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Is there a way to make a bash script which scans a directory dir0 for folders, such that for all folders dir1 which are found in dir0 a command dir1-command is created with the following definition:

dir1-command() {
 cd dir0/dir1
 echo "inside dir0/dir1"
 cd -
}

EDIT: The specific problem I'm trying to solve is this: I have a folder called scripts which has subfolders like scripts/haskell_script1 and scripts/haskell_script2. The way to launch, say, scripts/haskell_script1 is as follows:

haskell_script1() {
 cd scripts/haskell_script1
 stack exec haskell_script1 -- $@
 cd -
}

And so what I'd like to do is, every time ~/.bashrc is sourced, scan my scripts folder and generate a bunch of commands of form haskell_scriptX to save me time. That way I don't have to manually create a command for every haskell project/folder folder that I create.

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  • What is the issue that you're trying to solve by this? Commented Jun 7, 2017 at 12:01
  • @Kusalananda: I provided more detail in my question above. Commented Jun 7, 2017 at 12:27
  • 1
    Wouldn't it be easier to make a function, run_haskel, that takes an argument and executes ( cd scripts/haskel_$arg && stack exec haskell_$arg ) ? Commented Jun 7, 2017 at 12:30

1 Answer 1

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1

Add this to your .bashrc:

for script in scripts/haskell_script*
do
  # strip off the leading "scripts/" portion
  script=${script##*/}
  eval "function ${script} { 
    cd scripts/${script}
    stack exec ${script} -- \$@
    cd -
  }"
done

You have to be careful about anything inside that definition that might get prematurely evaluated, which is why I escaped the $ in $@.

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