仅从 Bash 脚本中的路径获取文件名 [英] Get just the filename from a path in a Bash script

问题描述

如何只获取没有扩展名和路径的文件名?

How would I get just the filename without the extension and no path?

以下没有给我扩展名，但我仍然附上了路径:

The following gives me no extension, but I still have the path attached:

source_file_filename_no_ext=${source_file%.*}

推荐答案

大多数类 UNIX 操作系统都有一个 basename 可执行文件，用于非常相似的目的(以及 dirname 用于路径):

Most UNIX-like operating systems have a basename executable for a very similar purpose (and dirname for the path):

pax> a=/tmp/file.txt
pax> b=$(basename $a)
pax> echo $b
file.txt

不幸的是，这只是给了你文件名，包括扩展名，所以你需要找到一种方法来去掉它.

That unfortunately just gives you the file name, including the extension, so you'd need to find a way to strip that off as well.

因此，既然您无论如何都必须这样做，那么您不妨找到一种可以去除路径和扩展名的方法.

So, given you have to do that anyway, you may as well find a method that can strip off the path and the extension.

一种方法(这是一个 bash-only 解决方案，不需要其他可执行文件):

One way to do that (and this is a bash-only solution, needing no other executables):

pax> a=/tmp/xx/file.tar.gz
pax> xpath=${a%/*} 
pax> xbase=${a##*/}
pax> xfext=${xbase##*.}
pax> xpref=${xbase%.*}
pax> echo;echo path=${xpath};echo pref=${xpref};echo ext=${xfext}

path=/tmp/xx
pref=file.tar
ext=gz

那个小片段设置了 xpath(文件路径)、xpref(文件前缀，你特别要求的)和 xfext(文件扩展名).

That little snippet sets xpath (the file path), xpref (the file prefix, what you were specifically asking for) and xfext (the file extension).

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