webpack try to require a module that might be missing

Question

I want to try to require a module in a webpack build and if that file is not found just do nothing, don't throw an error etc.

I tried doing it this way:

try {
  const local = require('./config-local.js');
  extend(config, local);
} catch (err) {
  // do nothing here
}

Basically what I want to do is to extend a config object with a local config if that file is found but if it isn't just don't extend it with anything.

Webpack throws an error that module is missing even though that require is wrapped in a try/catch clause.

How to tell webpack to ignore it?

Answer 1

Based on @trysis comment, I ended up doing this in my project:

const REQUIRE_TERMINATOR = '';
try {
  const local = require(`./config-local.js${REQUIRE_TERMINATOR}`);
  extend(config, local);
} catch (err) {
  // do nothing here
}

This will give WebPack enough information to include all files starting with config-local.js (which will be exactly the file you want), but also confuse it enough that it won't try to verify file's existence ahead of time, so your try/catch block will trigger during runtime.