shelldirectorymovebasename

Why is basename in this shell script necessary?

I found this shell script($1,$2 are directories):

mv "$2" "$1"  ||  exit               # Make $2 a subdirectory of $1
cd "$1/$(basename "$2")"  ||  exit   # Change directories for simplicity
for f in *; do
    mv "$f" "${f%.*}.txt"            # Add or change the extension
done

it moves the second directory in the first one (so the second directory becomes a subdirectory of the first one) and all the files from the second directory will have a ".txt" extension. I can't figure out why cd "$1/$2" won't work doesn't basename strip the filename from a given directory? Could someone explain how this part: cd "$1/$(basename "$2")" works? Also $2 needs to be in separate quotes("$2") because of basename?

Solution

  • The basename removes the leading directory information. Assume that the sript is called with arguments like this:

    $$> script usr/local/foo usr/local/bar

    Now the $2 is equal to usr/local/bar.

    If you only do cd $1/$2, the command will look like this:

    cd usr/local/foo/usr/local/bar

    But in reality, it should be

    cd usr/local/foo/bar

    It is removing this directory prefix from usr/local/bar is whhat is done by basename.

    https://www.geeksforgeeks.org/basename-command-in-linux-with-examples/