Angular 2 Routing在新选项卡中运行

Question

我正在使用带有angular2的asp.net核心应用程序,我的路由工作正常.

<a target="target" routerLink="/page1" routerLinkActive="active">Associates</a> 
<a routerLink="/page2" routerLinkActive="active">Account managers</a> 

我想在新标签页中打开每个页面链接(routerLink).是否可能在新选项卡中打开每个链接,而无需刷新页面？

我试图替换routerLink="/Page2",target="target" href="/associates"但页面刷新所有引用.

Answer 1

试试这个, <a target="_blank" routerLink="/Page2">

Update1:救援的自定义指令!完整代码在这里:https://github.com/anandchakru/angular2-olnw

import { Directive, ElementRef, HostListener, Input, Inject } from '@angular/core';

@Directive({ selector: '[olinw007]' })
export class OpenLinkInNewWindowDirective {
    //@Input('olinwLink') link: string; //intro a new attribute, if independent from routerLink
    @Input('routerLink') link: string;
    constructor(private el: ElementRef, @Inject(Window) private win:Window) {
    }
    @HostListener('mousedown') onMouseEnter() {
        this.win.open(this.link || 'main/default');
    }
}

注意,提供了Window,并在下面声明了OpenLinkInNewWindowDirective:

import { AppAboutComponent } from './app.about.component';
import { AppDefaultComponent } from './app.default.component';
import { PageNotFoundComponent } from './app.pnf.component';
import { OpenLinkInNewWindowDirective } from './olinw.directive';
import { BrowserModule } from '@angular/platform-browser';
import { NgModule } from '@angular/core';
import { FormsModule } from '@angular/forms';
import { HttpModule } from '@angular/http';
import { RouterModule, Routes } from '@angular/router';

import { AppComponent } from './app.component';


const appRoutes: Routes = [
  { path: '', pathMatch: 'full', component: AppDefaultComponent },
  { path: 'home', component: AppComponent },
  { path: 'about', component: AppAboutComponent },
  { path: '**', component: PageNotFoundComponent }
];

@NgModule({
  declarations: [
    AppComponent, AppAboutComponent, AppDefaultComponent, PageNotFoundComponent, OpenLinkInNewWindowDirective
  ],
  imports: [
    BrowserModule,
    FormsModule,
    HttpModule,
    RouterModule.forRoot(appRoutes)
  ],
  providers: [{ provide: Window, useValue: window }],
  bootstrap: [AppComponent]
})
export class AppModule { }

第一个链接在新窗口中打开,第二个链接不会:

<h1>
    {{title}}
    <ul>
        <li><a routerLink="/main/home" routerLinkActive="active" olinw007> OLNW</a></li>
        <li><a routerLink="/main/home" routerLinkActive="active"> OLNW - NOT</a></li>
    </ul>
    <div style="background-color:#eee;">
        <router-outlet></router-outlet>
    </div>
</h1>

田田!..欢迎你=)

Update2:从v2.4.10 <a target="_blank" routerLink="/Page2">起作用

Answer 2

迟到了，但我刚刚发现了一种替代方法：

在你的模板上，

<a (click)="navigateAssociates()">Associates</a> 

而在你的component.ts上，你可以使用serializeUrl将路由转换成一个字符串，它可以与window.open()

navigateAssociates() {
  const url = this.router.serializeUrl(
    this.router.createUrlTree(['/page1'])
  );

  window.open(url, '_blank');
}

Answer 3

这似乎有点困惑.

在另一个窗口或选项卡中打开您的应用程序将需要重新引导整个应用程序,然后为您的路由器...拿起该URL,将其转换为路由,并加载适当的组件.

如果您只使用链接,这正是会发生的情况.事实上,这就是发生的一切.

路由器的要点是将组件交换进出路由器插座,这是已经被引导的并且存在于正在运行的应用程序的范围内,并且不在多个窗口之间共享.

Answer 4

该指令用作[routerLink]替代品.您所要做的就是用[link]替换[routerLink]用法.它适用于ctrl +单击,cmd +单击,中键单击.

import {Directive, HostListener, Input} from '@angular/core'
import {Router} from '@angular/router'
import _ from 'lodash'
import qs from 'qs'

@Directive({
  selector: '[link]'
})
export class LinkDirective {
  @Input() link: string

  @HostListener('click', ['$event'])
  onClick($event) {
    // ctrl+click, cmd+click
    if ($event.ctrlKey || $event.metaKey) {
      $event.preventDefault()
      $event.stopPropagation()
      window.open(this.getUrl(this.link), '_blank')
    } else {
      this.router.navigate(this.getLink(this.link))
    }
  }

  @HostListener('mouseup', ['$event'])
  onMouseUp($event) {
    // middleclick
    if ($event.which == 2) {

      $event.preventDefault()
      $event.stopPropagation()
      window.open(this.getUrl(this.link), '_blank')
    }
  }

  constructor(private router: Router) {}

  private getLink(link): any[] {
    if ( ! _.isArray(link)) {
      link = [link]
    }

    return link
  }

  private getUrl(link): string {
    let url = ''

    if (_.isArray(link)) {
      url = link[0]

      if (link[1]) {
        url += '?' + qs.stringify(link[1])
      }
    } else {
      url = link
    }

    return url
  }
}