用Ruby解析字符串路径名

Question

我有一个类似于以下的路径字符串：

/some/long/path/filename.extension

我需要在Ruby中解析出“文件名”部分。

- sysconfig

Answer 1

使用File.basename方法的后缀参数：

# irb
irb(main):001:0> File.basename('/some/long/path/filename.extension', '.*')
=> "filename"
irb(main):002:0> File.basename('/some/long/path/filename.v1.extension', '.*')
=> "filename.v1"

参考文献：http://www.ruby-doc.org/core/classes/File.html#M000026

该链接提供了Ruby编程语言中File类的文档。其中，方法M000026是File类的实例方法之一。这个方法可以将文件名作为参数，并返回一个布尔值，表示该文件是否存在。

- qingbo

Answer 2

还有Pathname类：

require 'pathname'

Pathname.new("/a/b/c/d.txt").basename.to_s
=> "d.txt"

- Patelify

Answer 3

1. if you are looking for regex solution (as in tags), here it is:

   irb> "/some/long/path/filename.ext1.ext2".gsub(%r{.*/|\..*$},'')
   => "filename"
   

2. or more effective solution without regexp:

   irb> path = "/some/long/path/filename.ext1.ext2"
   => "/some/long/path/filename.extension"
   irb> filename = path[path.rindex('/')+1..-1]
   => "filename.ext1.ext2"
   

   and to crop the extension(s):

   • if you want to crop the last one:

     irb> filename[0,filename.rindex('.')]
     => "filename.ext1"
     

   • if you want to strip all the extensions (the same behaviour like the regex solution):

     irb> filename[0,filename.index('.')]
     => "filename"
     

- geronime