Bash variable assignment

Question

I have a basic script to edit config files in ~/.config - it works with the cd lines, but that seems redundant:

dir=$HOME/.config/$1
if [ ! -d "$dir" ]; then :
    else 
    cd "$dir" && 
    for file in * ; do
        case "$file" in
        conf | config | *.cfg | *rc)  $EDITOR "$file" ;;
        *)  :  ;;
        esac
    done
    cd - 1>/dev/null;
fi

Changing it to use the variable "$dir" fails. What am I doing wrong?

dir=$HOME/.config/$1
if [ ! -d "$dir" ]; then :
    else 
    for file in "$dir" ; do
        case "$file" in
        conf | config | *.cfg | *rc)  $EDITOR "$file" ;;
        *)  :  ;;
        esac
    done;
fi

Why on earth do you write if [ ! -d "$dir" ]; then :; else cd "$dir" ... instead of simply if [ -d "$dir" ]; then cd "$dir" ...? — glenn jackman
– glenn jackman, Commented Jul 18, 2011 at 3:30

Hello71 · Accepted Answer · 2011-07-17 21:20:46Z

1

You're not globbing the files inside $dir, merely listing $dir itself. Try $dir/*.

answered Jul 17, 2011 at 21:20

Hello71

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1 Comment

Peter K. Over a year ago

Precisely! Beat me to it. :-)

Ryan Stewart · Accepted Answer · 2011-07-17 21:33:24Z

0

You can't use just "$dir" because that gives just a single item: the directory. You need $dir/*, but that includes the path as well, so you have to strip that off to compare just the file name:

...
for file in $dir/*; do
    filename=`basename $file`
    case $filename in
    ...

answered Jul 17, 2011 at 21:33

Ryan Stewart

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2 Comments

Edouard Over a year ago

Thank you - that worked. Is there anything I could read up on to more fully understand this?

Ryan Stewart Over a year ago

@Edouard: I'm afraid I can't really point you to a great resource. Most of what I know of shell scripting, I've picked up on the fly. I had to do a google search to find the "basename" command because I couldn't remember it :) Basically, the "for" iterates over all the files that come after "in". The $dir/* expands to a list of all the files that are in whatever directory "$dir" is. If $dir is /foo/bar, and that has a file named "baz" in it, then inside the loop, $file will have the value /foo/bar/baz. The basename command will strip off the path and give you "baz". Does that explain it?

Peter K. · Accepted Answer · 2011-07-17 21:22:35Z

0

Your first version does a * on the directory, yielding a list of all the files in the directory. Your second version just has one entry in the list --- the directory itself, not its contents.

answered Jul 17, 2011 at 21:22

Peter K.

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