回声＆QUOT; -n＆QUOT;不会打印-n？ [英] echo "-n" will not print -n?

问题描述

非常相似这个问题。

我通过几件事迭代与BASH自动化脚本。有时，脚本会遇到-n和echo将尝试间preT这一点。

I'm iterating through a few things with an automated script in BASH. Occasionally the script will come across "-n" and echo will attempt to interpret this.

此尝试：

 $ POSIXLY_CORRECT=1 /bin/echo -n

和

 $ POSIXLY_CORRECT=1 /bin/echo "-n"

但是，除$ P $每次PTED的说法。

But it interpreted the argument each time.

然后这一点，这工作，但它可能在弦打转义字符，这就是为什么我不想空字符适用于所有的输入，并使用-e。

Then this, which works but it's possible to hit escaped characters in the strings, which is why I don't want to apply a null character to all input and use -e.

$ echo -e "\x00-n"
-n

printf的是可能的，但是要避免的，除非没有其他选项（并非所有的机器都printf的作为实用程序）。

printf is possible, but is to be avoided unless there are no other options (Not all machines have printf as a utility).

$printf "%s" "-n"
-n

那么，有没有办法让回声打印-n？

So is there a way to get echo to print "-n"?

推荐答案

的printf 应该是一个内置于所有的炮弹，除非你的一些机器有很老外壳的版本。这是一个内置的庆典很长一段时间。这也可能是更便携比回声-e 。

printf should be a built-in in all your shells, unless some of your machines have very old shell versions. It's been a built-in in bash for a long time. It's probably more portable than echo -e.

否则，真的没有办法让回声选项它关心。

Otherwise, there's really no way to get echo options it cares about.

修改：从<一个href=\"http://stackoverflow.com/questions/13042734/parsing-command-line-arguments-in-bash-strange-behavior/13042777#13042777\">an回答另一个类似的问题;避免使用这个方便的无数字版权保留包装报价与的printf 问题：

Edit: from an answer to another similar question; avoid quoting issues with printf by using this handy no-digital-rights-retained wrapper:

ech-o() { printf "%s\n" "$*"; }

（这是ECH-O，如不带选项）

(That's ech-o, as in "without options")

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