bash 如何在一行上定义一个函数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/38595559/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to define a function on one line
提问by Steffen
Often when moving files around, I need to do the opposite later. So in my .bashrc I included this working code:
通常在移动文件时，我需要稍后做相反的事情。所以在我的 .bashrc 中我包含了这个工作代码：
rmv() {
mv /${1##*/} ${1%/*}
}
Now I wonder why I can't write this as a single liner. This is what I tried:
现在我想知道为什么我不能把它写成单行。这是我尝试过的：
rmv() {mv /${1##*/} ${1%/*}}
If I do so, I get this error:
如果我这样做，我会收到此错误：
-bash: .bashrc: line 1: syntax error near unexpected token `{mv'
回答by ruakh
In Bash, {is not automatically recognized as a special/separate token from what's around it. So you need whitespace between {and mv.
在 Bash 中，{不会自动将其识别为与周围事物不同的特殊/单独标记。所以你需要在{和之间留空格mv。
Additionally:
此外：
}needs to be the start of a command; so if it's not on its own line, you need;to terminate the previous command.- It's a best practice to always use double-quotes around any parameter expansion, since otherwise you'll get bizarre behaviors when the parameters include whitespace or special characters.
}需要是命令的开始；因此，如果它不在自己的行上，则需要;终止上一个命令。- 最佳做法是始终在任何参数扩展周围使用双引号，否则当参数包含空格或特殊字符时，您会出现奇怪的行为。
So:
所以：
rmv() { mv "/${1##*/}" "${1%/*}" ; }
