Angular 2：从父组件获取RouteParams-Java 学习之路

如何从父组件获取RouteParams？

App.ts ：

@Component({
  ...
})

@RouteConfig([
  {path: '/', component: HomeComponent, as: 'Home'},
  {path: '/:username/...', component: ParentComponent, as: 'Parent'}
])

export class HomeComponent {
  ...
}

然后，在 ParentComponent 中，我可以轻松获取用户名参数并设置子路由 .

Parent.ts ：

@Component({
  ...
})

@RouteConfig([
  { path: '/child-1', component: ChildOneComponent, as: 'ChildOne' },
  { path: '/child-2', component: ChildTwoComponent, as: 'ChildTwo' }
])

export class ParentComponent {

  public username: string;

  constructor(
    public params: RouteParams
  ) {
    this.username = params.get('username');
  }

  ...
}

但是，如何在这些子组件中获得相同的“用户名”参数？做与上面相同的技巧，不会这样做 . 因为这些参数是在ProfileComponent上定义的还是什么？

@Component({
  ...
})

export class ChildOneComponent {

  public username: string;

  constructor(
    public params: RouteParams
  ) {
    this.username = params.get('username');
    // returns null
  }

  ...
}

12 回答

使用RxJS的 Observable.combineLatest ，我们可以得到一些接近惯用的params处理：

import 'rxjs/add/operator/combineLatest';

import {Component} from '@angular/core';
import {ActivatedRoute, Params} from '@angular/router';
import {Observable} from 'rxjs/Observable';

@Component({ /* ... */ })
export class SomeChildComponent {
  email: string;
  id: string;

  constructor(private route: ActivatedRoute) {}

  ngOnInit() {
    Observable.combineLatest(this.route.params, this.route.parent.params)
        .forEach((params: Params[]) => {
          this.id = params[0]['id'];
          this.email = params[1]['email'];
        });
  }
}

回复于 2024-04-29T10:10:14+08:00

UPDATE:

现在Angular2 final正式发布，正确的方法如下：

export class ChildComponent {

    private sub: any;

    private parentRouteId: number;

    constructor(private route: ActivatedRoute) { }

    ngOnInit() {
        this.sub = this.route.parent.params.subscribe(params => {
            this.parentRouteId = +params["id"];
        });
    }

    ngOnDestroy() {
        this.sub.unsubscribe();
    }
}

ORIGINAL:

以下是我使用“@ angular / router”：“3.0.0-alpha.6”包的方法：

export class ChildComponent {

    private sub: any;

    private parentRouteId: number;

    constructor(
        private router: Router,
        private route: ActivatedRoute) {
    }

    ngOnInit() {
        this.sub = this.router.routerState.parent(this.route).params.subscribe(params => {
            this.parentRouteId = +params["id"];
        });
    }

    ngOnDestroy() {
        this.sub.unsubscribe();
    }
}

在此示例中，路由具有以下格式：/ parent /：id / child /：childid

export const routes: RouterConfig = [
    {
        path: '/parent/:id',
        component: ParentComponent,
        children: [
            { path: '/child/:childid', component: ChildComponent }]
    }
];

回复于 2024-04-29T10:10:14+08:00

您不应该尝试在 ChildOneComponent 中使用 RouteParams .

请改用RouteRegistry！

@Component({
  ...
})

export class ChildOneComponent {

  public username: string;

  constructor(registry: RouteRegistry, location: Location) {
    route_registry.recognize(location.path(), []).then((instruction) => {
      console.log(instruction.component.params['username']);
    })
  }


  ...
}

UPDATE: 从此拉取请求（角度β9）：https://github.com/angular/angular/pull/7163

您现在可以在不使用 recognize(location.path(), []) 的情况下访问当前指令 .

例：

@Component({
  ...
})

export class ChildOneComponent {

  public username: string;

  constructor(_router: Router) {
    let instruction = _router.currentInstruction();
    this.username = instruction.component.params['username'];
  }

  ...
}

I haven't tried it, yet

UPDATE 2: 从角度2.0.0.beta15开始的小变化：

现在 currentInstruction 不再是一个功能了 . 此外，您必须加载 root 路由器 . （感谢@ Lxrd-AJ报道）

@Component({
  ...
})

export class ChildOneComponent {

  public username: string;

  constructor(_router: Router) {
    let instruction = _router.root.currentInstruction;
    this.username = instruction.component.params['username'];
  }

  ...
}

回复于 2024-04-29T10:10:14+08:00

2
如GünterZöchbauer所述，我使用了https://github.com/angular/angular/issues/6204#issuecomment-173273143的评论来解决我的问题 . 我使用 angular2/core 中的 Injector 类来获取父级的routeparams . 原来角度2不能处理深层嵌套的路线 . 也许他们将来会加上这个 .
```
constructor(private _issueService: IssueService,
            private _injector: Injector) {}

getIssues() {
    let id = this._injector.parent.parent.get(RouteParams).get('id');
    this._issueService.getIssues(id).then(issues => this.issues = issues);
}
```
回复于 2024-04-29T10:10:14+08:00

我找到了一个丑陋但有效的解决方案，通过请求父母（正是第二个祖先）注射器，并从这里获取 RouteParams .

就像是

@Component({
  ...
})
export class ChildOneComponent {
  public username: string;

  constructor(injector: Injector) {
    let params = injector.parent.parent.get(RouteParams);

    this.username = params.get('username');
  }
}

回复于 2024-04-29T10:10:14+08:00

RC5 + @angular/router": "3.0.0-rc.1 SOLUTION: 似乎 this.router.routerState.queryParams 已被弃用 . 您可以通过以下方式获取父路线参数：

constructor(private activatedRoute: ActivatedRoute) {
}    

this.activatedRoute.parent.params.subscribe(
  (param: any) => {
    let userId = param['userId'];
    console.log(userId);
  });

回复于 2024-04-29T10:10:14+08:00

您可以从注入器中获取子组件内部父路径的组件，然后从子组件中获取任何组件 . 在你的情况下这样

@Component({
  ...
})

export class ChildOneComponent {

  public username: string;

  constructor(
    public params: RouteParams
    private _injector: Injector

  ) {
    var parentComponent = this._injector.get(ParentComponent)

    this.username = parentComponent.username;
    //or
    this.username = parentComponent.params.get('username');
  }

  ...
}

回复于 2024-04-29T10:10:14+08:00

如果要为代码编写单元测试，将Injector实例传递给子组件中的构造函数可能不太好 .

解决此问题的最简单方法是在父组件中创建一个服务类，您可以在其中保存所需的参数 .

@Component({
    template: `<div><router-outlet></router-outlet></div>`,
    directives: [RouterOutlet],
    providers: [SomeServiceClass]
})
@RouteConfig([
    {path: "/", name: "IssueList", component: IssueListComponent, useAsDefault: true}
])
class IssueMountComponent {
    constructor(routeParams: RouteParams, someService: SomeServiceClass) {
        someService.id = routeParams.get('id');
    }
}

然后，您只需向子组件注入相同的服务并访问参数 .

@Component({
    template: `some template here`
})
class IssueListComponent implements OnInit {
    issues: Issue[];
    constructor(private someService: SomeServiceClass) {}

    getIssues() {
        let id = this.someService.id;
        // do your magic here
    }

    ngOnInit() {
        this.getIssues();
    }
}

请注意，您应该使用父组件装饰器中的“providers”将此类服务的范围限定为父组件及其子组件 .

我推荐这篇关于Angular 2中DI和范围的文章：http://blog.thoughtram.io/angular/2015/08/20/host-and-visibility-in-angular-2-dependency-injection.html

回复于 2024-04-29T10:10:14+08:00

1

In RC6, router 3.0.0-rc.2 （也可能在RC5中工作），你可以从URL中获取路径参数作为快照，以防params不会改变，没有这个衬管的可观察量：

this.route.snapshot.parent.params['username'];

不要忘记按如下方式注入ActivatedRoute：

constructor(private route: ActivatedRoute) {};

回复于 2024-04-29T10:10:14+08:00

我最终为Angular 2 rc.1写了这种黑客

import { Router } from '@angular/router-deprecated';
import * as _ from 'lodash';

interface ParameterObject {
  [key: string]: any[];
};

/**
 * Traverse route.parent links until root router and check each level
 * currentInstruction and group parameters to single object.
 *
 * e.g.
 * {
 *   id: [314, 593],
 *   otherParam: [9]
 * }
 */
export default function mergeRouteParams(router: Router): ParameterObject {
  let mergedParameters: ParameterObject = {};
  while (router) {
    let currentInstruction = router.currentInstruction;
    if (currentInstruction) {
      let currentParams = currentInstruction.component.params;
      _.each(currentParams, (value, key) => {
        let valuesForKey = mergedParameters[key] || [];
        valuesForKey.unshift(value);
        mergedParameters[key] = valuesForKey;
      });
    }
    router = router.parent;
  }
  return mergedParameters;
}

现在在视图中我收集视图中的参数而不是读取 RouteParams 我只是通过路由器获取它们：

@Component({
  ...
})

export class ChildishComponent {

  constructor(router: Router) {
    let allParams = mergeRouteParams(router);
    let parentRouteId = allParams['id'][0];
    let childRouteId = allParams['id'][1];
    let otherRandomParam = allParams.otherRandomParam[0];
  }

  ...
}

回复于 2024-04-29T10:10:14+08:00

6
在 FINAL 中，在 RXJS 的帮助下，您可以组合两个 Map （来自孩子和父母）：
```
(route) => Observable
    .zip(route.params, route.parent.params)
    .map(data => Object.assign({}, data[0], data[1]))
```
可能有的其他问题：
- 使用上面是否真的是一个好主意 - 因为耦合（将子组件与父组合's param' s - 不在api级别 - 隐藏耦合），
- 对于RXJS而言它是否是正确的方法（它需要硬核RXJS用户反馈;）
回复于 2024-04-29T10:10:14+08:00

您可以使用以下内容在快照上执行此操作，但如果更改，则不会更新 id 属性 .

此示例还说明了如何通过合并所有参数observable来订阅所有祖先参数更改并查找您感兴趣的参数更改 . 但是，请注意此方法，因为可能有多个祖先具有相同的参数键/名称 .

import { Component } from '@angular/core';
import { ActivatedRoute, Params, ActivatedRouteSnapshot } from '@angular/router';
import { Observable } from 'rxjs/Observable';
import { Subscription } from 'rxjs/Subscription';
import 'rxjs/add/observable/merge';

// This traverses the route, following ancestors, looking for the parameter.
function getParam(route: ActivatedRouteSnapshot, key: string): any {
  if (route != null) {
    let param = route.params[key];
    if (param === undefined) {
      return getParam(route.parent, key);
    } else {
      return param;
    }
  } else {
    return undefined;
  }
}

@Component({ /* ... */ })
export class SomeChildComponent {

  id: string;

  private _parameterSubscription: Subscription;

  constructor(private route: ActivatedRoute) {
  }

  ngOnInit() {
    // There is no need to do this if you subscribe to parameter changes like below.
    this.id = getParam(this.route.snapshot, 'id');

    let paramObservables: Observable<Params>[] =
      this.route.pathFromRoot.map(route => route.params);

    this._parametersSubscription =
      Observable.merge(...paramObservables).subscribe((params: Params) => {
        if ('id' in params) {
          // If there are ancestor routes that have used
          // the same parameter name, they will conflict!
          this.id = params['id'];
        }
      });
  }

  ngOnDestroy() {
    this._parameterSubscription.unsubscribe();
  }
}

回复于 2024-04-29T10:10:14+08:00

Angular 2：从父组件获取RouteParams

12 回答

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