I'm trying to write a very simple mkcd command:
#!/bin/bash
mkdir $1
cd $1
The directory is created but the change directory part doesn't seem to run.
Update based on comment:
mkcd () {
mkdir "$1"
cd "$1"
}
I'm trying to run it first as a local file:
./mkcd
My end location is /opt/bin, neither location seems to work.
It needs to be a function:
mkcd() { mkdir -p "$1" && cd "$1"; }
A script will get run inside its own separate process. Changing directories there will have no effect on the parent shell (neither will changing directories inside a subshell as in (cd /tmp)).
mkcd() { if [ ! -e "$@" ];then mkdir -p "$@" && cd "$@";fi; } .
[ -d "$1" ] || mkdir -p "$1" is good for saving time on process start overheads in scripts, though. (I think -d is better than -e because then you'll run mkdir iff the arg isn't a directory and mkdir will then generate an error message for you).
mkcd() { if [ ! -d "$@" ];then mkdir -p "$@" ;fi; cd "$@"; } .
cddoes run - but its effect is lost as soon as the subshell running the script dies.mkcd() { if [ ! -d "$@" ];then mkdir -p "$@" ;fi; cd "$@"; }(Posted as comment because answering is disabled here.).