Bash：“printf %q $str”在脚本中删除空格。（备择方案？）
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Bash: "printf %q $str" deletes spaces when in scripts. (Alternatives?)
提问by Elena
printf %qshould quote a string. However, when executed into a script, it deletes the spaces.
printf %q应该引用一个字符串。但是，当在脚本中执行时，它会删除空格。
This command:
这个命令：
printf %q "hello world"
outputs:
输出：
hello\ world
which is correct.
哪个是正确的。
This script:
这个脚本：
#!/bin/bash
str="hello world"
printf %q $str
outputs:
输出：
helloworld
which is wrong.
这是错误的。
If such behavior is indeed expected, what alternative exists in a script for quoting a string containing any character in a way that it can be translated back to the original by a called program?
如果确实预料到了这种行为，那么脚本中存在什么替代方法来引用包含任何字符的字符串，以便可以通过被调用的程序将其翻译回原始字符串？
Thanks.
谢谢。
Software: GNU bash, version 4.1.5(1)-release (i486-pc-linux-gnu)
软件：GNU bash，版本 4.1.5(1)-release (i486-pc-linux-gnu)
EDITED: Solved, thanks.
编辑：已解决，谢谢。
回答by Susam Pal
You should use:
你应该使用：
printf %q "$str"
Example:
例子：
susam@nifty:~$ cat a.sh
#!/bin/bash
str="hello world"
printf %q "$str"
susam@nifty:~$ ./a.sh
hello\ world
When you run printf %q $str, the shell expands it to:
运行时printf %q $str，shell 将其扩展为：
printf %q hello world
So, the strings helloand worldare supplied as two separate arguments to the printfcommand and it prints the two arguments side by side.
因此，字符串hello和world作为printf命令的两个单独参数提供，并并排打印这两个参数。
But when you run printf %q "$str", the shell expands it to:
但是当您运行时printf %q "$str"，shell 将其扩展为：
printf %q "hello world"
In this case, the string hello worldis supplied as a single argument to the printfcommand. This is what you want.
在这种情况下，字符串hello world作为printf命令的单个参数提供。这就是你想要的。
Here is something you can experiment with to play with these concepts:
以下是您可以尝试使用这些概念的方法：
susam@nifty:~$ showargs() { echo "COUNT: $#"; printf "ARG: %s\n" "$@"; }
susam@nifty:~$ showargs hello world
COUNT: 2
ARG: hello
ARG: world
susam@nifty:~$ showargs "hello world"
COUNT: 1
ARG: hello world
susam@nifty:~$ showargs "hello world" "bye world"
COUNT: 2
ARG: hello world
ARG: bye world
susam@nifty:~$ str="hello world"
susam@nifty:~$ showargs $str
COUNT: 2
ARG: hello
ARG: world
susam@nifty:~$ showargs "$str"
COUNT: 1
ARG: hello world
回答by Zsolt Botykai
Try
尝试
printf %q "${str}"
in your script.
在你的脚本中。
回答by user3606895
This worked for me. Satisfies these requirements
这对我有用。满足这些要求
- Accept arbitrary input that may include shell special characters
- Does not output escape character, the "\"
- 接受可能包含 shell 特殊字符的任意输入
- 不输出转义字符，“\”
#! /bin/bash FOO='myTest3$; t%^&;frog now! and *()"' FOO=`printf "%q" "$FOO"` # Has \ chars echo $FOO # Eat all the \ chars FOO=$(printf "%q" "$FOO" | sed "s/\\//g") # Strip \ chars echo $FOO
