35

使用 Apple 的新组合框架，我想从列表中的每个元素发出多个请求。然后我想从减少所有响应中得到一个单一的结果。基本上，我想从发布者列表转到拥有响应列表的单个发布者。

我曾尝试制作出版商列表，但我不知道如何将该列表简化为单个出版商。而且我尝试制作一个包含列表的发布者，但我无法平面映射发布者列表。

请看“createIngredients”函数

func createIngredient(ingredient: Ingredient) -> AnyPublisher<CreateIngredientMutation.Data, Error> {
    return apollo.performPub(mutation: CreateIngredientMutation(name: ingredient.name, optionalProduct: ingredient.productId, quantity: ingredient.quantity, unit: ingredient.unit))
            .eraseToAnyPublisher()
}

func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<[CreateIngredientMutation.Data], Error> {
    // first attempt
    let results = ingredients
            .map(createIngredient)
    // results = [AnyPublisher<CreateIngredientMutation.Data, Error>]

    // second attempt
    return Publishers.Just(ingredients)
            .eraseToAnyPublisher()
            .flatMap { (list: [Ingredient]) -> Publisher<[CreateIngredientMutation.Data], Error> in
                return list.map(createIngredient) // [AnyPublisher<CreateIngredientMutation.Data, Error>]
            }
}

我不确定如何获取一组发布者并将其转换为包含数组的发布者。

“[AnyPublisher]”类型的结果值不符合闭包结果类型“Publisher”

4

4 回答 4

46

本质上，在您的特定情况下，您正在查看以下内容：

func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<[CreateIngredientMutation.Data], Error> {
    Publishers.MergeMany(ingredients.map(createIngredient(ingredient:)))
        .collect()
        .eraseToAnyPublisher()
}

这会“收集”上游发布者生成的所有元素，并在它们全部完成后生成一个包含所有结果的数组并最终自行完成。

请记住，如果上游发布者之一失败 - 或产生多个结果 - 元素的数量可能与订阅者的数量不匹配，因此您可能需要额外的运营商来缓解这种情况，具体取决于您的情况。

更通用的答案，您可以使用EntwineTest 框架对其进行测试：

import XCTest
import Combine
import EntwineTest

final class MyTests: XCTestCase {
    
    func testCreateArrayFromArrayOfPublishers() {

        typealias SimplePublisher = Just<Int>

        // we'll create our 'list of publishers' here. Each publisher emits a single
        // Int and then completes successfully – using the `Just` publisher.
        let publishers: [SimplePublisher] = [
            SimplePublisher(1),
            SimplePublisher(2),
            SimplePublisher(3),
        ]

        // we'll turn our array of publishers into a single merged publisher
        let publisherOfPublishers = Publishers.MergeMany(publishers)

        // Then we `collect` all the individual publisher elements results into
        // a single array
        let finalPublisher = publisherOfPublishers.collect()

        // Let's test what we expect to happen, will happen.
        // We'll create a scheduler to run our test on
        let testScheduler = TestScheduler()

        // Then we'll start a test. Our test will subscribe to our publisher
        // at a virtual time of 200, and cancel the subscription at 900
        let testableSubscriber = testScheduler.start { finalPublisher }

        // we're expecting that, immediately upon subscription, our results will
        // arrive. This is because we're using `just` type publishers which
        // dispatch their contents as soon as they're subscribed to
        XCTAssertEqual(testableSubscriber.recordedOutput, [
            (200, .subscription),            // we're expecting to subscribe at 200
            (200, .input([1, 2, 3])),        // then receive an array of results immediately
            (200, .completion(.finished)),   // the `collect` operator finishes immediately after completion
        ])
    }
}
于 2019-06-27T07:51:49.070 回答
7

我认为这Publishers.MergeMany可能会有所帮助。在您的示例中，您可以像这样使用它：

func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<CreateIngredientMutation.Data, Error> {
    let publishers = ingredients.map(createIngredient(ingredient:))
    return Publishers.MergeMany(publishers).eraseToAnyPublisher()
}

这将为您提供一个发布者，向您发送Output.

但是，如果您特别希望Output在所有发布者完成后一次全部放入数组中，则可以使用collect()with MergeMany：

func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<[CreateIngredientMutation.Data], Error> {
    let publishers = ingredients.map(createIngredient(ingredient:))
    return Publishers.MergeMany(publishers).collect().eraseToAnyPublisher()
}

如果您愿意，上述任何一个示例都可以简化为一行，即：

func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<CreateIngredientMutation.Data, Error> {
    Publishers.MergeMany(ingredients.map(createIngredient(ingredient:))).eraseToAnyPublisher()
}

您还可以定义自己的自定义merge()扩展方法Sequence并使用它来稍微简化代码：

extension Sequence where Element: Publisher {
    func merge() -> Publishers.MergeMany<Element> {
        Publishers.MergeMany(self)
    }
}

func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<CreateIngredientMutation.Data, Error> {
    ingredients.map(createIngredient).merge().eraseToAnyPublisher()
}
于 2020-06-10T21:45:27.547 回答
2

为了补充 Tricky 的答案，这是一个保留数组中元素顺序的解决方案。它通过整个链传递每个元素的索引，并按索引对收集的数组进行排序。

由于排序，复杂度应该是 O(n log n)。

import Combine

extension Publishers {

    private struct EnumeratedElement<T> {
        let index: Int
        let element: T

        init(index: Int, element: T) {
            self.index = index
            self.element = element
        }

        init(_ enumeratedSequence: EnumeratedSequence<[T]>.Iterator.Element) {
            index = enumeratedSequence.offset
            element = enumeratedSequence.element
        }
    }

    static func mergeMappedRetainingOrder<InputType, OutputType>(
        _ inputArray: [InputType],
        mapTransform: (InputType) -> AnyPublisher<OutputType, Error>
    ) -> AnyPublisher<[OutputType], Error> {

        let enumeratedInputArray = inputArray.enumerated().map(EnumeratedElement.init)

        let enumeratedMapTransform: (EnumeratedElement<InputType>) -> AnyPublisher<EnumeratedElement<OutputType>, Error> = { enumeratedInput in
            mapTransform(enumeratedInput.element)
                .map { EnumeratedElement(index: enumeratedInput.index, element: $0)}
                .eraseToAnyPublisher()
        }

        let sortEnumeratedOutputArrayByIndex: ([EnumeratedElement<OutputType>]) -> [EnumeratedElement<OutputType>] = { enumeratedOutputArray in
            enumeratedOutputArray.sorted { $0.index < $1.index }
        }

        let transformToNonEnumeratedArray: ([EnumeratedElement<OutputType>]) -> [OutputType] = {
            $0.map { $0.element }
        }

        return Publishers.MergeMany(enumeratedInputArray.map(enumeratedMapTransform))
            .collect()
            .map(sortEnumeratedOutputArrayByIndex)
            .map(transformToNonEnumeratedArray)
            .eraseToAnyPublisher()
    }
}

解决方案的单元测试：

import XCTest
import Combine

final class PublishersExtensionsTests: XCTestCase {

    // MARK: - Private properties

    private var cancellables = Set<AnyCancellable>()

    // MARK: - Tests

    func test_mergeMappedRetainingOrder() {
        let expectation = expectation(description: "mergeMappedRetainingOrder publisher")

        let numbers = (1...100).map { _ in Int.random(in: 1...3) }

        let mapTransform: (Int) -> AnyPublisher<Int, Error> = {
            let delayTimeInterval = RunLoop.SchedulerTimeType.Stride(Double($0))
            return Just($0)
                .delay(for: delayTimeInterval, scheduler: RunLoop.main)
                .setFailureType(to: Error.self)
                .eraseToAnyPublisher()
        }

        let resultNumbersPublisher = Publishers.mergeMappedRetainingOrder(numbers, mapTransform: mapTransform)

        resultNumbersPublisher.sink(receiveCompletion: { _ in }, receiveValue: { resultNumbers in
            XCTAssertTrue(numbers == resultNumbers)
            expectation.fulfill()
         }).store(in: &cancellables)

        waitForExpectations(timeout: 5)
    }
}
于 2021-06-02T09:32:58.160 回答
1

您可以在一行中完成：

.flatMap(Publishers.Sequence.init(sequence:))
于 2021-12-17T11:29:34.307 回答