如何处理角度5递归未知确切数字路由器参数?

问题描述

有没有办法处理递归未知的确切数量的路由器参数?

Is there a way to handle recursively unknown exact number of router parameters?

例如:

我们有产品类别，其中可以有子类别，子类别可以有自己的子类别等等.有几个主要条件:

We have products categories, which can have subcategories, subcategories can have it's own subcategories and so on. There are a few main conditions:

• 如果此类类别没有子类别，我们将重定向到 /categories/{id}/items，这将打开项目列表组件.
• 如果类别有子类别，它应该被重定向到下一个嵌套树级别 /categories/{id}/{id}/.../{id} 这应该打开最后一个 categoryId 子类别列表组件.
• 在到达最后一个没有子类别项目列表组件的类别后，将显示 /categories/{id}/{id}/.../{id}/items.

• if a such category has no subcategories we redirect to /categories/{id}/items that will open items list component.
• if category has subcategory it should be redirected to next nested tree level /categories/{id}/{id}/.../{id} which should open the last categoryId subcategories list component.
• after getting to the last category which doesn't has subcategories items list component will be shown /categories/{id}/{id}/.../{id}/items.

检查和重定向的解决方案是使用路由器解析器.但是如何在路由模块中跟踪这些 url 呢?

The solutions to check and redirect is to have router resolver. But how track those urls in routing module ?

在我看来，路线应该是这样的:

From my perspective the routes should look something like this:

{
  path: '/categories/:id',
  component: SubcategoriesListComponent
},
{
  path: '/categories/:id/**/:id',
  component: SubcategoriesListComponent,
},
{
  path: '/categories/:id/**/:id/items',
  component: CategoryItemsListComponent
}

可以这样实现吗?

推荐答案

无组件路由的可能解决方案:在路线配置中

Possible solution with componentless routes: in routes config

{
    path: 'categories/:id', children: [
    {path: '**', component: SubcategoriesListComponent}
]
}

在组件文件中:

export class SubcategoriesListComponent {

  constructor(aroute: ActivatedRoute) {
    aroute.url.subscribe((data) => {
      console.log('params ', data); //contains all the segments so put logic here of determining what to do according to nesting depth
    });

  }

}

这里是 输出示例(我在我的项目 ErrorComponent 上测试过，所以不要混淆)

Here is the output example (i tested on my project ErrorComponent so don't be confused)

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