我想使用Ruby从一个文件夹中获得所有文件名。


当前回答

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

其他回答

Dir.entries(folder)

例子:

Dir.entries(".")

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

这是一个在目录中查找文件的解决方案:

files = Dir["/work/myfolder/**/*.txt"]

files.each do |file_name|
  if !File.directory? file_name
    puts file_name
    File.open(file_name) do |file|
      file.each_line do |line|
        if line =~ /banco1/
          puts "Found: #{line}"
        end
      end
    end
  end
end
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

如果你用空格创建目录:

mkdir "a b"
touch "a b/c"

你不需要转义目录名,它会自动完成:

p Dir["a b/*"] # => ["a b/c"]

您还有快捷方式选项

Dir["/path/to/search/*"]

如果你想在任何文件夹或子文件夹中找到所有Ruby文件:

Dir["/path/to/search/**/*.rb"]