2

我有一个 main router-outlet，用于显示登录屏幕（/login）和主内容屏幕（登录后显示）（/main）。

一旦用户在内容屏幕上，我想在顶部显示一个导航栏，有 2 个选项（例如，“概述”、“见解”）。此导航栏常见于OverviewComponent和InsightsComponent

在此导航栏下方，我想显示另一个路由器插座，它将根据用户在导航栏中单击的内容加载OverviewComponent或。InsightsComponent如果我将“/overview”和/“insights”作为路线，它将直接向我显示相应的组件，而不是导航栏。

以下是我当前的路由配置（这是不对的）：

const appRoutes: Routes = [
  { path: 'main', component:  MainComponent},
  { path: 'overview', component:  OverviewComponent},
  { path: 'insights', component: InsightsComponent },
  { path: 'login', component: LoginComponent },
  { path: '',
    redirectTo: '/login',
    pathMatch: 'full'
  },
  { path: '**', component: PageNotFoundComponent }
];

请让我知道我们是否可以在 angular2 angular4 中实现这一点。我正在使用以下版本：

"@angular/core": "^4.0.0"
"@angular/router": "^4.0.0"
"@angular/cli": "1.0.1"

******************尝试2 - 仍然不工作******************

const appRoutes: Routes = [
  { path: 'main',
    children: [
      { path: '', component: MainComponent },
      { path: 'overview', component:  OverviewComponent },
      { path: 'insights', component: InsightsComponent },
    ]

  },
  { path: 'login', component: LoginComponent },
  { path: '',
    redirectTo: '/login',
    pathMatch: 'full'
  },
  { path: '**', component: PageNotFoundComponent }
];

*******尝试 2 - 包含所有组件的 Sudo 代码 - 仍然无法正常工作*******

//app.component.html

<router-outlet></router-outlet>


//app.module.ts

const appRoutes: Routes = [
  { path: 'main',
    children: [
      { path: '', component: MainComponent },
      { path: 'overview', component:  OverviewComponent },
      { path: 'insights', component: InsightsComponent },
    ]

  },
  { path: 'login', component: LoginComponent },
  { path: '',
    redirectTo: '/login',
    pathMatch: 'full'
  },
  { path: '**', component: PageNotFoundComponent }
];

//login.component.html
<div class="text-center vertical-center">
    <form>
        <div class="horizontal">
            <label for="">Email</label>
            <input class="form-control" type="text" name="" value="">
        </div>
        <div class="horizontal">
            <label for="">Password</label>
            <input class="form-control" type="password" name="" value="">
        </div>
        <button class="btn btn-primary" (click)="navigate()">Login</button>
    </form>
</div>

//login.component.ts
navigate() {
    this.router.navigate(['./main']);
  }

//main.component.html

<app-header></app-header>
<router-outlet></router-outlet>


//app.header.html

<ul class="nav navbar-nav">
               <li class=""><a routerLink="/main/overview" routerLinkActive="active">OVERVIEW</a></li>
                <li class=""><a routerLink="/main/insights" routerLinkActive="active">INSIGHTS</a></li>
            </ul>

//overview.html
<p>This is overview section</p>

//insights.html
<p>This is insights section</p>

********尝试3 - 工作********

const appRoutes: Routes = [
  { path: 'main', component: MainComponent,
    children: [
      { path: '', component: MainComponent },
      { path: 'overview', component:  OverviewComponent },
      { path: 'insights', component: InsightsComponent },
    ]

  },
  { path: 'login', component: LoginComponent },
  { path: '',
    redirectTo: '/login',
    pathMatch: 'full'
  },
  { path: '**', component: PageNotFoundComponent }
];
4

2 回答 2

6

因此，如果我的问题是正确的，您最初希望有登录屏幕，并且在用户登录后，您希望他看到 /main 显示导航的位置。登录屏幕和主应用程序都应该有不同的布局。

我们有一个类似的案例并使用 LayoutComponent。这是简化的示例。

// This is main component that get's bootstrapped that has 'top-level' router.
@Component({selector: 'app', template: '<router-outlet></router-outlet>'})
class AppComponent {}

// main router config
// Here AuthModule has router with login and logout configured and LoginGuard
// redirect the user to /auth/login when she is not authenticated.
// We're using lazy-loading but you can use direct component instead
export const APP_ROUTES: Routes = [
  {path: '', redirectTo: 'main', pathMatch: 'full'},
  {path: 'auth', loadChildren: '../modules/+auth/auth.module#AuthModule'},
  {
    path: '',
    component: LayoutComponent,
    canActivate: [LoginGuard],
    children: [
      {path: 'main', loadChildren: '../modules/+main/main.module#MainModule'}
    ]
  }
];

// AuthModule/LoginComponent has own template and it will be rendered
// into 'top-level' router-outlet.

// LayoutComponent
// Here you define your main application layout that can include navigation
// and anything else that are global to the app. It has another router-outlet
// that get rendered when the layout is accessible (which in this case when the user is authenticated).
@Component({
  selector: 'app-layout',
  template: `
    <div id="wrapper">
      <app-sidebar></app-sidebar>
      <div id="page-wrapper" class="gray-bg dashboard-1" adjust-content-height>
        <router-outlet></router-outlet>
      </div>
    </div>
    <notifications></notifications>
    <error-modal></error-modal>
  `
})
export class LayoutComponent {}

// Auth/LoginComponent can have its own template that will have different layout from the main application

所以流程是这样的：

  • 当用户尝试加载 / 然后她重定向到 /main
  • 如果用户未通过身份验证，她将重定向到 /auth/login 否则她加载 /main

希望有帮助。

编辑：使用示例应用程序更新了 Sickelap/ng-starter存储库，该应用程序具有：

  • 延迟加载路由
  • 布局
  • 和其他东西
于 2017-05-03T07:28:30.793 回答
0

我想我收集了你想要实现的目标。我可以建议您使用变量来模拟某种状态变化，并将其分配给组件视图。让你的 app.component.html 只包含一个路由器插座。创建一个复制现有 component.html 的新 main.component.html

`<app-header></app-header>`

用*替换href(click)="handleChange(<linkValue>)'"

所以每个链接如下所示。

<ul class="nav navbar-nav"> <li class=""><a href="/main/overview">OVERVIEW</a></li>

handleChange方法：声明 currentLink -public currentLink string; // or public currentLink: string = '<a default value>'; public handleChange(link: string) { this.currentLink = link; }

创建一个view.component。

示例选择器<view [link]='currentLink'></view>

给视图组件一个 @Input() public link: string;

back to view.component.html

<div id="overview" *ngIf="link = 'overview'">overview content</div> <div id="main" *ngIf="link = 'main'">overview content</div>

然后，您可以将它们重构为单独的子组件。

概述：您正在使 app-header 成为处理“链接”变量的全局组件。我建议查看 ngRx 或一般的应用程序状态方法。因为这可能是管理 UI 的好方法。

于 2017-05-03T05:45:08.283 回答