 2016-07-06 at 14:29 -0600, Charlie Kravetz wrote:
> On Wed, 6 Jul 2016 13:48:24 -0600
> Glenn English  wrote:
> 
> > 
> > I put wheezy on a 386 computer last night ('aptitude dist-upgrade'
> > from squeeze -- it'd been in the junk box for a while), and when I
> > hit tab, bash just gives me a tab -- I have to type the whole
> > command manually. This happens only for the user; root works fine.
> > 
> > I've copied the .* scripts (the ones that looked relevant to me,
> > anyway) from /root and from a user directory on one of the (wheezy)
> > servers, and accomplished nothing. 
> > 
> > I looked at the scripts in /etc, but saw nothing that looked
> > promising -- there was some talk about bash-completion, but I could
> > see nothing in those scripts about a difference between user and
> > root.
> > 
> > A solution or a link to a solution would be much appreciated. 
> > 
> There should be a set of commands towards the bottom
> of /etc/bash.bashrc to enable completion. The commands are:
> 
> # enable bash completion in interactive shells
> #if ! shopt -oq posix; then
> #  if [ -f /usr/share/bash-completion/bash_completion ]; then
> #. /usr/share/bash-completion/bash_completion
> #  elif [ -f /etc/bash_completion ]; then
> #. /etc/bash_completion
> #  fi
> #fi

That's for the bash-completion package.

A normal bash shell should give completion for anything in PATH without
it, and it sounds like even this doesn't work for the OP?

Are you sure that your user uses bash for the login shell? There was a
transition from bash to dash some releases ago.

-- 
Cheers,
Sven Arvidsson
http://www.whiz.se



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Re: Bash command completion

[Charlie Kravetz]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA256

On Wed, 6 Jul 2016 13:48:24 -0600
Glenn English  wrote:

>I put wheezy on a 386 computer last night ('aptitude dist-upgrade' from 
>squeeze -- it'd been in the junk box for a while), and when I hit tab, bash 
>just gives me a tab -- I have to type the whole command manually. This happens 
>only for the user; root works fine.
>
>I've copied the .* scripts (the ones that looked relevant to me, anyway) from 
>/root and from a user directory on one of the (wheezy) servers, and 
>accomplished nothing. 
>
>I looked at the scripts in /etc, but saw nothing that looked promising -- 
>there was some talk about bash-completion, but I could see nothing in those 
>scripts about a difference between user and root.
>
>A solution or a link to a solution would be much appreciated. 
>

There should be a set of commands towards the bottom
of /etc/bash.bashrc to enable completion. The commands are:

# enable bash completion in interactive shells
#if ! shopt -oq posix; then
#  if [ -f /usr/share/bash-completion/bash_completion ]; then
#. /usr/share/bash-completion/bash_completion
#  elif [ -f /etc/bash_completion ]; then
#. /etc/bash_completion
#  fi
#fi

You usually have to uncomment these lines. 

- -- 
Charlie Kravetz
Linux Registered User Number 425914
[http://linuxcounter.net/user/425914.html]
Never let anyone steal your DREAM.   [http://keepingdreams.com]
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Bash command completion

[Glenn English]

I put wheezy on a 386 computer last night ('aptitude dist-upgrade' from squeeze 
-- it'd been in the junk box for a while), and when I hit tab, bash just gives 
me a tab -- I have to type the whole command manually. This happens only for 
the user; root works fine.

I've copied the .* scripts (the ones that looked relevant to me, anyway) from 
/root and from a user directory on one of the (wheezy) servers, and 
accomplished nothing. 

I looked at the scripts in /etc, but saw nothing that looked promising -- there 
was some talk about bash-completion, but I could see nothing in those scripts 
about a difference between user and root.

A solution or a link to a solution would be much appreciated. 

-- 
Glenn English

Re: bash command

[Darac Marjal]

On Fri, Oct 21, 2011 at 03:42:36PM -0400, Stefan Monnier wrote:
  Sadly, this can't be done in-place, so you'll either need to use mv to
  replace /etc/conf.file with /etc/conf.file.new or repeat the loop (with
  no substitution) to copy /etc/conf.file.new into /etc/conf.file.
 
 It can be done inplace with `rm' in place or `mv':
 
 (rm /etc/conf.file;
  while read line; do
  echo ${line/old_word/new_word}
  done /etc/conf.file) /etc/conf.file

True, but that's using external tools. My point was that it's difficult
to do simply with bash (which is what the OP asked for). If you're
allowing the use of external tools, sed is much easier.


-- 
Darac Marjal


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Re: bash command

[Harry Putnam]

Darac Marjal mailingl...@darac.org.uk writes:

 Sadly, this can't be done in-place, so you'll either need to use mv to
 replace /etc/conf.file with /etc/conf.file.new or repeat the loop (with
 no substitution) to copy /etc/conf.file.new into /etc/conf.file.

Maybe now with bash, but with perl it can be done in place quite
handily.

 perl -i -pe 's/WORDS/REPLACEMENT/' my.conf


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Re: bash command

[Stefan Monnier]

 Sadly, this can't be done in-place, so you'll either need to use mv to
 replace /etc/conf.file with /etc/conf.file.new or repeat the loop (with
 no substitution) to copy /etc/conf.file.new into /etc/conf.file.

It can be done inplace with `rm' in place or `mv':

(rm /etc/conf.file;
 while read line; do
 echo ${line/old_word/new_word}
 done /etc/conf.file) /etc/conf.file


-- Stefan


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bash command

[Jesus arteche]

Hey guys,

I want to create  a script to change some words in some sonf files at the
start up of the system...do you know the command in bash for search the word
and replace it??

Thanks

Re: bash command

[Raf Czlonka]

On Thu, Oct 20, 2011 at 09:29:35AM BST, Jesus arteche wrote:
 I want to create  a script to change some words in some sonf files at the
 start up of the system...do you know the command in bash for search the word
 and replace it??

You don't need bash for it, sed's your friend, e.g.:

% sed -i 's/old_word/new_word/' /etc/conf.file

I highly recommend.

% man sed

Regards,
-- 
Raf


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Re: bash command

[kuLa]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 20/10/11 09:29, Jesus arteche wrote:
 Hey guys,
 
 I want to create  a script to change some words in some sonf files at
 the start up of the system...do you know the command in bash for search
 the word and replace it??

Well, I don't know about bash command but I know about sed.

- -- 

|_|0|_|  |
|_|_|0| Heghlu'Meh QaQ jajVam  |
|0|0|0|  kuLa -  |

gpg --keyserver pgp.mit.edu --recv-keys 0xC100B4CA
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Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/

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=XBdl
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Re: bash command

[Teemu Likonen]

* 2011-10-20T09:29:35+01:00 * Jesus arteche wrote:

 I want to create a script to change some words in some sonf files at
 the start up of the system...do you know the command in bash for
 search the word and replace it??

Sounds like you need sed command and its s/.../.../ command. Probably
sed's --in-place option will be useful.


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Re: bash command

[Darac Marjal]

On Thu, Oct 20, 2011 at 09:36:50AM +0100, Raf Czlonka wrote:
 On Thu, Oct 20, 2011 at 09:29:35AM BST, Jesus arteche wrote:
  I want to create  a script to change some words in some sonf files at the
  start up of the system...do you know the command in bash for search the word
  and replace it??
 
 You don't need bash for it, sed's your friend, e.g.:
 
 % sed -i 's/old_word/new_word/' /etc/conf.file

Proving that TIMTOWTDI (or, You don't need sed for it, bash's your
friend):

while read line; do
   echo ${line/old_word/new_word}
done  /etc/conf.file  /etc/conf.file.new

Sadly, this can't be done in-place, so you'll either need to use mv to
replace /etc/conf.file with /etc/conf.file.new or repeat the loop (with
no substitution) to copy /etc/conf.file.new into /etc/conf.file.

By the way, I don't recommend this method. sed IS your friend, but bash
IS capable and that's what you asked for :)

-- 
Darac Marjal


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Re: bash command

[Arno Schuring]

kuLa (deb...@kulisz.net on 2011-10-20 09:37 +0100):
 On 20/10/11 09:29, Jesus arteche wrote:
  Hey guys,
  
  I want to create  a script to change some words in some sonf files
  at the start up of the system...do you know the command in bash for
  search the word and replace it??
 
 Well, I don't know about bash command but I know about sed.

Theoretically it's possible with bash (=3) as well:
$ VAR=oldword
$ echo $VAR
oldword
$ echo ${VAR/old/new}
newword

And use a read loop over all the lines in the file :)


Regards,
Arno


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Re: bash command

[Doug]

On 10/20/2011 04:37 AM, kuLa wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 20/10/11 09:29, Jesus arteche wrote:

Hey guys,

I want to create  a script to change some words in some sonf files at
the start up of the system...do you know the command in bash for search
the word and replace it??

Well, I don't know about bash command but I know about sed.

- -- 


|_|0|_|  |
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|0|0|0|  kuLa -  |



What the heck is that, Klingon?
What does it mean?  --doug

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--
Blessed are the peacemakers...for they shall be shot at from both sides. --A. 
M. Greeley


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Re: bash command

[Raf Czlonka]

On Thu, Oct 20, 2011 at 06:31:16PM BST, Doug wrote:
 |_|0|_|  |
 |_|_|0| Heghlu'Meh QaQ jajVam  |
 |0|0|0|  kuLa -  |
 
 What the heck is that, Klingon?
 What does it mean?  --doug

http://en.wikiquote.org/wiki/Klingon_proverbs

-- 
Raf


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Re: bash command

[kuLa]

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Hash: SHA1

On 20/10/11 18:31, Doug wrote:

 - -- 
 |_|0|_|  |
 |_|_|0| Heghlu'Meh QaQ jajVam  |
 |0|0|0|  kuLa -  |

 What the heck is that, Klingon?
 What does it mean?  --doug

lol, yes it is :-)
http://www.google.co.uk/search?aq=fgcx=csourceid=chromeclient=ubuntuchannel=csie=UTF-8q=Heghlu'Meh+QaQ+jajVam

- -- 

|_|0|_|  |
|_|_|0| Heghlu'Meh QaQ jajVam  |
|0|0|0|  kuLa -  |

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Re: ^C show up on bash command line

[Sven Joachim]

On 2010-09-28 03:25 +0200, T o n g wrote:

 Now,when pressing ^C to abolish changes to bash command, the actual ^C 
 will show up where the cursor is. IIRC, previously it wasn't like this -- 
 editing aborted with actual command intact, without being messed up with 
 ^C.

This looked like a bug to me, but it was actually added deliberately.

 Is it possible to get back the old bash ^C behaviour? 

$ echo 'set echo-control-characters off'  ~/.inputrc

See the following thread:

http://lists.gnu.org/archive/html/bug-bash/2010-02/threads.html#00115

Sven


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Re: need help with bash command

[s. keeling]

François Cerbelle [EMAIL PROTECTED]:
  Bob McGowan a écrit :
  
  Why do this with a `ls` when a simple * will work?
  for a in *; do ...
 
  Hum... Because ! ;-)
 
  In fact, first, I used to do this, because I seldomly use all files in a 
  single directory, but often a pattern in a whole tree with find, second 
  because there was a ls in the original problem and third because I did 

Ah, the original homework problem?


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Re: need help with bash command

[François Cerbelle]

Le Ven 21 novembre 2008 02:10, Ding Honghui a écrit :.jpg
 [EMAIL PROTECTED]:~/test/jpg$ ls | awk -F_ '{print $2}'
 3563
 3616
 3620

IMHO, awk is quite oversized just to cut fields... It is typically a cut
job : cut -d_ -f2 ;-)

Fanfan
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need help with bash command

[tôba]

Hello,

Can you help me with a bash command?
I a directory, I want to extract with a command line for jpg name files.
I have a lot of:

icon_3620_0_1.jpg
icon_3616_0_1.jpg
icon_3563_0_1.jpg

I want to extract the number betwen icon_*_0_1.jpg AND I want to use
the entire filename too in the same command.

The goal is to create a directory called 3620 and move the
icon_3620_0_1.jpg file into this directory.

So, I did like this for test:

# for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
icon_$a_0_1.jpg; done

I hope it should give me:

icon_3620_0_1.jpg
icon_3616_0_1.jpg
icon_3563_0_1.jpg

like the original

BUT in my surprise, it shows:

icon_.jpg
icon_.jpg
icon_.jpg
icon_.jpg

Why?
Where is the $a_0_1 gone? Surely I missed something but I cannot
find it.
Can you tell me how to reproduce the exact filename so?

Best regards,

--
Tôba


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Re: need help with bash command

[Eduardo M KALINOWSKI]

tôba escreveu:
 Hello,

 Can you help me with a bash command?
 I a directory, I want to extract with a command line for jpg name files.
 I have a lot of:

 icon_3620_0_1.jpg
 icon_3616_0_1.jpg
 icon_3563_0_1.jpg

 I want to extract the number betwen icon_*_0_1.jpg AND I want to use
 the entire filename too in the same command.

 The goal is to create a directory called 3620 and move the
 icon_3620_0_1.jpg file into this directory.

 So, I did like this for test:

 # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
 icon_$a_0_1.jpg; done
   

Try icon_${a}_0_1.jpg

Why it works is left as an exercise for the reader. (Hint: man bash).

 I hope it should give me:

 icon_3620_0_1.jpg
 icon_3616_0_1.jpg
 icon_3563_0_1.jpg

 like the original

 BUT in my surprise, it shows:

 icon_.jpg
 icon_.jpg
 icon_.jpg
 icon_.jpg

 Why?
 Where is the $a_0_1 gone? Surely I missed something but I cannot
 find it.
 Can you tell me how to reproduce the exact filename so?
   

-- 
Eduardo M Kalinowski
[EMAIL PROTECTED]


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[SOLVED] Re: need help with bash command

[tôba]

On Thu, 20 Nov 2008 13:28:17 -0200
Eduardo M KALINOWSKI [EMAIL PROTECTED] wrote:

 tôba escreveu:
  Hello,
 
  Can you help me with a bash command?
  I a directory, I want to extract with a command line for jpg name
  files. I have a lot of:
 
  icon_3620_0_1.jpg
  icon_3616_0_1.jpg
  icon_3563_0_1.jpg
 
  I want to extract the number betwen icon_*_0_1.jpg AND I want to use
  the entire filename too in the same command.
 
  The goal is to create a directory called 3620 and move the
  icon_3620_0_1.jpg file into this directory.
 
  So, I did like this for test:
 
  # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
  icon_$a_0_1.jpg; done

 
 Try icon_${a}_0_1.jpg
 
 Why it works is left as an exercise for the reader. (Hint: man bash).


Ok,
Thank you Eduardo, it works,

Regards,

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Tôba


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Re: need help with bash command

[François Cerbelle]

Le Jeu 20 novembre 2008 16:18, tôba a écrit :
 icon_3620_0_1.jpg
 icon_3616_0_1.jpg
 icon_3563_0_1.jpg

 The goal is to create a directory called 3620 and move the
 icon_3620_0_1.jpg file into this directory.

 So, I did like this for test:
 # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
 icon_$a_0_1.jpg; done

for a in `ls`; do DIR=`echo $a | cut -d_ -f2`;  echo  mkdir -p $DIR
/$a; echo rmdir $DIR/$a; echo mv $a $DIR/$a;  done

First, compute the dirname to put the file in.
Second, create a directory corresponding to the full destination path
including the filename to create the parent directory if it does not
exists
Third, remove the directory with the name of the file
Fourth, move the file

The command line includes echo for you to see what will be done. When
you want the job to be done, just remove the echo before mkdir, rmdir
and mv.




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Re: need help with bash command

[Bob McGowan]

On Thu, 2008-11-20 at 17:27 +0100, François Cerbelle wrote:
 Le Jeu 20 novembre 2008 16:18, tôba a écrit :
  icon_3620_0_1.jpg
  icon_3616_0_1.jpg
  icon_3563_0_1.jpg
 
  The goal is to create a directory called 3620 and move the
  icon_3620_0_1.jpg file into this directory.
 
  So, I did like this for test:
  # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
  icon_$a_0_1.jpg; done
 
 for a in `ls`; do DIR=`echo $a | cut -d_ -f2`;  echo  mkdir -p $DIR
 /$a; echo rmdir $DIR/$a; echo mv $a $DIR/$a;  done
...

Why do this with a `ls` when a simple * will work?

for a in *; do ...

 
-- 
Bob McGowan


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Re: need help with bash command

[H.S.]

tôba wrote:
 Hello,
 
 Can you help me with a bash command?
 I a directory, I want to extract with a command line for jpg name files.
 I have a lot of:
 
 icon_3620_0_1.jpg
 icon_3616_0_1.jpg
 icon_3563_0_1.jpg
 
 I want to extract the number betwen icon_*_0_1.jpg AND I want to use
 the entire filename too in the same command.
 
 The goal is to create a directory called 3620 and move the
 icon_3620_0_1.jpg file into this directory.
 

I am sure somebody would have solved this already, but here what I do in
such a situation:

#in the directory with the images. This will create dirs based on the
#numbers in the image filenames and move that file to its corresponding
#directory
for f in icon__0_1.jpg; do
  num=`echo $f | sed -e 's/icon_\([0-9]*\)_0_1\.jpg/\1/'`; #get number
  mkdir ${num};
  mv $f ${num};
done



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Re: need help with bash command

[François Cerbelle]

Bob McGowan a écrit :

for a in `ls`; do DIR=`echo $a | cut -d_ -f2`;  echo  mkdir -p $DIR
/$a; echo rmdir $DIR/$a; echo mv $a $DIR/$a;  done

...
Why do this with a `ls` when a simple * will work?
for a in *; do ...


Hum... Because ! ;-)

In fact, first, I used to do this, because I seldomly use all files in a 
single directory, but often a pattern in a whole tree with find, second 
because there was a ls in the original problem and third because I did 
not optimize the command ;-)


I'm sure it is possible to replace the mkdir -p xxx; rmdir xxx with 
something better. For exemple in the problem, it was not needed, a 
simple mkdir would do the job.


Fanfan


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Re: need help with bash command

[Ding Honghui]

tôba wrote:
 Hello,
 
 Can you help me with a bash command?
 I a directory, I want to extract with a command line for jpg name files.
 I have a lot of:
 
 icon_3620_0_1.jpg
 icon_3616_0_1.jpg
 icon_3563_0_1.jpg
 
 I want to extract the number betwen icon_*_0_1.jpg AND I want to use
 the entire filename too in the same command.
 
 The goal is to create a directory called 3620 and move the
 icon_3620_0_1.jpg file into this directory.
 
 So, I did like this for test:
 
 # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
 icon_$a_0_1.jpg; done
 
 I hope it should give me:
 
 icon_3620_0_1.jpg
 icon_3616_0_1.jpg
 icon_3563_0_1.jpg
 
 like the original
 
 BUT in my surprise, it shows:
 
 icon_.jpg
 icon_.jpg
 icon_.jpg
 icon_.jpg
 
 Why?
 Where is the $a_0_1 gone? Surely I missed something but I cannot
 find it.
 Can you tell me how to reproduce the exact filename so?
 
 Best regards,
 
 --
 Tôba
 
 

Hello, do you means like that?
for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
icon_${a}_0_1.jpg; done

A better way maybe is:
[EMAIL PROTECTED]:~/test/jpg$ touch icon_3563_0_1.jpg  icon_3616_0_1.jpg
icon_3620_0_1.jpg
[EMAIL PROTECTED]:~/test/jpg$ ls
icon_3563_0_1.jpg  icon_3616_0_1.jpg  icon_3620_0_1.jpg
[EMAIL PROTECTED]:~/test/jpg$ ls | awk -F_ '{print $2}'
3563
3616
3620
[EMAIL PROTECTED]:~/test/jpg$


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Re: need help with bash command

[tôba]

On Fri, 21 Nov 2008 09:10:08 +0800
Ding Honghui [EMAIL PROTECTED] wrote:

 tôba wrote:
  Hello,
  
  Can you help me with a bash command?
  I a directory, I want to extract with a command line for jpg name
  files. I have a lot of:
  
  icon_3620_0_1.jpg
  icon_3616_0_1.jpg
  icon_3563_0_1.jpg
  
  I want to extract the number betwen icon_*_0_1.jpg AND I want to use
  the entire filename too in the same command.
  
  The goal is to create a directory called 3620 and move the
  icon_3620_0_1.jpg file into this directory.
  
  So, I did like this for test:
  
  # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
  icon_$a_0_1.jpg; done
  
  I hope it should give me:
  
  icon_3620_0_1.jpg
  icon_3616_0_1.jpg
  icon_3563_0_1.jpg
  
  like the original
  
  BUT in my surprise, it shows:
  
  icon_.jpg
  icon_.jpg
  icon_.jpg
  icon_.jpg
  
  Why?
  Where is the $a_0_1 gone? Surely I missed something but I cannot
  find it.
  Can you tell me how to reproduce the exact filename so?
  
  Best regards,
  
  --
  Tôba
  
  
 
 Hello, do you means like that?
 for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
 icon_${a}_0_1.jpg; done
 
 A better way maybe is:
 [EMAIL PROTECTED]:~/test/jpg$ touch icon_3563_0_1.jpg  icon_3616_0_1.jpg
 icon_3620_0_1.jpg
 [EMAIL PROTECTED]:~/test/jpg$ ls
 icon_3563_0_1.jpg  icon_3616_0_1.jpg  icon_3620_0_1.jpg
 [EMAIL PROTECTED]:~/test/jpg$ ls | awk -F_ '{print $2}'
 3563
 3616
 3620
 [EMAIL PROTECTED]:~/test/jpg$

Yes, I did like you suggested and it's already done now.

Best regards,

--
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Re: bash command timer

[Hugo Vanwoerkom]

Roberto C. Sanchez wrote:

On Tue, Sep 26, 2006 at 05:19:47PM -0500, Hugo Vanwoerkom wrote:

Ron Johnson wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 09/25/06 13:54, Hugo Vanwoerkom wrote:

Hi,

I am looking for a way to time scripts, how long they take to execute.

That has appeared in the list before, but now I can't find it.

I could write it myself, but I bet it exists already.

Are you referring to time(1)?

 $ time ./testmem
 Out of memory at 2802 M

 real0m0.020s
 user0m0.002s
 sys 0m0.017s


Aha! I am and it is a package that I did not have installed ;-)



Just remember that if you run time without a patch qualification, you
get the shell's builtin time function.  If you want to use the time
binary, then be sure to specify /usr/bin/time.

Regards,

-Roberto


I ran right into this when I did not specify /usr/bin/time.
However... I do not think time is what I want. What I want is a 
chronometer function that works just like one: you click and it shows 
0:00, click and it starts running, click and it shows how far it got...


I wrote a command to do that and keep the elapsed time in a environment 
variable, but setenv(CHRONO, msecs.data(), 1)); does not set the 
variable $CHRONO, so I have to rethink.


H

















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Re: bash command timer

[Hugo Vanwoerkom]

Ron Johnson wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 09/25/06 13:54, Hugo Vanwoerkom wrote:

Hi,

I am looking for a way to time scripts, how long they take to execute.

That has appeared in the list before, but now I can't find it.

I could write it myself, but I bet it exists already.


Are you referring to time(1)?

  $ time ./testmem
  Out of memory at 2802 M

  real0m0.020s
  user0m0.002s
  sys 0m0.017s



Aha! I am and it is a package that I did not have installed ;-)

Thanks!

H


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Re: bash command timer

[Roberto C. Sanchez]

On Tue, Sep 26, 2006 at 05:19:47PM -0500, Hugo Vanwoerkom wrote:
 Ron Johnson wrote:
 -BEGIN PGP SIGNED MESSAGE-
 Hash: SHA1
 
 On 09/25/06 13:54, Hugo Vanwoerkom wrote:
 Hi,
 
 I am looking for a way to time scripts, how long they take to execute.
 
 That has appeared in the list before, but now I can't find it.
 
 I could write it myself, but I bet it exists already.
 
 Are you referring to time(1)?
 
   $ time ./testmem
   Out of memory at 2802 M
 
   real0m0.020s
   user0m0.002s
   sys 0m0.017s
 
 
 Aha! I am and it is a package that I did not have installed ;-)
 

Just remember that if you run time without a patch qualification, you
get the shell's builtin time function.  If you want to use the time
binary, then be sure to specify /usr/bin/time.

Regards,

-Roberto
-- 
Roberto C. Sanchez
http://people.connexer.com/~roberto
http://www.connexer.com


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Description: Digital signature

bash command timer

[Hugo Vanwoerkom]

Hi,

I am looking for a way to time scripts, how long they take to execute.

That has appeared in the list before, but now I can't find it.

I could write it myself, but I bet it exists already.

Thanks

H


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Re: bash command timer

[Roberto C. Sanchez]

On Mon, Sep 25, 2006 at 01:54:53PM -0500, Hugo Vanwoerkom wrote:
 Hi,
 
 I am looking for a way to time scripts, how long they take to execute.
 
 That has appeared in the list before, but now I can't find it.
 
 I could write it myself, but I bet it exists already.
 

Use /usr/bin/time, as the built-in time in bash is not very good.

Regards,

-Roberto

-- 
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http://people.connexer.com/~roberto
http://www.connexer.com


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Re: bash command timer

[Alexander Schmehl]

* Hugo Vanwoerkom [EMAIL PROTECTED] [060925 20:54]:

 I am looking for a way to time scripts, how long they take to execute.

time?


Yours sincerely,
  Alexander

-- 
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http://www.catb.org/~esr/faqs/smart-questions.html


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Re: bash command timer

[Ron Johnson]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 09/25/06 13:54, Hugo Vanwoerkom wrote:
 Hi,
 
 I am looking for a way to time scripts, how long they take to execute.
 
 That has appeared in the list before, but now I can't find it.
 
 I could write it myself, but I bet it exists already.

Are you referring to time(1)?

  $ time ./testmem
  Out of memory at 2802 M

  real0m0.020s
  user0m0.002s
  sys 0m0.017s

- --
Ron Johnson, Jr.
Jefferson LA  USA

Is common sense really valid?
For example, it is common sense to white-power racists that
whites are superior to blacks, and that those with brown skins
are mud people.
However, that common sense is obviously wrong.
-BEGIN PGP SIGNATURE-
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Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org

iD8DBQFFGEtRS9HxQb37XmcRAm+7AJ9o9PSis2SQRzTZ6F9J9Mz1TMD8zwCg7ETn
+0igi/pNCo84tO4xlNnDD28=
=5+5p
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bash command substitution problem

[Pollywog]

I have a function defined in my .bashrc as:

function lf { /bin/ls -l | grep ^- ; }

It prints the files in the CWD without listing other directories.

Suppose I want to 'chmod 600' all the files in the CWD without affecting 
directories, I try this:

chmod 600 `lf`

but I get this error:
chmod: invalid character `w' in mode string `-rw---'


What am I doing wrong?


thanks  8)






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Re: bash command substitution problem

[s. keeling]

Incoming from Pollywog:
 I have a function defined in my .bashrc as:
 
 function lf { /bin/ls -l | grep ^- ; }
 
 It prints the files in the CWD without listing other directories.
 
 Suppose I want to 'chmod 600' all the files in the CWD without affecting 
 directories, I try this:
 
 chmod 600 `lf`
 
 but I get this error:
 chmod: invalid character `w' in mode string `-rw---'
 
 
 What am I doing wrong?

You're passing it /bin/ls -l instead of /bin/ls.  -l works in the
function to pick up files only, but fails in chmod (you can't chmod 600
-rw-r--r--1 keeling  keeling  5973 Oct  7  2004 .emacs.  You
have to chmod 600 .emacs.


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Re: bash command substitution problem

[Pollywog]

On Sunday 08 May 2005 04:52 pm, s. keeling wrote:


 You're passing it /bin/ls -l instead of /bin/ls.  -l works in the
 function to pick up files only, but fails in chmod (you can't chmod 600
 -rw-r--r--1 keeling  keeling  5973 Oct  7  2004 .emacs.  You
 have to chmod 600 .emacs.

Back to the drawing board then.  

thanks

8)


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Re: bash command substitution problem

[Phil Dyer]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Pollywog said:
 I have a function defined in my .bashrc as:
 
 function lf { /bin/ls -l | grep ^- ; }
 
 It prints the files in the CWD without listing other directories.
 
 Suppose I want to 'chmod 600' all the files in the CWD without affecting 
 directories, I try this:
 
 chmod 600 `lf`
 
 but I get this error:
 chmod: invalid character `w' in mode string `-rw---'

yeah, the ls -l listing is getting you when you display.

try this one.

function lsf {
for i in *; do
if [ -f $i ]; then
echo $i
fi;
done;
}

lsf | xargs chmod 600


- --

/phil


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Re: bash command substitution problem

[Pollywog]

On Sunday 08 May 2005 05:08 pm, Phil Dyer wrote:

 try this one.

 function lsf {
 for i in *; do
 if [ -f $i ]; then
 echo $i
 fi;
 done;
 }

 lsf | xargs chmod 600

Thanks, that works.

8)


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bash command history

[Paul Miller]

Is there any way to get bash to work like 4dos? -- If you type the first
couple of characters of a command in history, it will only scroll through
those commands begining with those characters...

-Paul


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Re: bash command history

[Philippe Troin]

On Sat, 09 Aug 1997 02:22:32 EDT Paul Miller ([EMAIL PROTECTED])
 wrote:

 Is there any way to get bash to work like 4dos? -- If you type the first
 couple of characters of a command in history, it will only scroll through
 those commands begining with those characters...

^R aka CTRL-R
man readline

Phil.



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