bash 中双引号字符串中的单引号不受支持 [英] Single quotes inside a double-quoted string in bash not honored

问题描述

我正在尝试在 bash 中分配一个变量:

I am trying to assign a variable in bash:

assignssid="airport -I | awk '/ SSID/ {print substr($0, index($0, $2))}'"

当我当前这样做时，然后运行 ​​echo "$assignssid"，结果类似于以下内容:

When I currently do this, and then run echo "$assignssid", the result is something like the following:

airport -I | awk '/ SSID/ {print substr(-bash, index(-bash, ))}'

如何让变量与引号内的代码一起工作?

How do I get the variable to work with the code inside the quotes?

推荐答案

问题:单引号在双引号内不符合语法

当你有 "'$0'" 时，只有那些外引号是语法上的(并控制 shell 如何扩展值:内引号只是字面值，不会改变 $0 被处理.

The Problem: Single Quotes Aren't Syntactic Within Double Quotes

When you have "'$0'", only those outer quotes are syntactic (and control how the shell expands the value: The inner quotes are simply literal values, and don't change how $0 is treated.

如果您不想使用扩展名，请不要使用双引号.

If you don't want expansions to be honored, don't use double quotes.

# assign the directory separately, just to make the line shorter for easier reading
dir=/System/Library/PrivateFrameworks/Apple80211.framework/Versions/Current/Resources/

# now, expand the directory name in double quotes, but use the $'' quoting style otherwise
assignssid="$dir"$'airport -I | awk \'/ SSID/ {print substr($0, index($0, $2))}\''

在 $'' 里面，\' 是一个文字单引号(同样，\n 是一个换行符，\t 是制表符等)，并且不会发生参数扩展($'$0' 与 '$0' 相同，而不是 "$0").

Inside $'', \' is a literal single quote (likewise, \n is a newline, \t is a tab, etc), and no parameter expansions take place ($'$0' is the same as '$0', not "$0").

执行此操作后，您可以看到 echo 变量发出带有完整单引号内容的命令:

After doing this, you can see that echoing the variable emits your command with single-quoted contents intact:

$ echo "$assignssid"
/System/Library/PrivateFrameworks/Apple80211.framework/Versions/Current/Resources/airport -I | awk '/ SSID/ {print substr($0, index($0, $2))}'

...或者，与您可能预期的用例更密切相关:

...or, more germane to your presumably-intended use case:

current_ssid=$(eval "$assignssid")

<小时>

字面答案(POSIX sh):'"'"'

让我们以一种不是 bash-only的方式来做第一个答案:

---

Literal Answer (POSIX sh): '"'"'

Let's do the first answer in a way that isn't bash-only:

dir=/System/Library/PrivateFrameworks/Apple80211.framework/Versions/Current/Resources/
assignssid="$dir"'airport -I | awk '"'"'/ SSID/ {print substr($0, index($0, $2))}'"'"''

这个字符串的作用如下:

What this string does is as follows:

• 第一个 ' 关闭单引号上下文
• 第一个 " 打开一个双引号上下文
• 下面的'是文字，因此成为你价值的一部分
• 以下 " 关闭双引号上下文
• 最后的 ' 恢复了原来的单引号上下文

• The first ' closes the single-quoted context
• The first " opens a double-quoted context
• The following ' is literal, thus becomes part of your value
• The following " closes the double-quoted context
• The final ' resumes the original single-quoted context

字符串变量根本不是这项工作的合适工具.相反，使用函数:

A string variable isn't the appropriate tool for this job at all. Instead, use a function:

airport_dir=/System/Library/PrivateFrameworks/Apple80211.framework/Versions/Current/Resources/
assignssid() {
    "$airport_dir"/airport -I | awk '/ SSID/ {print substr($0, index($0, $2))}'
}

...以及此后:

current_ssid=$(assignssid)

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