击：阅读报价/正确地从一个字符串参数逃脱 [英] Bash: Reading quoted/escaped arguments correctly from a string

问题描述

我遇到一个Bash脚本传递参数给命令的问题。

I'm encountering an issue passing an argument to a command in a Bash script.

poc.sh：

#!/bin/bash

ARGS='"hi there" test'
./swap ${ARGS}

调剂：

#!/bin/sh
echo "${2}" "${1}"

电流输出是：

there" "hi

仅仅改变poc.sh（我相信换做我想要的，正确），我怎么poc.sh通过您好，并作为测试两个参数，用你好有周围没有引号呢？

Changing only poc.sh (as I believe swap does what I want it to correctly), how do I get poc.sh to pass "hi there" and test as two arguments, with "hi there" having no quotes around it?

推荐答案

嵌入式报价不保护的空白;他们从字面上对待。在使用数组庆典：

Embedded quotes do not protect whitespace; they are treated literally. Use an array in bash:

args=( "hi there" test)
./swap "${args[@]}"

在POSIX shell中，您使用评估卡住（这就是为什么大多数的炮弹支持数组）。

In POSIX shell, you are stuck using eval (which is why most shells support arrays).

args='"hi there" test'
eval "./swap $args"

像往常一样，是的非常的一定要知道的$ args 的内容并了解使用之前生成的字符串将如何解析评估。

As usual, be very sure you know the contents of $args and understand how the resulting string will be parsed before using eval.

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