在 bash 中嵌套引号
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Nesting quotes in bash
提问by Tyilo
I want to something like this in bash:
我想在 bash 中做这样的事情：
alias foo='bar="$(echo hello world | grep \"hello world\")"; echo $bar;'; foo
Expected output: hello world
预期输出：你好世界
Ouput: grep: world": No such file or directory
输出：grep：world”：没有那个文件或目录
The outer quotes have to be single quotes, with double quotes $bar would be empty.
The next quotes have to be double quotes, with single quotes $() wouldn't expand.
The inner quotes could be both type of quotes, but single quotes doesn't allow single quotes inside of them.
外部引号必须是单引号，双引号 $bar 将为空。
下一个引号必须是双引号，单引号 $() 不会扩展。
内引号可以是两种类型的引号，但单引号不允许在其中包含单引号。
How to I achieve this?
我如何实现这一目标？
回答by glenn Hymanman
The stuff inside $()represents a subshell, so you are allowed to place un-escaped double quotes inside
里面的东西$()代表一个子shell，所以你可以在里面放置未转义的双引号
alias foo='bar="$(echo testing hello world | grep "hello world")"; echo "$bar"'
回答by l0b0
It's a bit unclear what "something like this" means, but the simplest way to achieve what seems to be the point here is a simple function:
有点不清楚“像这样的东西”是什么意思，但实现这里似乎是重点的最简单方法是一个简单的函数：
foo() {
echo 'hello world' | grep 'hello world'
}
foo
- There's no need for an intermediate variable assignment (it will be lost anyway).
- Functions are generally preferred over aliases because of more flexibility (parameter handling) and readability (multiple lines; less escaping).
- Always use the simplest solution which could possibly work.
- 不需要中间变量赋值（无论如何它都会丢失）。
- 函数通常比别名更受欢迎，因为它具有更大的灵活性（参数处理）和可读性（多行；更少的转义）。
- 始终使用可能有效的最简单的解决方案。
回答by Karoly Horvath
Escape the spaces
逃离空间
alias foo='bar="$(echo hello world | grep hello\ world)"; echo $bar;'
回答by bmk
The double quotes around $()are not necessary:
周围的双引号$()不是必需的：
alias foo='bar=$(echo hello world | grep "hello world"); echo $bar;'
foo
# Output:
hello world
