On 19Feb2017 05:22, Mike Wright <nob...@nospam.hostisimo.com> 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 <c...@zip.com.au>
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