如何将带引号的参数从变量传递到bash脚本 [英] How to pass quoted arguments from variable to bash script

问题描述

我尝试在变量中构建一组参数，并将其传递给脚本，但是行为与预期不同.

I tried building a set of arguments in a variable and passing that to a script but the behavior different from what I expected.

test.sh

#!/bin/bash

for var in "$@"; do
  echo "$var"
done

输入

usr@host$ ARGS="-a \"arg one\" -b \"arg two\""
usr@host$ ./test.sh $ARGS

输出

-a
"arg
one"
-b
"arg
two"

预期

-a
arg one
-b
arg two

请注意，如果将带引号的参数直接传递给脚本，它将起作用.我也可以使用eval解决此问题，但我想了解为什么第一种方法失败了.

Note if you pass the quoted arguments directly to the script it works. I also can work around this with eval but I wanted to understand why the first approach failed.

解决方法

ARGS="./test.sh -a "arg one" -b "arg two""
eval $ARGS

推荐答案

您应该使用一个数组，从某种意义上讲，它提供了第二级引用:

You should use an array, which in some sense provides a 2nd level of quoting:

ARGS=(-a "arg one" -b "arg two")
./test.sh "${ARGS[@]}"

数组扩展会为数组的每个元素产生一个单词，因此在构造要传递给test.sh的参数列表时，创建数组时引用的空格不会被视为单词分隔符.

The array expansion produces one word per element of the array, so that the whitespace you quoted when the array was created is not treated as a word separator when constructing the list of arguments that are passed to test.sh.

请注意，POSIX shell不支持数组，但这正是POSIX shell中引入数组的确切缺点.

Note that arrays are not supported by the POSIX shell, but this is the precise shortcoming in the POSIX shell that arrays were introduced to correct.

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