使用空格将参数传递给 Bash 脚本中的命令
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Passing arguments to a command in Bash script with spaces
提问by Dougnukem
I'm trying to pass 2 arguments to a command and each argument contains spaces, I've tried escaping the spaces in the args, I've tried wrapping in single quotes, I've tried escaping \" but nothing will work.
我试图将 2 个参数传递给命令，并且每个参数都包含空格，我试过转义参数中的空格，我试过用单引号括起来，我试过转义 \" 但没有任何效果。
Here's a simple example.
这是一个简单的例子。
#!/bin/bash -xv
ARG="/tmp/a b/1.txt"
ARG2="/tmp/a b/2.txt"
ARG_BOTH="\"$ARG\" \"$ARG2\""
cat $ARG_BOTH
I'm getting the following when it runs:
运行时我得到以下信息：
ARG_BOTH="$ARG $ARG2"
+ ARG_BOTH='/tmp/a\ b/1.txt /tmp/a\ b/2.txt'
cat $ARG_BOTH
+ cat '/tmp/a\' b/1.txt '/tmp/a\' b/2.txt
cat: /tmp/a\: No such file or directory
cat: b/1.txt: No such file or directory
cat: /tmp/a\: No such file or directory
cat: b/2.txt: No such file or directory
回答by SiegeX
See http://mywiki.wooledge.org/BashFAQ/050
见http://mywiki.wooledge.org/BashFAQ/050
TLDR
TLDR
Put your args in an array and call your program as myutil "${arr[@]}"
将您的 args 放在一个数组中并将您的程序称为 myutil "${arr[@]}"
#!/bin/bash -xv
file1="file with spaces 1"
file2="file with spaces 2"
echo "foo" > "$file1"
echo "bar" > "$file2"
arr=("$file1" "$file2")
cat "${arr[@]}"
Output
输出
file1="file with spaces 1"
+ file1='file with spaces 1'
file2="file with spaces 2"
+ file2='file with spaces 2'
echo "foo" > "$file1"
+ echo foo
echo "bar" > "$file2"
+ echo bar
arr=("$file1" "$file2")
+ arr=("$file1" "$file2")
cat "${arr[@]}"
+ cat 'file with spaces 1' 'file with spaces 2'
foo
bar
回答by DigitalRoss
This might be a good use-case for the generic "set"command, which sets the top-level shell parameters to a word list. That is, $1, $2, ... and so also $* and $@ get reset.
这可能是通用“set”命令的一个很好的用例，它将顶级 shell 参数设置为单词列表。也就是说，$1, $2, ... 以及 $* 和 $@ 也会被重置。
This gives you some of the advantages of arrays while also staying all-Posix-shell-compatible.
这为您提供了数组的一些优点，同时还保持所有 Posix-shell 兼容。
So:
所以：
set "arg with spaces" "another thing with spaces"
cat "$@"
回答by zwol
The most straightforward revision of your example shell scriptthat will work correctly is:
可以正常工作的示例 shell 脚本的最直接的修订是：
#! /bin/sh
ARG="/tmp/a b/1.txt"
ARG2="/tmp/a b/2.txt"
cat "$ARG" "$ARG2"
However, if you need to wrap up a whole bunch of arguments in one shell variable, you're up a creek; there is no portable, reliable way to do it. (Arrays are Bash-specific; the only portable options are setand eval, both of which are asking for grief.) I would consider a need for this as an indication that it was time to rewrite in a more powerful scripting language, e.g. Perl or Python.
但是，如果您需要将一大堆参数包含在一个 shell 变量中，那您就麻烦了；没有便携、可靠的方法来做到这一点。（数组是 Bash 特定的；唯一的可移植选项是setand eval，这两个选项都令人痛苦。）我认为需要这样做来表明是时候用更强大的脚本语言重写，例如 Perl 或 Python .
