Re: bash help

Wednesday, 22 February 2017

On 19Feb2017 05:22, Mike Wright <nobody(a)nospam.hostisimo.com&gt; wrote:
...
My brain cell ran away from home.  I have an incredibly simple script
that 
doesn't do what I expect.  I use "mkdir DIR; cd DIR" a lot so I'm trying
to 
put it in a script: "~/bin/mdcd".

After checking that $1 exists:

dir="$1"
mkdir -p "$dir"
cd "$dir"   <------ never executes

The directory is created so there is no error there.

Huh? Insight anyone? 
Run you script with -x:

  sh -x your-script.sh args...

My suspicion is that your script work.

_However_, "cd" is a shell builtin because it affects the _current_ process. If

you cd inside a shell script it moves the process running the shell script, 
_not_ your parent shell that invoked the script.

If you want a shell convenience to make a directory an cd into it you need a 
shell function: effectively a piece of script that runs in you current shell 
instead of as a subprocess.

For example, I have a 'cdf" function defined like this:

  cdf(){
    [ $# = 1 ] || { echo "Usage: cdf dir" >&2; return 2; }
    needdir "$1" && cd "$1" && pwd && {
prepdir; L; }
  }

For context, "needdir" is a personal script that effectively does what your 
"mkdir -p" does, except that it checks first and tells me if it needed to make 
the directory; "prepdir" is a personal shell function that is intended to do 
some personal setup (but presently does nothing:-); "L" is a personal shell 
alias for "ls -la".

One important thing to note when writing shell functions: you can leave a shell 
function early (and set its "exit status") with "return". This is like
"exit", 
but it terminates the shell function, not your shell!

Cheers,
Cameron Simpson <cs(a)zip.com.au&gt;