${var} 参数扩展表达式可以嵌套在 bash 中吗? [英] Can ${var} parameter expansion expressions be nested in bash?

问题描述

我拥有的是这个:

progname=${0%.*}
progname=${progname##*/}

这是否可以嵌套(或不嵌套)为一行，即单个表达式?

Can this be nested (or not) into one line, i.e. a single expression?

我试图从脚本名称中去除路径和扩展名，以便只保留基本名称.以上两行工作正常.我的C"天性只是驱使我更加混淆这些.

I'm trying to strip the path and extension off of a script name so that only the base name is left. The above two lines work fine. My 'C' nature is simply driving me to obfuscate these even more.

推荐答案

如果说嵌套，你的意思是这样的:

If by nest, you mean something like this:

#!/bin/bash

export HELLO="HELLO"
export HELLOWORLD="Hello, world!"

echo ${${HELLO}WORLD}

那么不，你不能嵌套 ${var} 表达式.bash 语法扩展器不会理解它.

Then no, you can't nest ${var} expressions. The bash syntax expander won't understand it.

但是，如果我理解您的问题，您可能会考虑使用 basename 命令 - 它从给定的文件名中去除路径，如果给定扩展名，也会去除它.例如，运行 basename/some/path/to/script.sh .sh 将返回 script.

However, if I understand your problem right, you might look at using the basename command - it strips the path from a given filename, and if given the extension, will strip that also. For example, running basename /some/path/to/script.sh .sh will return script.

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