Angular 角度2获取父路径
是否仍然可以使用提供的激活路由获取父路由集合 如果我有嵌套的管线结构,如:Angular 角度2获取父路径,angular,angular2-routing,Angular,Angular2 Routing,是否仍然可以使用提供的激活路由获取父路由集合 如果我有嵌套的管线结构,如: [ { path: 'search', component: SearchComponent, children: [ { path: 'view/:id', component: ViewSearchComponent, children: [ { pa
[
{
path: 'search', component: SearchComponent, children:
[
{
path: 'view/:id', component: ViewSearchComponent, children:
[
{ path: 'person/:id', component: PersonComponent }
]
},
]
}
]
[
{ part: 'search' }
{ part: 'view/3' }
{ part: 'person/5' }
]
我的url如下所示:/search/view/3/person/5
我如何将其转化为某种结构,如:
[
{
path: 'search', component: SearchComponent, children:
[
{
path: 'view/:id', component: ViewSearchComponent, children:
[
{ path: 'person/:id', component: PersonComponent }
]
},
]
}
]
[
{ part: 'search' }
{ part: 'view/3' }
{ part: 'person/5' }
]
我已经查看了URLTree和urlsecgments,但是看起来它无法区分参数和路径 您可以注入
ActivatedRoute
,然后迭代parent
属性,直到没有
您可以运行以下代码,它将返回您想要的相同对象
import {ActivatedRoute} from '@angular/router';
export class Component {
constructor(private route: ActivatedRoute){
let pathroots = this.route.pathFromRoot;
let arr = [];
pathroots.forEach(path => {
let obj: any = {};
let pathurl = '';
path.url.subscribe(url => {
url.forEach(e => {
pathurl += e + '/';
});
});
obj['part'] = pathurl;
arr.push(obj);
});
console.log(arr,'*******************');
}
}
多亏了冈特,我自己想出了办法。这应该足以在组件上实现某种面包屑特性。它与Vikash几乎相同,但有点“rxjs'y”
当我看到你的答案时,我已经写了一些东西,但是它们看起来几乎一样。谢谢