我有一个带有过滤器的主要组件,并加载项目列表。 在其中一项上单击时,将加载我的详细信息组件。现在,我想在详细信息组件中创建一个按钮以切换到下一个或上一个项目。
有人可以指出我正确的方向吗?
我在main.component.html中的列表:
<table class="table table-striped">
<thead>
<tr>
<td><strong>Nr</strong></td>
<td><strong>Name</strong></td>
</tr>
</thead>
<tbody style="overflow:scroll;">
<tr routerLink="./{{item.id}}" <!-- this goes to my detail-component -->
ngFor="let item of listItems; let i = index;"
(click)="setClickedRow(i)"
[class.active]="i == selectedRow">
<td >{{item.number}}</td>
<td>{{item.description}}</td>
</tr>
</tbody>
</table>
<div class="col-sm-9 col-pd-ow grid-wo">
<div class="row">
<div class="col-sm-12 col-pd-ow">
<!-- List view component -->
<router-outlet></router-outlet>
</div>
</div>
</div>
我的路线:
const routes: Routes = [
{
path: '',
component: ItemComponent,
children: [
{ path: '', redirectTo: 'list' },
{ path: 'list', component: ListViewComponent},
{ path: ':id', component: DetailsViewComponent }
]
}
];
在我的详细信息部分:
export class DetailsViewComponent implements OnInit, OnDestroy {
subscription: Subscription;
public id: number;
public item = new Item();
constructor(private route: ActivatedRoute, private itemService: ItemService) {}
ngOnInit() {
this.route.params.subscribe(
params => {
this.id = params['id'];
this.itemService.get(this.id).subscribe(item => {
this.item = item;
});
});
}