必须在 Windows 7 上运行 ruby​​ 脚本并获得权限被拒绝 EACCES [英] Have to run ruby script on Windows 7 and got Permission denied EACCES

问题描述

我必须在 Windows 7 上运行 ruby​​ 脚本(我知道这是个坏主意).我的脚本创建文件夹(如果它们不存在)并将文件复制到其中.我正在使用 FileUtils lib 来执行这项工作，例如:

I have to run ruby script on windows 7 (I know that it's bad idea). My script creates folders (if they is not exist) and copies files into them. I'm using FileUtils lib to perform this job, like:

FileUtils.mkdir_p(path)
FileUtils.cp_r(file.path, path)
FileUtils.touch(file)

在 ubuntu 和 mac 上一切正常，但在 Windows 7 机器上我遇到了下一个错误:

On ubuntu and mac everything is ok, but on windows 7 machine I got next error:

Permission denied - ./program_folder/input/. (Errno::EACCES)

在此代码行上:

Dir.entries('./program_folder/input').map { |file_name| File.new("./program_folder/input/#{file_name}") }.compact

有什么想法可以解决吗?

Any ideas how can I fix it?

我尝试以管理员访问权限运行 ruby​​ 和 irb termianl 并尝试在所有路径上执行 FileUtils.chmod_R(0777, @path) 但仍然没有任何更改...

I have tried to run ruby and irb termianl with administrator access and tried to do FileUtils.chmod_R(0777, @path) on all paths but still no changes...

推荐答案

你的命令

Dir.entries('./program_folder/input').map { |file_name| 
  File.new("./program_folder/input/#{file_name}")
}.compact

尝试创建一个与您之前阅读的文件/文件夹同名的文件.

tries to create a File with the same name as the file/folder you read before.

详细说明:

1. Dir.entries('.') 找到的第一个文件是实际目录 (.).
2. "./program_folder/input/#{file_name}" 是 ./program_folder/input/.(现有目录).
3. 这个目录路径应该是一个新文件的路径.
4. 使用 File.new，您无法将目录作为文件打开.

1. The first file found by Dir.entries('.') is the actual directory (.).
2. "./program_folder/input/#{file_name}" is ./program_folder/input/. (an existing directory).
3. This directory path should be the path of a new file.
4. With File.new you can't open a directory as a File.

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评论后备注:

---

Remark after comment:

在 Dir.entries 中调用 File.new - 创建文件句柄.如果没有模式，它会尝试打开现有文件(文件，而不是目录！).. 是一个不能作为文件打开的目录.

Inside the Dir.entries you call File.new - that creates a file handle. Without a mode, it tries to open an existing File (File, not Directory!). . is a directory which can't be opended as a file.

如果您只想要文件名，则不需要 File.new，字符串 "./program_folder/input/#{file_name}" 将是足够的.更好的解决方案是 File.join 方法:

If you only want the filename, you don't need the File.new, the string "./program_folder/input/#{file_name}" would be enough. A better solution would be the File.join method:

File.join("./program_folder/input", file_name)

或

File.join(".", "program_folder", "input", file_name)

如果你需要真实的文件名，你可以检查目录:

If you need on real filename, you can check for directories:

Dir.entries('./program_folder/input').map { |file_name|
  "./program_folder/input/#{file_name}" unless File.directory?("./program_folder/input/#{file_name}")
}.compact

或者更好的是，您删除目录:

or better, you remove the directories:

Dir.entries('.').delete_if{|file_name| 
  File.directory?(file_name)
}

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