Swift Combine:如何从发布者列表中创建单个发布者? [英] Swift Combine: How to create a single publisher from a list of publishers?
问题描述
使用Apple的新的Combine框架,我希望从列表中的每个元素发出多个请求.然后,我希望减少所有响应得到一个结果.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.
Using Apple's new Combine framework I want to make multiple requests from each element in a list. Then I want a single result from a reduction of all the the responses. Basically I want to go from list of publishers to a single publisher that holds a list of responses.
我曾尝试制作一个发布者列表,但我不知道如何将列表缩小为一个发布者.而且我尝试过使发布者包含一个列表,但是我无法平面映射发布者列表.
I've tried making a list of publishers, but I don't know how to reduce that list into a single publisher. And I've tried making a publisher containing a list but I can't flat map a list of publishers.
请查看"createIngredients"功能
Please look at the "createIngredients" function
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>]
}
}
我不确定如何获取发布者数组并将其转换为包含数组的发布者.
I'm not sure how to take an array of publishers and convert that to a publisher containing an array.
类型"[AnyPublisher]"的结果值与关闭结果类型"Publisher"不符
Result value of type '[AnyPublisher]' does not conform to closure result type 'Publisher'
推荐答案
本质上,在您的特定情况下,您正在寻找的是这样的东西:
Essentially, in your specific situation you're looking at something like this:
func createIngredients(ingredients: [Ingredient]) -> AnyPublisher<[CreateIngredientMutation.Data], Error> {
Publishers.MergeMany(ingredients.map(createIngredient(ingredient:)))
.collect()
.eraseToAnyPublisher()
}
这将收集"上游发布者产生的所有元素,并在它们全部完成之后生成包含所有结果的数组,并最终完成自身.
This 'collects' all the elements produced by the upstream publishers and – once they have all completed – produces an array with all the results and finally completes itself.
请记住,如果上游发布者之一失败-或产生多个结果-元素数量可能与订阅者数量不匹配,因此您可能需要其他运营商来缓解这种情况.
Bear in mind, if one of the upstream publishers fails – or produces more than one result – the number of elements may not match the number of subscribers, so you may need additional operators to mitigate this depending on your situation.
更通用的答案,您可以使用 EntwineTest框架对其进行测试:>
The more generic answer, with a way you can test it using the EntwineTest framework:
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
])
}
}
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