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I want to be able to start my express server directly via:
$ 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. Then I can start later it with some options.
app = require './app'
app.listen options.someCustomPort
I'm basically looking for the equivalent of this ruby snippet, but in node.js.
if __FILE__ == $0
app.listen options[:some_custom_port]
end
Is there an idiom for this?
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The usual way to run a Node. js program is to run the globally available node command (once you install Node. js) and pass the name of the file you want to execute. While running the command, make sure you are in the same directory which contains the app.
Method 1: Using ctrl+C key: When running a program of NodeJS in the console, you can close it with ctrl+C directly from the console with changing the code shown below: Method 2: Using process. exit() Function: This function tells Node. js to end the process which is running at the same time with an exit code.
In windows you can simply go to the Task Manager and check for node in the application list. If it is there then it is running in the machine.
Check
module.parent
If it's null or undefined, you're the main file. If not, you've been required. Your module.parent is the module object of the module that required you.
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