Ah, the type-level equality operator. @MattMcCutchen has come up with a solution involving generic conditional types which does a decent job of detecting when two types are exactly equal, as opposed to just mutually assignable. In a perfectly sound type system, “mutually assignable” and “equal” would probably be the same thing, but TypeScript isn’t perfectly … Read more
canActivate needs to return an Observable that completes: @Injectable() export class AuthGuard implements CanActivate { constructor(private auth: AngularFireAuth, private router: Router) {} canActivate(route:ActivatedRouteSnapshot, state:RouterStateSnapshot):Observable<boolean>|boolean { return this.auth.map((auth) => { if (auth) { console.log(‘authenticated’); return true; } console.log(‘not authenticated’); this.router.navigateByUrl(‘/login’); return false; }).first(); // this might not be necessary – ensure `first` is imported if you … Read more
To augment Ryan’s answer with guidance from the TypeScript Do’s and Don’ts: Don’t ever use the types Number, String, Boolean, Symbol, or Object These types refer to non-primitive boxed objects that are almost never used appropriately in JavaScript code. /* WRONG */ function reverse(s: String): String; Do use the types number, string, boolean, and symbol. … Read more
You will get ” Unexpected anonymous System.register call” because the references are not being loaded in the correct order. I use JSPM to properly build my angular app for production. There are 4 parts to the process. Part 1: Compile your typescript files var ts = require(“gulp-typescript”); var tsProject = ts.createProject(“./App/tsconfig.json”); gulp.task(“compile:ts”, function () { … Read more
found what I was looking for: Declare vs. var var creates a new variable. declare is used to tell TypeScript that the variable has been created elsewhere. If you use declare, nothing is added to the JavaScript that is generated – it is simply a hint to the compiler. For example, if you use an … Read more
TypeScript allows you to declare overloads but you can only have one implementation and that implementation must have a signature that is compatible with all overloads. In your example, this can easily be done with an optional parameter as in, interface IBox { x : number; y : number; height : number; width : number; … Read more
var stuff: { [key: string]: string; } = {}; stuff[‘a’] = ”; // ok stuff[‘a’] = 4; // error // … or, if you’re using this a lot and don’t want to type so much … interface StringMap { [key: string]: string; } var stuff2: StringMap = { }; // same as above
You have to narrow down the type. You can do so by using the in operator. const getText = (obj: Obj1 | Obj2): string => { if (“message” in obj) { return obj.message } return obj.text }
A few examples follow, going from basic through to adding transformations after the request and/or error handling: Basic: // Implementation code where T is the returned data shape function api<T>(url: string): Promise<T> { return fetch(url) .then(response => { if (!response.ok) { throw new Error(response.statusText) } return response.json<T>() }) } // Consumer api<{ title: string; message: … Read more
There are actually a number of modules which attempt to translate TypeScript type information into runtime information that can be used to validate data structures. I’ll attempt to list and compare the various solutions here. (ordered roughly by how effective/general-purpose I estimate them to be; yes, this is somewhat subjective!) Core features: (marked with ✔️yes, … Read more