代码之家 › 专栏 › 技术社区 › wonderfulthunk

相对于绝对路径转换(对于perforce-depot路径),我能做得更好吗?

absolute-path relative-path algorithm ruby

wonderfulthunk · 技术社区 · 14 年前

A.任何关于改进它的建议都是受欢迎的,特别是缺乏非连续的dotdot支持。

我发现了很多使用其他语言转换其他类型的相对路径的例子,但是没有一个适合我的情况。

以下是我需要转换的示例路径:

//depot/foo/../bar/single.c

//depot/foo/docs/../../other/double.c

//depot/foo/usr/bin/../../../else/more/triple.c

收件人:

//depot/bar/single.c

//depot/other/double.c

//depot/else/more/triple.c

begin

paths = File.open(ARGV[0]).readlines

puts(paths)

new_paths = Array.new

paths.each { |path|
  folders = path.split('/')
  if ( folders.include?('..') )
    num_dotdots = 0
    first_dotdot = folders.index('..')
    last_dotdot = folders.rindex('..')
    folders.each { |item|
      if ( item == '..' )
        num_dotdots += 1
      end
    }
    if ( first_dotdot and ( num_dotdots > 0 ) ) # this might be redundant?
      folders.slice!(first_dotdot - num_dotdots..last_dotdot) # dependent on consecutive dotdots only
    end
  end

  folders.map! { |elem| 
    if ( elem !~ /\n/ )
      elem = elem + '/' 
    else
      elem = elem
    end
  }
  new_paths << folders.to_s

}

puts(new_paths)


end

3 回复 | 直到 14 年前

Marc-André Lafortune 14 年前

S File.expand_path 对你来说:

[
  '//depot/foo/../bar/single.c',
  '//depot/foo/docs/../../other/double.c',
  '//depot/foo/usr/bin/../../../else/more/triple.c'
].map {|p| File.expand_path(p) }
# ==> ["//depot/bar/single.c", "//depot/other/double.c", "//depot/else/more/triple.c"]

Arkku 14 年前

为什么不用呢 File.expand_path

irb(main):001:0> File.expand_path("//depot/foo/../bar/single.c")
=> "//depot/bar/single.c"
irb(main):002:0> File.expand_path("//depot/foo/docs/../../other/double.c")
=> "//depot/other/double.c"
irb(main):003:0> File.expand_path("//depot/foo/usr/bin/../../../else/more/triple.c")
=> "//depot/else/more/triple.c"

absolute = []
relative = "//depot/foo/usr/bin/../../../else/more/triple.c".split('/')
relative.each { |d| if d == '..' then absolute.pop else absolute.push(d) end }
puts absolute.join('/')

compie 14 年前

Python代码:

paths = ['//depot/foo/../bar/single.c',
         '//depot/foo/docs/../../other/double.c',
         '//depot/foo/usr/bin/../../../else/more/triple.c']

def convert_path(path):
    result = []
    for item in path.split('/'):
        if item == '..':
            result.pop()
        else:
            result.append(item)
    return '/'.join(result)

for path in paths:
    print convert_path(path)

印刷品:

//depot/bar/single.c
//depot/other/double.c
//depot/else/more/triple.c

您可以在Ruby中使用相同的算法。