6

是否可以使用 NGXS 存储用户路由解析器？

我有一个这样的测试，但我不知道这是否是正确的方法：

import {ActivatedRouteSnapshot, Resolve} from "@angular/router";
import {Todo} from "./todos.models";
import {Observable} from "rxjs/Observable";
import {Select, Store} from "@ngxs/store";
import {GetTodo, TodosState} from "./todos.state";
import {Injectable} from "@angular/core";

@Injectable()
export class TodoResolver implements Resolve<Todo> {

    constructor(
        private store:Store
    ) {}

    @Select(TodosState.getTodo)
    private todo$:Observable<Todo>;

    resolve(route:ActivatedRouteSnapshot): Observable<Todo>
    {
        const id = <number><any> route.paramMap.get('id');
        this.store.dispatch(new GetTodo(id));
        return this.todo$;
    }
}

当我尝试这个时，应用程序只是挂起。没有显示错误。

欢迎大家帮忙。谢谢

4

2 回答 2

5

请忽略，我自己找到了解决方案...

将代码更改为：

@Injectable()
export class TodoResolver implements Resolve<Todo> {

    constructor(
        private store:Store
    ) {}

    resolve(route:ActivatedRouteSnapshot): Observable<Todo>
    {
        const id = <number><any> route.paramMap.get('id');
        this.store.dispatch(new GetTodo(id));

        return this.store.selectOnce(TodosState.getTodo);
    }
}
于 2018-11-09T11:32:01.123 回答
2

我认为 rauwebieten 的解决方案仅适用于同步操作。异步操作将在 selectOnce 执行后修改存储。

我必须设置两个状态来处理异步操作。一个是 Todos 状态，另一个是 RequestingTodo 状态。

代码如下所示：

@Injectable()
export class AsyncTodoResolver implements Resolve<Todo> {
    @Select(RequestingTodoState)
    private todo$:Observable<Todo>;

    constructor(
        private store:Store
    ) {}

    resolve(route:ActivatedRouteSnapshot): Observable<Todo> {
        const id = <number><any> route.paramMap.get('id');
        return this.store.dispatch(new GiveOrFetchTodo(id))
           .pipe(
              withLatestFrom(todo$),
              map([any, todo]) => {return todo;}
            );
    }
}
于 2019-02-08T03:58:22.883 回答