bash 回声退格
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Echoing a backspace
提问by sidyll
Is it possible to echo a backspace in bash?
是否可以在 bash 中回显退格？
Something like
就像是
echo $'stack\b'
Shouldn't output stac? Or I'm missing something?
不应该输出stac吗？或者我错过了什么？
More specifically, I'd like to use that in:
更具体地说，我想将其用于：
ls | wc -l; echo $'\b items'
回答by Ignacio Vazquez-Abrams
\bmakes the cursor move left, but it does not erase the character. Output a space if you want to erase it.
\b使光标向左移动，但不会擦除字符。如果要删除它，请输出一个空格。
For some distributions you may also need to use -eswitch of echo:
对于某些发行版，您可能还需要使用以下-e开关echo：
-e enable interpretation of backslash escapes
-e enable interpretation of backslash escapes
So it will look like
所以它看起来像
echo -e 'stack\b '
Also, files=(*) ; echo "${#files[@]} items".
还有，files=(*) ; echo "${#files[@]} items"。
回答by rawbone
So to answer the actual question about backspaces this will simulate a backspace:
因此，要回答有关退格的实际问题，这将模拟退格：
echo -e "\b \b"
It will move the character back one, then echo a space overwriting whatever character was there, then moving back again - in effect deleting the previous character. It won't go back up a line though so the output before that should not create a new line:
它会将字符移回一个，然后回显一个空格覆盖那里的任何字符，然后再次移回 - 实际上删除前一个字符。它不会返回一行，因此之前的输出不应创建新行：
echo -n "blahh"; echo -e "\b \b"
回答by Robert Vila
It is not exactly what you're asking for, but also, in the line of Ignacio's answer, you could use for this case:
这不完全是您所要求的，而且，在 Ignacio 的回答中，您可以在这种情况下使用：
echo "$(ls | wc -l) items"
AFAIK you cannot print a character that deletes the one before, not even printing the char whose hexadecimanl number correspoonds to backspace. You can move back and print a blank space to delete, though. With cputyou can do many things and print wherever you want in the screen.
AFAIK 你不能打印一个删除之前的字符，甚至不能打印十六进制数字对应于退格的字符。不过，您可以向后移动并打印一个空白区域进行删除。使用cput，您可以做很多事情并在屏幕上的任何位置打印。
