替换bash变量中的换行符？

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.