bash 替换bash变量中的换行符？
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replace newline character in bash variable?
提问by yorua007
I am trying to understand the "cdargs-bash.sh" script with cdargs packages. And I have a question about in the following function:
我试图理解带有 cdargs 包的“cdargs-bash.sh”脚本。我对以下功能有疑问：
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 "$HOME/.cdargs" ]; then
dir=`/bin/grep "^ " "$HOME/.cdargs"`
if [ -z "$dir" ]; then
bookmark="${1/\/*/}"
if [ "$bookmark" != "" ]; then
dir=`/bin/grep "^$bookmark " "$HOME/.cdargs"`
extrapath=`echo "" | /bin/sed 's#^[^/]*/#/#'`
fi
fi
[ -n "$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 '$dir'" >&2
return 2
fi
}
}
What's the purpose of the testing:
测试的目的是什么：
"$dir" != "${dir/
/}"
Here the testing span over two lines; does it want to remove the newline character in $diror maybe for some other reason? I am just starting to learn bash scripting and I have Googled some time but couldn't find any usage like this.
这里的测试跨越两行；它是要删除换行符$dir还是出于其他原因？我刚刚开始学习 bash 脚本，我在谷歌上搜索了一段时间，但找不到任何这样的用法。
回答by Mu Qiao
Yes you are right, it removes the newline character. I think the purpose of the test is to make sure $dirdoesn't contain multiple lines.
是的，你是对的，它删除了换行符。我认为测试的目的是确保$dir不包含多行。
Alternatively, you can remove \newlineby
或者，您可以\newline通过删除
${dir/$'\n'/}
This doesn't require two lines so I think it looks better.
这不需要两行，所以我认为它看起来更好。
