我如何获得$(/bin/printf -6)返回-6却不认为-6是一个选择 [英] How do I get $(/bin/printf -6) to return -6 and not think -6 is an option

问题描述

我有一个bash shell脚本，其内容如下:

I have a bash shell script which has the line:

g=$(/bin/printf ${i})

当${i}包含类似-6的内容时，printf认为它被传递了一个选项.它无法识别该选项，因此会产生错误.

when ${i} contains something like -6, printf thinks its being passed an option. It does not recognize the option so produces an error.

如果将${i}用引号引起来，则printf仍认为它被传递了一个选项.

if wrap ${i} in quotes, printf still thinks its being passed an option.

g=$(/bin/printf "${i}")

如果我不使用引号，则变量$g会保留"-6"，这也不是我想要的.

if I escape the quotes, variable $g then holds "-6" which is not what I want either.

g=$(/bin/printf \"${i}\") 

那里是逃脱破折号(-)的地方.

Is there away to escape the dash (-).

printf是BusyBox应用

printf is a BusyBox app

推荐答案

如果您使用实际的格式字符串调用printf怎么办?

What if you called printf with an actual format string?

$ printf "%d\n" -6
-6
$ /sbin/busybox printf "%d\n" -6
-6
$

这显然适用于GNU coreutils和busybox的printf.

This works with both GNU coreutils' and busybox' printf, apparently.

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