TSLint角度错误2

时间:2016-11-23 12:31:04

标签: angular typescript tslint

我的代码中有几个错误。我使用的是Angular 2 + TSLint:

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

在currentUser中我有这个错误: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;
        }
    });
}

password: password, username: username这个:message: 'Expected property shorthand in object literal. 我真的明白这一点。而finalli我可以写一个简单的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;
            }
        });
}

任何 - Type declaration of 'any' is forbidden;

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

1 个答案:

答案 0 :(得分:1)

对于expected variable-declaration: 'currentUser' to have a typedef,您可以为自定义类型定义interface

export interface User {
  token: string;
}

并使用它来设置类型。

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

对于Expected property shorthand in object literal,当密钥名称与变量名称匹配时,您可以使用简写语法。

JSON.stringify({token, username})

对于Type declaration of 'any' is forbidden,您可以尝试将类型更改为Object。如果不是,您需要为您的模型声明另一个interface

public model: Object = {};

对于expected arrow-parameter: 'result' to have a typedef,您需要设置参数的类型。

.subscribe((result: boolean) => {