将JS函数实现为typescript

Question

I'm implementing a JS function into typescript, it works fine on JS, but on TS, I get:

[ts] Cannot find name 'PasscodeAuth'. Did you mean 'passcodeAuth'?

function passcodeAuth() {
  return PasscodeAuth.isSupported()
    .then(() => {
      return PasscodeAuth.authenticate()
    })
    .then(success => {
      AlertIOS.alert('Authenticated Successfully');
    })
    .catch(error => {
      console.log(error)
      AlertIOS.alert(error.message);
    });
}

So is not an import, or defined anywhere, but it works on JS, so how can I make it recognizable for TS? On JS, I can name my parameter as any word on JS and it will have isSupported and authenticate, even though is now a real parameter?

thanks!

Answer 1

Putting aside the issue of if PasscodeAuth actually exists, you can inform typescript of it by using a declare statement. For your example:

// Put this at the root level of the file that uses PasscodeAuth
declare const PasscodeAuth: any;

That will declare PasscodeAuth as a variable with type any . You can then use it in your file without errors related to that variable, but since it's typed as an any , you won't be getting any type safety from it. If you happen to know the type for it, you can specify a more specific type instead of any .

A declare statement is only a type annotation, so that line won't actually have any effect on the emitted javascript code. You mentioned that your code already works as-is, despite the typescript errors, so a declaration sounds like exactly what you need.