有点奇怪。在我的Angular应用程序上,我有一个refreshUser
方法来在显示所有屏幕数据之前加载可能已登录的用户。为此,我有以下ngOnInit
方法:
async ngOnInit() {
await this.authService.refreshUser();
}
refreshUser方法具有以下代码:
public refreshUser(): Promise<void> {
return this.manager.signinSilent()
.then(user => {
this.user = user;
})
.catch(e => {
console.log('User not logged in', e);
});
}
在全新的导航上,它可以顺利运行,并且出现错误:
User not logged in ErrorResponse: login_required
at new e (main.2a53201b6340bff6258a.js:1)
at t._processSigninParams (main.2a53201b6340bff6258a.js:1)
at t.validateSigninResponse (main.2a53201b6340bff6258a.js:1)
at main.2a53201b6340bff6258a.js:1
at e.invoke (polyfills.3efb6da5a4e032fb1cf9.js:1)
at Object.onInvoke (main.2a53201b6340bff6258a.js:1)
at e.invoke (polyfills.3efb6da5a4e032fb1cf9.js:1)
at t.run (polyfills.3efb6da5a4e032fb1cf9.js:1)
at polyfills.3efb6da5a4e032fb1cf9.js:1
at e.invokeTask (polyfills.3efb6da5a4e032fb1cf9.js:1)
(用户从未登录过)。
一种奇怪的情况是,如果我有已登录的用户,则关闭Web浏览器并再次打开它,屏幕挂起大约10秒钟,然后出现另一个错误:
main.2a53201b6340bff6258a.js:1 User not logged in Error: Frame window timed out
at t._error (main.2a53201b6340bff6258a.js:1)
at t._timeout (main.2a53201b6340bff6258a.js:1)
at e.invokeTask (polyfills.3efb6da5a4e032fb1cf9.js:1)
at Object.onInvokeTask (main.2a53201b6340bff6258a.js:1)
at e.invokeTask (polyfills.3efb6da5a4e032fb1cf9.js:1)
at t.runTask (polyfills.3efb6da5a4e032fb1cf9.js:1)
at t.invokeTask (polyfills.3efb6da5a4e032fb1cf9.js:1)
at invoke (polyfills.3efb6da5a4e032fb1cf9.js:1)
但是,当屏幕加载时,我单击“登录”(因为没有登录),因此在加载组件之前会触发“经典” AuthGuard,因此我运行标准的startAuthentication()
:
canActivate(route: ActivatedRouteSnapshot, state: RouterStateSnapshot): boolean | Observable<boolean> | Promise<boolean> {
if (this.authService.isLoggedIn()) {
return true;
}
this.authService.startAuthentication();
return false;
}
快速登录用户。我的问题是:就用户登录而言,该功能和signingSilent功能之间有什么区别?但是,由于确实如此,因此至少在时间和用户体验方面,行为不是预期的。我希望能够“如果可能的话登录用户,但不提示登录屏幕”。这有道理吗?
谢谢!