typescript 从 Observable<T> 过滤所有“空”值
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Filter all 'null' values from an Observable<T>
提问by fasth
I have a service with a subject:
我有一个主题的服务：
@Injectable() export class UserService() {
private currentUserSubject = new BehaviorSubject<User>(null);
public currentUser = this.currentUserSubject.asObservable().distinctUntilChanged();
... // emitting new User
}
Have a component I inject this service into and subscribing on updates:
有一个组件，我将此服务注入并订阅更新：
@Component() export class UserComponent {
constructor(private userService: UserService) {
this.userService.currentUser
.subscribe((user) => {
// I want to see not null value here
})
}
}
I want to apply something to Observable<User>to filter all null values and get into subscribe only when Useris actually loaded.
我想应用一些东西来Observable<User>过滤所有空值并仅在User实际加载时才进入订阅。
回答by snorkpete
Add a filter operator to your observable chain. You can filter nulls explicitly or just check that your user is truthy - you will have to make that call depending on your needs.
将过滤器运算符添加到您的可观察链。您可以显式过滤空值或仅检查您的用户是否真实 - 您必须根据您的需要进行该调用。
Filtering out null users only:
仅过滤掉空用户：
public currentUser = this.currentUserSubject
.asObservable()
.filter(user => user !== null)
.distinctUntilChanged();
回答by ackuser
Anoter way to check the value exists
检查值是否存在的另一种方法
public currentUser = this.currentUserSubject
.asObservable()
.filter(Boolean)
.distinctUntilChanged();
回答by Amit Portnoy
with rxjs@6and typescript the recommended (readable/maintainable) way is to define the following type guard:
使用rxjs@6和打字稿推荐的（可读/可维护）方式是定义以下类型保护：
export function isNonNull<T>(value: T): value is NonNullable<T> {
return value != null;
}
and augment a subject with pipeand filter:
并用pipe和扩充主题filter：
subject.pipe(filter(isNonNull))
回答by Wilt
It's as simple as:
这很简单：
filter(user => !!user),
If user evaluates to false, the filter will not be passed.
如果用户评估为 false，则不会通过过滤器。
So would filter these falsy values (link to Falsy on MDN).
所以会过滤这些假值（链接到 MDN 上的 Falsy）。
