Extracting words in quotes in shell script

Question

Hi!

I am making a shell script that will automate the install process of Arch Linux AUR packages. I need to list all dependencies of the package (to check if they are installed), they appear like this in install script:

depends=('sdl' 'libvorbis' 'openal')

The easiest way (or the only idea) that I could come up with is something like this:

grep "depends" PKGBUILD | awk -F"'" '{print $2 "\n" $4 "\n" $6;}'

But the dependency count varies from package to package. So, how I output the names in quotes if the word count is varying?

Thanks in advance,

-skazhy

Answer 1

You could do something like:

grep "depends" PKGBUILD | perl -e "while(<>){print \"\$1\\n\" while m/'.{-}'/g;}"

Answer 2

Try this on for size:

grep "depends" PKGBUILD > /tmp/depends
. /tmp/depends

echo ${depends[@]}

Hey look, is that an array? Yes it is.

for d in "${depends[@]}" ; do
    printf '"%s"' "$d"
done

Note: In a real script you'd want to be more careful with the naming of the temporary file.

Answer 3

If the depends is just one line, one thing you may try is to evaluate the line in bash itself... This will lead to an array called "depends" that holds all the values. Seems tricky, but not with dynamic languages:

depend_line=`grep depends $PKGBUILD`
eval "${depend_line}"
echo ${depend[0]} # Will print sdl in your example

Answer 4

awk -F"'" '{for(i=2;i<=NF;i+=2) print($i)}'

Answer 5

You can avoid the security issues of using eval or sourcing a temporary file by using declare:

declare -a "$(grep "depends" PKGBUILD)"

This will create an array called "depends" containing "sdl", "libvorbis" and "openal" based on the example data.