庆典最后覆盖终端线 [英] bash overwrite last terminal line

问题描述

我的bash脚本看起来如下：

My bash-script looks as following:

echo "Description:"
while [ $finishInput -eq 0 ]; do
read tmp
desc="$desc"$'\n'"$tmp"
if [ -z "$tmp" ]; then
finishInput="1"
fi
done
echo -n "Maintainer:"
read maintainer

直到空行传递它读给递减变种。在那之后，我想其他的东西阅读。

It reads to the desc var until a empty line is passed. After that, i want to read in other stuff.

当我执行当前脚本，它看起来是这样的：

When executing my current script it looks like this:

Description:
Line 1
Line 2

Maintainer:

我想覆盖的最后一个空行维护者。

I would like to overwrite the last empty line with the "Maintainer:".

我搜索了一个解决方案，但只发现了其中像

I searched for a solution but only found suggestions which were like

echo -n "Old line"
echo -e "\r new line"

它可以停留在该行并将其覆盖。这是不可能在我的情况。

which stays on the line and overwrites it. This is not possible in my case.

推荐答案

只需使用\\ E [1A。

在您的例子中，你在同一行中删除文本：

In your example you delete the text at the same line:

$ echo -n "Old line"; echo -e "\e[0K\r new line"
 new line

当您想返回到previous行使用\\ E [1A：

When you want to return to the previous line use \e[1A:

$ echo "Old line"; echo -en "\e[1A"; echo -e "\e[0K\r new line"
 new line

当你想要去N行了，使用 \\ E [＆LT; N方式＆gt; A

When you want to go N lines up, use \e[<N>A.

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