Mixing 合并

Mixing指的是将多个publishers合并

combineLatest

将2个或多个publishers合并

示例1

let firstPublisher = PassthroughSubject<Int, Never>()
let secondPublisher = PassthroughSubject<String, Never>()

firstPublisher
    .combineLatest(secondPublisher)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

firstPublisher.send(1)
secondPublisher.send("a")
firstPublisher.send(2)

输出：

(1, "a")
(2, "a")

注意，combineLatest会读取最新的数据或者数据序列，如果更改以上代码如下：

示例2

let firstPublisher = PassthroughSubject<Int, Never>()
let secondPublisher = PassthroughSubject<String, Never>()

firstPublisher
    .combineLatest(secondPublisher)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

firstPublisher.send(1)
firstPublisher.send(2)
secondPublisher.send("a")

输出：

(2, "a")

问题，为甚不是输出

(1, "a")
(2, "a")

重点在于理解Latest:
firstPublisher会等待secondPublisher最新的数据进行组合成元祖，上例中firstPublisher第一个消息（1）并没有等到secondPublisher的数据就被其他数据（2）覆盖，则不会产生配对输出

示例3

let firstPublisher = PassthroughSubject<Int, Never>()
let secondPublisher = PassthroughSubject<String, Never>()

firstPublisher
    .combineLatest(secondPublisher)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

firstPublisher.send(1)
secondPublisher.send("a")
firstPublisher.send(2)
secondPublisher.send("b")
secondPublisher.send("c")

输出：

(1, "a")
(2, "a")
(2, "b")
(2, "c")

类似的函数还有CombineLatest3,CombineLatest4，用法如下：

let firstPublisher = PassthroughSubject<Int, Never>()
let secondPublisher = PassthroughSubject<String, Never>()
let threePublisher = PassthroughSubject<String, Never>()

Publishers
    .CombineLatest3(firstPublisher, secondPublisher, threePublisher)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

firstPublisher.send(1)
secondPublisher.send("a")
threePublisher.send("b")
firstPublisher.send(2)

输出：

(1, "a", "b")
(2, "a", "b")

merge

merge只要接受到新数据就输出，适用于合并多个无序的publishers

let pub1 = PassthroughSubject<Int, Never>()
let pub2 = PassthroughSubject<Int, Never>()
let pub3 = PassthroughSubject<Int, Never>()
let pub4 = PassthroughSubject<Int, Never>()
let pub5 = PassthroughSubject<Int, Never>()

pub1
    .merge(with: pub2, pub3, pub4, pub5)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

pub1.send(1)
pub2.send(2)
pub3.send(3)
pub4.send(4)
pub5.send(5)
pub1.send(1)
pub3.send(3)
pub5.send(5)

同样，可使用Publishers.MergeMany合并

zip

zip同combineLatest在结果上一样输出元祖，不同在于zip并不使用Latest最新的数据

例combineLatest

let firstPublisher = PassthroughSubject<Int, Never>()
let secondPublisher = PassthroughSubject<String, Never>()

firstPublisher
    .combineLatest(secondPublisher)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

firstPublisher.send(1)
firstPublisher.send(2)
secondPublisher.send("a")

输出

(2, "a")

例 zip

let firstPublisher = PassthroughSubject<Int, Never>()
let secondPublisher = PassthroughSubject<String, Never>()

firstPublisher
    .zip(secondPublisher)
    .sink(receiveCompletion: { _ in
        print("Finished")
    }, receiveValue: { someValue in
        print(someValue)
    })

firstPublisher.send(1)
firstPublisher.send(2)
secondPublisher.send("a")

输出

(1, "a")

zip数据序列中，最先进入的值优先组成元祖，而combineLatest则是最新的值