Bash parameter expansion... how can I remove two substrings? [SOLVED]

graysky · 2014-09-08 21:21:58

I want to take the output of a find command and:

1) Remove the preceding ./
2) Remove the file extension

Example:

myvar=./running_with_scissors.txt
echo ${myvar%.*}
  ./running_with_scissors
echo ${myvar#./}
  running_with_scissors.txt

I need to combine the two so that my echo commend returns simply: running_with_scissors

Is this possible without defining a 2nd var?

Last edited by graysky (2014-09-08 22:16:11)

nomorewindows · 2014-09-08 21:25:36

Wouldn't this be a job for sed/awk/etc. ?

graysky · 2014-09-08 21:26:22

@nomorewindows - I can do it with sed but I want to learn how to use shell expansion if possible.

ninian · 2014-09-08 21:36:03

I think you need two assignments:

myvar=./running_with_scissors.txt; myvar=${myvar%.*}; myvar=${myvar#.*/}; echo "$myvar"

Or am I missing something?!
Edit: yup, you want to retain the original value of $myvar - sorry!

Last edited by ninian (2014-09-08 21:37:33)

Saint0fCloud · 2014-09-08 21:38:48

If you're using zsh, this will work, although it's not very elegant or flexible.

$ a="./hello.txt"
$ print ${a[(ws./.)2]%.*}
hello

Edit: I got so excited about the problem that I failed to read the thread title properly

For future posterity, a much better zsh solution (in addition to nesting the substitutions) would be:

$ setopt hist_subst_pattern
$ print ${myvar:t:s/.*//}

Last edited by Saint0fCloud (2014-09-08 22:16:20)

graysky · 2014-09-08 21:38:55

Perhaps it cannot be done (googling suggests this is the case). Again, I can do it with a temp var but we'll see what others come up with.

jasonwryan · 2014-09-08 21:40:39

You can't nest parameter expansions in bash (you can in zsh: `${${myvar%.*}#.*/}`).

graysky · 2014-09-08 22:15:53

@jason - Nice! Thanks for the tip. I use zsh as my default anyway.

myvar=./running_with_scissors.txt
% echo ${${myvar%.*}#.*/}
running_with_scissors

Trilby · 2014-09-08 22:27:06

For bashers:

echo ${myvar:2:${#myvar}-6}

Though this assumes the "./" is always there, and the extension is always 3 characters.

aiBo · 2014-09-09 05:26:35

If Trilby's assumptions apply, the follwing code works as well in bash:

myvar=./running_with_scissors.txt
echo ${myvar:2:-4}                  # => running_with_scissors

Or, if you don't mind some regular expression:

myvar=./running_with_scissors.txt
[[ $myvar =~ ^\./(.*)\.[^.]+$ ]]
echo ${BASH_REMATCH[1]}             # => running_with_scissors

Last edited by aiBo (2014-09-09 05:46:00)

nomorewindows · 2014-10-01 22:27:11

Didn't anyone ever tell you not to run with scissors?

jasonwryan · 2014-10-01 23:02:18

nomorewindows: please don't emptybump, especially on this board https://wiki.archlinux.org/index.php/Fo … mpty_Posts

Arch Linux

#1 2014-09-08 21:21:58

Bash parameter expansion... how can I remove two substrings? [SOLVED]

#2 2014-09-08 21:25:36

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#3 2014-09-08 21:26:22

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#4 2014-09-08 21:36:03

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#5 2014-09-08 21:38:48

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#6 2014-09-08 21:38:55

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#7 2014-09-08 21:40:39

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#8 2014-09-08 22:15:53

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#9 2014-09-08 22:27:06

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#10 2014-09-09 05:26:35

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#11 2014-10-01 22:27:11

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

#12 2014-10-01 23:02:18

Re: Bash parameter expansion... how can I remove two substrings? [SOLVED]

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