[英]How do I make node.js execute some code only if my file is the running file?
I want to be able to start my express server directly via: 我希望能够通过以下方式直接启动我的快速服务器:
$ node app.js
But I also want to able to require that file, and have it return the app instance but actually not start the server. 但我也希望能够要求该文件,让它返回应用程序实例但实际上不启动服务器。 Then I can start later it with some options.
然后我可以稍后用一些选项开始。
app = require './app'
app.listen options.someCustomPort
I'm basically looking for the equivalent of this ruby snippet, but in node.js. 我基本上在寻找相当于这个ruby片段的东西,但在node.js中。
if __FILE__ == $0
app.listen options[:some_custom_port]
end
Is there an idiom for this? 这有成语吗?
Check 校验
module.parent
If it's null
or undefined
, you're the main file. 如果它为
null
或undefined
,则您是主文件。 If not, you've been require
d. 如果没有,你一直
require
d。 Your module.parent
is the module
object of the module that require
d you. 你的
module.parent
是require
你的模块的module
对象。
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