ngrx商店4选择不工作

问题描述:

我一整天都在努力让我的Angular应用程序与ngrx/store 4一起工作。在经历了很多不成功的时间之后,我决定创建一个新的绝对最小应用程序来检查是否有其他问题。但最小的应用程序根本不工作:(ngrx商店4选择不工作

在Chrome Redux devtools中,我可以看到,商店已经正确更新,但UI没有更新,而且app.component.ts中的日志语句this.age$.subscribe(console.log))永远不会执行。我真的不知道我错过了什么。

下面是相关的代码。

的package.json

{ 
    "dependencies": { 
    "@angular/animations": "^4.2.4", 
    "@angular/common": "^4.2.4", 
    "@angular/compiler": "^4.2.4", 
    "@angular/core": "^4.2.4", 
    "@angular/forms": "^4.2.4", 
    "@angular/http": "^4.2.4", 
    "@angular/platform-browser": "^4.2.4", 
    "@angular/platform-browser-dynamic": "^4.2.4", 
    "@angular/router": "^4.2.4", 
    "@ngrx/store": "^4.0.2", 
    "@ngrx/store-devtools": "^4.0.0", 
    "rxjs": "^5.4.2", 
    "zone.js": "^0.8.14" 
    }, 
    "devDependencies": { 
    "@types/node": "^8.0.22", 
    "typescript": "^2.4.2" 
    } 
} 

个app.module.ts

import { AppState } from './state'; 
import { BrowserModule } from '@angular/platform-browser'; 
import { NgModule } from '@angular/core'; 
import { StoreModule } from '@ngrx/store'; 
import { StoreDevtoolsModule } from '@ngrx/store-devtools' 
import { reducer } from './reducer'; 

import { AppComponent } from './app.component'; 

@NgModule({ 
    declarations: [ 
    AppComponent 
    ], 
    imports: [ 
    BrowserModule, 
    StoreModule.forRoot({ reducer }), 
    StoreDevtoolsModule.instrument(), 
    ], 
    providers: [], 
    bootstrap: [AppComponent] 
}) 
export class AppModule { } 

state.ts

export interface AppState { 
    age: number 
} 

export const INITIAL_STATE: AppState = { 
    age: 1 
} 

actions.ts

import { Action } from '@ngrx/store'; 

export const ACTION_TYPE = 'CHANGE_AGE_ACTION'; 

export class ChangeAgeAction implements Action { 
    readonly type: string = ACTION_TYPE; 

    constructor(public payload?: number) { 
    } 
} 

reducer.ts

import { ChangeAgeAction } from './actions'; 
import { Action } from '@ngrx/store'; 
import { AppState, INITIAL_STATE } from './state'; 
import { ACTION_TYPE } from './actions'; 

function changeName(state: AppState, action: ChangeAgeAction): AppState { 
    return { 
    age: action.payload 
    } 
} 

export function reducer(state: AppState = INITIAL_STATE, action: Action) { 
    switch (action.type) { 
    case ACTION_TYPE: 
     return changeName(state, action); 

    default: 
     return state; 
    } 
} 

app.component.ts

import { ChangeAgeAction } from './actions'; 
import { AppState } from './state'; 
import { Component } from '@angular/core'; 
import { Observable } from 'rxjs/Observable'; 
import { Store } from '@ngrx/store'; 
import { map } from 'rxjs/operator/map'; 

@Component({ 
    selector: 'app-root', 
    templateUrl: './app.component.html', 
    styleUrls: ['./app.component.css'] 
}) 
export class AppComponent { 
    age$: Observable<number>; 

    constructor(private store: Store<AppState>) { 
    this.age$ = this.store.select<number>(state => state.age); 
    this.age$.subscribe(console.log); 
    } 

    changeAge() { 
    this.store.dispatch(new ChangeAgeAction(Math.random())); 
    } 
} 

app.component.html

<h3>Age: {{ age$ | async }}</h3> 
<button (click)="changeAge()">Change Age</button> 

您的应用程序状态不正确建模。实际上有一个包含你年龄的减速器对象。下面是我所做的更改,以使应用程序的工作:

在state.ts

export interface AppState { 
    reducer: { 
    age: number 
    } 
} 

在应用组件:

export class AppComponent { 
    age$: Observable<number>; 

    constructor(private store: Store<AppState>) { 
    this.age$ = this.store.select<number>(state => state.reducer.age); 
    this.age$.subscribe(console.log); 
    } 

    changeAge() { 
    this.store.dispatch(new ChangeAgeAction(Math.random())); 
    } 
} 

注意,您必须在“减速”对象的原因,你的AppState是因为当你设置这个时:StoreModule.forRoot({reducer})。它从减速器获得“减速器”的名称。

在,如果你不喜欢的东西StoreModule.forRoot({} ageReducer),那么你的应用程序的状态应该是这样的。换句话说:

export interface AppState { 
    ageReducer: { 
    age: number 
    } 
} 

下面的代码的工作版本:

https://stackblitz.com/edit/angular-jxszwh

+0

哇,这是诀窍!非常感谢你!我想我会浪费几天时间来发现问题。在我看来,这在ngrx/store文档中没有足够的表达能力。我从来不会得出结论,减速器名称必须与州名相同。 –

+0

我同意这是令人困惑的。我不止一次地绊倒了自己。 – seescode

另请注意,如果您使用角度异步管道,则无需在您的可观察值上调用.subscribe。异步管道处理订阅/取消订阅。