replace newline character in bash variable?

Question

I am trying to understand the "cdargs-bash.sh" script with cdargs packages. And I have a question about in the following function:

function _cdargs_get_dir () { local bookmark extrapath # if there is one exact match (possibly with extra path info after it), # then just use that match without calling cdargs if [ -e "$HOME/.cdargs" ]; then     dir=`/bin/grep "^$1 " "$HOME/.cdargs"`     if [ -z "$dir" ]; then         bookmark="${1/\/*/}"         if [ "$bookmark" != "$1" ]; then             dir=`/bin/grep "^$bookmark " "$HOME/.cdargs"`             extrapath=`echo "$1" | /bin/sed 's#^[^/]*/#/#'`         fi     fi     [ -n "$dir" ] && dir=`echo "$dir" | /bin/sed 's/^[^ ]* //'` fi if [ -z "$dir" -o "$dir" != "${dir/ /}" ]; then     # okay, we need cdargs to resolve this one.     # note: intentionally retain any extra path to add back to selection.     dir=     if cdargs --noresolve "${1/\/*/}"; then         dir=`cat "$HOME/.cdargsresult"`         /bin/rm -f "$HOME/.cdargsresult";     fi fi if [ -z "$dir" ]; then     echo "Aborted: no directory selected" >&2     return 1 fi [ -n "$extrapath" ] && dir="$dir$extrapath" if [ ! -d "$dir" ]; then     echo "Failed: no such directory '$dir'" >&2     return 2 fi 

}

What's the purpose of the testing:

"$dir" != "${dir/ /}" 

Here the testing span over two lines; does it want to remove the newline character in $dir or maybe for some other reason? I am just starting to learn bash scripting and I have Googled some time but couldn't find any usage like this.

Answer 1

Yes you are right, it removes the newline character. I think the purpose of the test is to make sure $dir doesn't contain multiple lines.

Alternatively, you can remove \newline by

${dir/$'\n'/} 

This doesn't require two lines so I think it looks better.