需要文件而不执行代码? [英] Require file without executing code?

问题描述

这里有两个文件:

def method
  puts "This won't be outputted."
end

puts "This will be outputted."

main.rb

require "./file"

运行main.rb时，它将在file.rb中加载所有代码，因此我将得到这将被输出".在屏幕上.

When running main.rb it will load all the code inside file.rb so I will get "This will be outputted." on the screen.

是否可以在不运行代码的情况下加载文件?

Is it possible to load a file without having it to run the code?

因为我想加载所有方法(也在模块和类中)，而不必执行这些范围之外的代码.

Cause I want to load all the methods (in modules and classes too) without having to execute code outside these scopes.

推荐答案

是否可以在不运行代码的情况下加载文件?

Is it possible to load a file without having it to run the code?

否，ruby文件中的所有内容都是可执行代码，包括类和方法定义(例如，当您尝试在if语句内定义方法时，您会看到此代码，它工作得很好).因此，如果您不执行文件中的任何内容，则不会定义任何内容.

No, everything in a ruby file is executable code, including class and method definitions (you can see this when you try to define a method inside an if-statement for example, which works just fine). So if you wouldn't execute anything in the file, nothing would be defined.

但是，您可以告诉ruby，某些代码仅在直接运行文件时才执行-不需要时才执行.为此，只需将有问题的代码放在if __FILE__ == $0块中.因此对于您的示例，这将起作用:

You can however tell ruby that certain code shall only execute if the file is run directly - not if it is required. For this simply put the code in question inside an if __FILE__ == $0 block. So for your example, this would work:

def method
  puts "This won't be outputted."
end
if __FILE__ == $0
  puts "This will not be outputted."
end

main.rb

require "./file"

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