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

就我个人而言，我发现这对于在文件夹中循环文件最有用，前瞻性安全:

Dir['/etc/path/*'].each do |file_name|
  next if File.directory? file_name 
end

Dir.entries(folder)

例子:

Dir.entries(".")

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

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

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

当加载操作目录中的所有文件名时，您可以使用

Dir.glob (*)

这将返回应用程序正在运行的上下文中的所有文件(注意，对于Rails，这是应用程序的顶级目录)

您可以在这里(https://ruby-doc.org/core-2.7.1/Dir.html#method-c-glob)进行额外的匹配和递归搜索