How to require module only if exist. React native

Question

Example:

let tmp;

try {
  tmp = require(\'module-name\');
} catch(e) {
  return;
}

I get error (react native Metro Bundler):

Answer 1

Use require.resolve which will return resolved file name.

function checkModuleAvailability (module) {
  try {
    require.resolve(module);
    return true
  } catch(e) {
    console.log(`${module} not found`);
  }
  return false
}

const moduleAvailable = checkModuleAvailability(MODULE_NAME) // true or false

Answer 2

That's what works for me:

let myPackage;
const myPackageToRequire = 'my-package-to-require';
try {
  myPackage = require.call(null, myPackageToRequire);
} catch (e) {}

The variable definition const myPackageToRequire = 'my-package-to-require'; is necessary here.

Hope I helped.

Answer 3

Loading optional dependencies via try-catch has been added in Metro 0.59, which in turn means that you should be able to use your original code in React Native 0.63 if you turn it on in metro.config.js:

module.exports = {
  transformer: {
    allowOptionalDependencies: true,
  },
}