如果在 Ruby 中使用 File 类不存在目录，如何创建目录？

How do I create directory if none exists using File class in Ruby?

提问by marcamillion

I have this statement:

我有这样的声明：

File.open(some_path, 'w+') { |f| f.write(builder.to_html)  }

Where

在哪里

some_path = "somedir/some_subdir/some-file.html"

What I want to happen is, if there is no directory called somediror some_subdiror both in the path, I want it to automagically create it.

我希望发生的是，如果有一个名为没有目录somedir或some_subdir路径或两个，我希望它自动地创建它。

How can I do that?

我怎样才能做到这一点？

回答by Eureka

You can use FileUtils to recursively create parent directories, if they are not already present:

您可以使用 FileUtils 递归创建父目录（如果它们尚不存在）：

require 'fileutils'

dirname = File.dirname(some_path)
unless File.directory?(dirname)
  FileUtils.mkdir_p(dirname)
end

Edit: Here is a solution using the core libraries only (reimplementing the wheel, not recommended)

编辑：这是一个仅使用核心库的解决方案（重新实现轮子，不推荐）

dirname = File.dirname(some_path)
tokens = dirname.split(/[\/\]/) # don't forget the backslash for Windows! And to escape both "\" and "/"

1.upto(tokens.size) do |n|
  dir = tokens[0...n]
  Dir.mkdir(dir) unless Dir.exist?(dir)
end

回答by Andrey Mikhaylov - lolmaus

For those looking for a way to create a directory if it doesn't exist, here's the simple solution:

对于那些正在寻找创建目录（如果目录不存在）的方法的人，这里有一个简单的解决方案：

require 'fileutils'

FileUtils.mkdir_p 'dir_name'

Based on Eureka's comment.

基于尤里卡的评论。

回答by Licysca

directory_name = "name"
Dir.mkdir(directory_name) unless File.exists?(directory_name)

回答by skplunkerin

Based on others answers, nothing happened (didn't work). There was no error, and no directory created.

根据其他人的答案，什么也没发生（不起作用）。没有错误，也没有创建目录。

Here's what I needed to do:

这是我需要做的：

require 'fileutils'
response = FileUtils.mkdir_p('dir_name')

I needed to create a variable to catch the response that FileUtils.mkdir_p('dir_name')sends back... then everything worked like a charm!

我需要创建一个变量来捕捉FileUtils.mkdir_p('dir_name')发回的响应......然后一切都像魅力一样！

回答by CopyLeft

The top answer's "core library" only solution was incomplete. If you want to only use core libraries, use the following:

最佳答案的“核心库”唯一解决方案不完整。如果您只想使用核心库，请使用以下命令：

target_dir = ""

Dir.glob("/#{File.join("**", "path/to/parent_of_some_dir")}") do |folder|
  target_dir = "#{File.expand_path(folder)}/somedir/some_subdir/"
end

# Splits name into pieces
tokens = target_dir.split(/\//)

# Start at '/'
new_dir = '/'

# Iterate over array of directory names
1.upto(tokens.size - 1) do |n|

  # Builds directory path one folder at a time from top to bottom
  unless n == (tokens.size - 1)
    new_dir << "#{tokens[n].to_s}/" # All folders except innermost folder
  else
    new_dir << "#{tokens[n].to_s}" # Innermost folder
  end

  # Creates directory as long as it doesn't already exist
  Dir.mkdir(new_dir) unless Dir.exist?(new_dir)
end

I needed this solution because FileUtils' dependency gem rmagick prevented my Rails app from deploying on Amazon Web Services since rmagick depends on the package libmagickwand-dev (Ubuntu) / imagemagick (OSX) to work properly.

我需要这个解决方案是因为 FileUtils 的依赖 gem rmagick 阻止了我的 Rails 应用程序在 Amazon Web Services 上部署，因为 rmagick 依赖包 libmagickwand-dev (Ubuntu) / imagemagick (OSX) 才能正常工作。

回答by ironsand

How about using Pathname?

怎么用Pathname？

require 'pathname'
some_path = Pathname("somedir/some_subdir/some-file.html")
some_path.dirname.mkdir_p
some_path.write(builder.to_html)

回答by Shell Bryson

Along similar lines (and depending on your structure), this is how we solved where to store screenshots:

沿着类似的路线（并且取决于您的结构），这就是我们解决存储屏幕截图的位置的方法：

In our env setup (env.rb)

在我们的环境设置 (env.rb)

screenshotfolder = "./screenshots/#{Time.new.strftime("%Y%m%d%H%M%S")}"
unless File.directory?(screenshotfolder)
  FileUtils.mkdir_p(screenshotfolder)
end
Before do
  @screenshotfolder = screenshotfolder
  ...
end

And in our hooks.rb

在我们的 hooks.rb

  screenshotName = "#{@screenshotfolder}/failed-#{scenario_object.title.gsub(/\s+/,"_")}-#{Time.new.strftime("%Y%m%d%H%M%S")}_screenshot.png";
  @browser.take_screenshot(screenshotName) if scenario.failed?

  embed(screenshotName, "image/png", "SCREENSHOT") if scenario.failed?