How to use Promise.all() with Typescript

Question

Here is what I want to do:

Promise.all([aurelia.start(), entityManagerProvider.initialize()])     .then((results:Array<any>) => {         let aurelia: any = results[0];         aurelia.setRoot();     }); 

aurelia.start() returns an Aurelia type, while initialize() returns void.

The compiler gives an error message that the type cannot be inferred from the usage.

What I am trying to achieve is to get them to run at the same time, as they are both very long processes, then run Aurelia.setRoot();

Answer 1

Its generally best to have arrays with consistent types. You can do the following manually though (passing in generic arguments):

Promise.all<Aurelia, void>(   [aurelia.start(), entityManagerProvider.initialize() ]) .then(results => {     let aurelia = results[0];     aurelia.setRoot(); });