bash script selecting the last path variable

Question

I am writing my first bash script. I am making it install all the repos I have on GitHub.

warpInLocations=("[email protected]:acc/toolkit.git" "[email protected]:acc/sms.git" "[email protected]:acc/boogle.git" "[email protected]:acc/cairo.git")

Those are them when I install them.

echo "warping in toolkit, sms, boogle and cairo"
for repo in "${warpInLocations[@]}"
do
  warpInDir=$(echo ${warpToLocation}${repo} | cut -d'.' -f1)
  if [ -d "$warpToLocation"]; then
    echo "somethings in the way.. $warpInDir all ready exists"
  else
    git clone $repo $warpInDir
  fi

done

This line here, I wanted it to give me a folder named toolkit or sms, so after the / and before the . in the warp in locations, but it's selecting git@github instead. I guess, because it's after the ..

How can I get it to select the name in the repo?

Answer 1

dir=$(basename [email protected]:acc/toolkit.git .git)

will set $dir to toolkit.

also useful is the dirname command.

Answer 2

You need to proceed in two steps:

[email protected]:acc/toolkit.git
dir=${dir#*/}                       # Remove everything up to /
dir=${dir%.*}                       # Remove everything from the .

Answer 3

in bash, you can also use a regular expression and capturing parentheses

for repo in "${warpInLocations[@]}"; do
    [[ $repo =~ /([^.]+)\. ]] && dir=${BASH_REMATCH[1]}
    warpInDir=${warpToLocation}$dir
    # ...