cutting the last N characters form a file name

Rindert · 08-16-2006, 06:02 AM

Hello,

I'm writing a bash script and I would like to do the following:
I have a directory with the following:
file001.jpg
file002.jpg
file003.jpg
Now i want a bash script that can give 'file' back to me.
The command: 'cut -c 1-4' doest work for me, cause i also want the same to work on
otherfile001.jpg
otherfiie002.jpg
otherfile003.jpg

So what i'm looking for is a command that can strip the last 7 characters for me.

Anyone know how this can be done?

Thanks,

Rindert

acid_kewpie · 08-16-2006, 06:19 AM

can i suggest using basename instead?

Code:

# basename /root/test.txt .txt
test

spirit receiver · 08-16-2006, 06:49 AM

How about this?

Code:

ada@barnabas:~> ( filename=otherfile002.jpg; echo ${filename/%???.jpg})
otherfile

acid_kewpie · 08-16-2006, 06:54 AM

Quote:

Originally Posted by spirit receiver

How about this?

Code:

ada@barnabas:~> ( filename=otherfile002.jpg; echo ${filename/%???.jpg})
otherfile

i'd still recommend basename over this approach, you could actually still use a cut, but use the . as a field seperator, and keep the first field only

Code:

# echo test.txt | cut -d. -f1
test

all depends exactly what your scenario is...

spirit receiver · 08-16-2006, 06:57 AM

But Rindert's trying to remove the last 7 characters which consist of varying digits and the extension .jpg. That's why basename won't work.

marozsas · 08-16-2006, 07:11 AM

The best (and easy) approach is to remove 7 characters (xxx.jpg) from the BEGINNING of the string. The command 'rev' is handy in this case.

Code:

$ echo otherfile001.jpg | rev | sed -e 's/.......//' | rev
otherfile
$ echo file001.jpg | rev | sed -e 's/.......//' | rev
file

marozsas · 08-16-2006, 07:15 AM

Of course, if you like cut, ou can use 'cut -c 8-', instead sed.

Code:

$ echo otherfile001.jpg | rev | cut -c 8- | rev
otherfile

acid_kewpie · 08-16-2006, 07:58 AM

best?? pah, not even close...

marozsas · 08-16-2006, 08:16 AM

Ok, not "The Best", but how about the "easy" part ?

Cheers,

acid_kewpie · 08-16-2006, 08:58 AM

easy? piping through 3 commands when 1 will do? keep trying

marozsas · 08-16-2006, 09:05 AM

Hey acid_kewpie !

All right, I will never more use the words "best" and "easy" in a post, specially if you are subscribed to it

cheers,

Rindert · 08-16-2006, 11:49 AM

Hello,

I've tried the
( filename=otherfile002.jpg; echo ${filename/%???.jpg})
command, it does work on the shell, but how can i use it in my bash script? I have to use it for mutpiple dirs so i think best way i to use with 'ls' right? But how do I get the result from 'ls' in the filename variable?
I'm kinda new to shell scripting. This is my first own script I'm writing.

Because I'm a newbie to shell scripts. Let say I want the easiest way of doing it, it doesn't have to be best way, as long as it is a good way.

Thanks, Rindert

spirit receiver · 08-16-2006, 01:18 PM

You could try something like

Code:

#! /bin/bash
cd /some/directory
for file in *.jpg
do
  echo "Found a filename that starts with ${file/%???.jpg}."
done

acid_kewpie · 08-16-2006, 01:19 PM

well it depends what your circumstances are, most of these ways will work, but do you have different length extensions? different test in extensions? if you do want to use that example there, then you'd simply replace "filename" in the echo part with the variable name you currently have the string assigned to. if you don't actaully want to echo it, but store it, then you'd do, for example:

Code:

original=test.txt
result=${original/%???.jpg}

Rindert · 08-16-2006, 07:48 PM

I'v tried some things, and for me this works good:
cat_profile= ls -1 *001.jpg | rev | cut -c 8- | rev
the result is get is: otherfile.
just what i wanted

thanks everyone for helping me