要合并它们，它们的Element需要具有相同的类型.

To merge them, they need to have the same type for their Element.

因此，一种选择是丢弃它们的类型信息并转换为 AnyObject.现在它们可以合并:

So, one option is to throw away their type information and cast to AnyObject. Now they can be merged:

let stringSubject = PublishSubject<String>()
let stringObservable = stringSubject.asObservable().map { $0 as AnyObject }

let intSubject = PublishSubject<Int>()
let intObservable = intSubject.asObservable().map { $0 as AnyObject }

Observable.of(stringObservable, intObservable).merge()
    .subscribeNext { print($0) }
    .addDisposableTo(disposeBag)

stringSubject.onNext("a")
stringSubject.onNext("b")
intSubject.onNext(1)
intSubject.onNext(2)
stringSubject.onNext("c")

输出:

一个
乙
1
2

a
b
1
2
c

另一种选择是将 then 包装在枚举中:

Another option would be to wrap then in an enum:

enum Container {
    case S(String)
    case I(Int)
}

let stringSubject = PublishSubject<String>()
let stringObservable = stringSubject.asObservable().map { Container.S($0) }

let intSubject = PublishSubject<Int>()
let intObservable = intSubject.asObservable().map { Container.I($0) }

Observable.of(stringObservable, intObservable).merge()
    .subscribeNext { e in
        switch e {
        case .S(let str):
            print("next element is a STRING: \(str)")
        case .I(let int):
            print("next element is an INT: \(int)")
        }
    }
    .addDisposableTo(disposeBag)

stringSubject.onNext("a")
stringSubject.onNext("b")
intSubject.onNext(1)
intSubject.onNext(2)
stringSubject.onNext("c")

输出:

下一个元素是一个字符串:a
下一个元素是一个字符串:b
下一个元素是 INT:1
下一个元素是一个 INT:2
下一个元素是一个字符串:c

next element is a STRING: a
next element is a STRING: b
next element is an INT: 1
next element is an INT: 2
next element is a STRING: c

至于可以组合不同类型的 Observable 的其他运算符(如 zip 和 combineLatest)，没有一个像 那样工作>合并代码>.但是，请检查这些.它们可能更适合您的要求.

As for the other operators that can combine Observables of varying types (like zip and combineLatest), none work quite like merge. However, check those out. They might be better suited to your requirements.