Grabbing the current directory in shell script

binarybob0001 · 11-02-2007, 12:12 PM

I'm trying to write a script to automatically package my kernel just the way Slackware like it. It would be nice though if I could get the current working directory someone. Is there a way I can do this? The code explains it all

Code:

mkdir $dest/build
mkdir $dest/build/boot
make -C $dest/$source \
  INSTALL_MOD_PATH=$dest/build \
  INSTALL_PATH=$dest/build/boot \
  install
cp $dest/$source/.config $dest/build/boot/config-$version
mv $dest/build/boot/vmlinuz $dest/build/boot/kernel-$version
cd $dest/build/boot
ln -s kernel-$version vmlinuz
ln -s config-$version config
cd -
cd $dest/build
makepkg -l y -c n kernel-$version-i686-1bob.tgz
mv kernel-$version-i686-1bob.tgz ..
cd -
rm -r $dest/build

Note how at on point I have cd - and then cd $dest/build just so I don't lose track of home base.

brazilnut · 11-02-2007, 12:19 PM

Code:

pwd

That Prints Working Directory

colucix · 11-02-2007, 12:24 PM

Or simply

Code:

echo $PWD

since the variable PWD is updated every time you change dir.

stocky021 · 11-02-2007, 12:37 PM

if you want to save pwd to a variable, use this
variable_name=`pwd`
the ` is next to the 1 key, it is not '
that way you can just user cd "$variable_name" to jump back to that directory...i think

binarybob0001 · 11-02-2007, 06:03 PM

Thanks guys, much appreciated