TSLint在Angle 2中的错误

Question

I have several mistakes in my code. I'm use Angular 2 + TSLint:

constructor(http: Http) {
    this.http = http;
--> let currentUser = JSON.parse(localStorage.getItem("currentUser"));
    this.token = currentUser && currentUser.token;
}

In currentUser i have this error: message: 'expected variable-declaration: 'currentUser' to have a typedef ;

public loginC (username: string, password: string): Observable<boolean> {
    return this.http.post( authURL + loginURL,
                     --> JSON.stringify({ password: password, username: username }))
    .map((response: Response) => {
        let token: string = response.json() && response.json().token;
        if (token) {
          this.token = token;
      --> localStorage.setItem("currentUser", JSON.stringify({token: token, username: username}));
            return true;
        } else {
            return false;
        }
    });
}

And in password: password, username: username this: message: 'Expected property shorthand in object literal. I really understand this. And finalli i can write a simple model: any {} ;

export class LoginComponent {
--> public model: any = {};
public loading: boolean = false;
public error: string = "";
constructor (private router: Router, private authenticationService: ServerDataComponent) {
    // 
}

public login(): void {
    this.loading = true;
    this.authenticationService.loginC(this.model.username, this.model.password)
        .subscribe(result => {
           --> if (result === true) {
                this.router.navigate(["/table_per"]);
            } else {
                this.error = "Введен неверный логин и/или пароль";
                this.loading = false;
            }
        });
}

For any - Type declaration of 'any' is forbidden ;

Ror result - expected arrow-parameter: 'result' to have a typedef

Answer 1

For expected variable-declaration: 'currentUser' to have a typedef you can define an interface for your custom type.

export interface User {
  token: string;
}

And use it to set the type.

let currentUser: User = JSON.parse(localStorage.getItem("currentUser"));

For Expected property shorthand in object literal you can use the shorthand syntax when the key name matches the name of the variable.

JSON.stringify({token, username})

For Type declaration of 'any' is forbidden you can try to change the type to Object . If not you'll need to declare another interface for your model.

public model: Object = {};

For expected arrow-parameter: 'result' to have a typedef you need to set the type of the parameter.

.subscribe((result: boolean) => {