仅在从命令行调用脚本时运行代码 [英] Run code only if script called from the command line

问题描述

所以，当我从命令行调用我的脚本，我想要它接受一个int，并做一些与值：

So when I call my script from the command line, I want it to take in an int and do something with the value:

ruby script.rb


puts ARGV[0], etc...

但是，无论何时加载或需要脚本，并且不从命令行调用，我想完全跳过这部分代码。如何检测脚本是通过命令行调用，还是刚刚加载？谢谢！

However, whenever the script is loaded or required and not called from command line, I want to completely skip this part of the code. How can I detect whether the script has been called via command line, or just loaded? Thanks!

推荐答案

通常将它放在脚本底部：

It is common to put this at the bottom of your script:

if __FILE__==$0
  # this will only run if the script was the main, not load'd or require'd
end

因为我喜欢看到文件顶部的主要操作，所以我通常把 def run！作为文件中的第一个方法，然后使用以下命令结束文件：

Because I like to see the main action at the top of my file, I usually put a def run! as the first method in the file and then end the file with:

run! if __FILE__==$0

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