def get_path_content(dir)
  queue = Queue.new
  result = []
  queue << dir
  until queue.empty?
    current = queue.pop
    Dir.entries(current).each { |file|
      full_name = File.join(current, file)
      if not (File.directory? full_name)
        result << full_name
      elsif file != '.' and file != '..'
          queue << full_name
      end
    }
  end
  result
end

返回文件在目录和所有子目录中的相对路径

Dir.entries(folder)

例子:

Dir.entries(".")

来源:http://ruby-doc.org/core/classes/Dir.html method-c-entries

def get_path_content(dir)
  queue = Queue.new
  result = []
  queue << dir
  until queue.empty?
    current = queue.pop
    Dir.entries(current).each { |file|
      full_name = File.join(current, file)
      if not (File.directory? full_name)
        result << full_name
      elsif file != '.' and file != '..'
          queue << full_name
      end
    }
  end
  result
end

返回文件在目录和所有子目录中的相对路径

这对我来说很管用:

如果你不想要隐藏文件[1]，使用Dir[]:

# With a relative path, Dir[] will return relative paths 
# as `[ './myfile', ... ]`
#
Dir[ './*' ].select{ |f| File.file? f } 

# Want just the filename?
# as: [ 'myfile', ... ]
#
Dir[ '../*' ].select{ |f| File.file? f }.map{ |f| File.basename f }

# Turn them into absolute paths?
# [ '/path/to/myfile', ... ]
#
Dir[ '../*' ].select{ |f| File.file? f }.map{ |f| File.absolute_path f }

# With an absolute path, Dir[] will return absolute paths:
# as: [ '/home/../home/test/myfile', ... ]
#
Dir[ '/home/../home/test/*' ].select{ |f| File.file? f }

# Need the paths to be canonical?
# as: [ '/home/test/myfile', ... ]
#
Dir[ '/home/../home/test/*' ].select{ |f| File.file? f }.map{ |f| File.expand_path f }

现在,Dir。条目将返回隐藏的文件，并且您不需要通配符asterix(您可以直接将目录名传递给变量)，但它将直接返回basename，因此File. xml将返回文件。XXX函数不能工作。

# In the current working dir:
#
Dir.entries( '.' ).select{ |f| File.file? f }

# In another directory, relative or otherwise, you need to transform the path 
# so it is either absolute, or relative to the current working dir to call File.xxx functions:
#
home = "/home/test"
Dir.entries( home ).select{ |f| File.file? File.join( home, f ) }

[1] .dotfile在unix上，我不知道Windows上

在Ruby 2.5中，你现在可以使用Dir.children。它以数组的形式获取文件名，除了"."和".."

例子:

Dir.children("testdir")   #=> ["config.h", "main.rb"]

http://ruby-doc.org/core-2.5.0/Dir.html#method-c-children

一个简单的方法是:

dir = './' # desired directory
files = Dir.glob(File.join(dir, '**', '*')).select{|file| File.file?(file)}

files.each do |f|
    puts f
end