基类是否有可能在允许子类订阅
Angular2 中的可观察对象之前捕获
某些错误。
例如
export class SomeBaseClass {
constructor(private _http: Http, private _location: Location) {}
protected _fetchData(url): Observable<any> {
const headers = new Headers();
headers.append('Authorization', 'Token foo');
return this._http.get(url, {headers})
.map(response => response.json())
.catch(response => this._handle401(error));
}
private _handle401(response: Response) {
if(response.status === 401) {
this._location.go('/login');
}
// What should this return?
}
}
export class SomeClass extends SomeBaseClass {
constructor( _http: Http, _location: Location) {
super(_http, _location);
}
doTheThing() {
this._fetchData('/someUrl')
.subscribe(
response => this._handleResponse(response),
error => this._handleErrorThatIsNot401(error));
}
private _handleResponse(response) {
// ...
}
private _handleErrorThatIsNot401(error) {
// ...
}
}
catch
是我要找的东西吗?我应该使用 map
(或其他东西)吗?还是我完全以错误的方式解决了这个问题?
更新
这两个答案(到目前为止)都让我走上了正确的轨道,最终 - 我是这样解决的:
protected _get(url: string, data?: any): Observable<any> {
return super._get(url, data, this._authorizationHeader)
.map(response => response.json())
.catch(response => this._handle401(response));
}
private _handle401(response: Response): Observable<any> {
try {
if(response.status === 401) {
this._router.navigateByUrl('/login');
return Observable.throw(response.status);
}
} catch(err) {
console.warn('AuthenticatedHttpService._handle401');
console.error(err);
}
return Observable.of(response);
}
最佳答案
单独使用 catch
没有多大帮助,因为您订阅了客户端代码并且您必须从 catch
返回 Observable
。
我会按如下方式实现它:
Rx.Observable.of(42)
.do(v=>{throw new Error('test')})
.catch(Rx.Observable.of(undefined))
.filter(v=>{return v !== undefined})
.subscribe(
(e)=>{console.log('next', e)},
(e)=>{console.log('error', e)},
()=>{console.log('complete')}
);
关于javascript - 我可以在 Angular2 中可观察到的 RXJS 中捕获 "subscribe()"之前的某些错误吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36666452/