在带有异步管道的模板中的多个位置使用相同的可观察值的性能

问题描述

在组件模板中，我在2个地方为相同的Observable调用async管道.

In my component template I am calling async pipe for same Observable in 2 places.

我应该订阅它并在模板中使用返回的数组，还是对模板的多个位置中的同一Observable使用async管道，对性能没有负面影响?

Shall I subscribe to it and use returned array in my template or using async pipe for same Observable in multiple places of template has no negative effect to performence?

推荐答案

每次使用observable$ | async都会为给定的observable$创建新的订阅(并因此生成单独的流)-如果此可观察项包含计算繁重的部分或重新调用，这些计算和重新调用是针对每个async单独执行的-所以-此可以会对性能产生影响.

Every use of observable$ | async will create a new subscription(and therefor an individual stream) to the given observable$ - if this observable contains parts with heavy calculations or rest-calls, those calculations and rest-calls are executed individually for each async - so yes - this can have performance implications.

但是，通过使用，以便在所有订阅者之间共享流，并为所有订阅者一次执行所有这些操作.不要忘记将share-运算符与import "rxjs/add/operator/share";

However this is easily fixed by extending your observable$ with .share(), to have a shared stream among all subscribers and execute all those things just once for all subscribers.Don't forget to add the share-operator with import "rxjs/add/operator/share";

默认情况下，异步管道不共享订阅的原因仅仅是灵活性和易用性:简单的.share()的编写速度比创建全新的流要快得多，如果要创建新的流，则必须这样做默认为共享.

The reason why async-pipes don't share subscriptions by default is simply flexibility and ease of use: A simple .share() is much faster to write than creating a completely new stream, which would be required if they were to be shared by default.

这是一个简单的例子

@Component({
    selector: "some-comp",
    template: `
        Sub1: {{squareData$ | async}}<br>
        Sub2: {{squareData$ | async}}<br>
        Sub3: {{squareData$ | async}}
    `
})
export class SomeComponent {
    squareData$: Observable<string> = Observable.range(0, 10)
        .map(x => x * x)
        .do(x => console.log(`CalculationResult: ${x}`)
        .toArray()
        .map(squares => squares.join(", "))
        .share();  // remove this line and the console will log every result 3 times instead of 1
}

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