WWDC 2019“Combine in Practice”代码示例中的错误 CombineLatest？

Question

我尝试关注 WWDC 2019 的 Combine in Practice 谈话（从 26:00 分钟开始）的代码 sn-p 或查看 slides 179 ff。但它不会编译并且查看 API 某些部分对我来说没有意义（例如，使用 closure 类型的第三个参数调用 CombineLatest.init(A, B)。我尝试调整示例以便它们编译。

问题第 1/2 部分：如果我对 WWDC 2019 代码 sn-ps 有误解，谁能帮助我并告诉我？

第一个代码 sn-p（幻灯片 179）

@Published var password: String = ""
@Published var passwordAgain: String = ""

var validatedPassword: CombineLatest<Published<String>, Published<String>, String?> {
    return CombineLatest($password, $passwordAgain) { password, passwordAgain in
        guard password == passwordAgain, password.count > 8 else { return nil }
        return password
    }
}

我只能让这个 sn-p 至少返回给我Publisher fromCombineLatest`

• 将Publishers 枚举添加到CombineLatest 的命名空间
• 移除尾随闭合
• 将.Publisher 添加到Published<String>

@Published var password: String = ""
@Published var passwordAgain: String = ""

var validatedPassword: Publishers.CombineLatest<Published<String>.Publisher, Published<String>.Publisher> {
    return Publishers.CombineLatest($password, $passwordAgain)
}

第二个代码 sn-p（幻灯片 185）

@Published var password: String = ""
@Published var passwordAgain: String = ""

var validatedPassword: Map<CombineLatest<Published<String>, Published<String>, String?>> {
    return CombineLatest($password, $passwordAgain) { password, passwordAgain in
        guard password == passwordAgain, password.count > 8 else { return nil }
        return password
    }
    .map { $0 == "password1" ? nil : $0 }
}

我可以让这个 sn-p 在以下情况下编译：

• 执行为第一个 sn-p 列出的所有步骤
• 在Map前面添加Publishers.
• 将<> 移动到正确的位置
• 显式返回Publishers.Map，并使用正确的参数upstream:

@Published var password: String = ""
@Published var passwordAgain: String = ""

var validatedPassword: Publishers.Map<Publishers.CombineLatest<Published<String>.Publisher, Published<String>.Publisher>, String?> {
        return Publishers.Map(upstream: Publishers.CombineLatest($password, $passwordAgain)) { password, passwordAgain in
            guard password == passwordAgain, password.count > 8 else { return nil }
            return password
    }

或者当包括幻灯片中的.map {} 时：

• 通过将 var 类型包装在另一个 Publishers.Map<..., String?>

@Published var password: String = ""
@Published var passwordAgain: String = ""
var validatedPassword: Publishers.Map<Publishers.Map<Publishers.CombineLatest<Published<String>.Publisher, Published<String>.Publisher>, String?>, String?> {
        return Publishers.Map(upstream: Publishers.CombineLatest($password, $passwordAgain)) { password, passwordAgain in
            guard password == passwordAgain, password.count > 8 else { return nil }
            return password
    }
    .map { $0 == "password1" ? nil : $0 }
}

问题第 2/2 部分：快速的方法是什么？例如，通过使用这样的东西（不能编译）：

@Published var password: String = ""
@Published var passwordAgain: String = ""

var validatedPassword: AnyPublisher<String?, Never> {
    return Just($password)
        .combineLatest($passwordAgain) { password, passwordAgain in
            guard password == passwordAgain, password.count > 8 else { return nil }
            return password
    }
    .map{ $0 == "password1" ? nil : $0 }
    .eraseToAnyPublisher()
}

Answer 1

在第一季度：

WWDC 演示使用当时存在的 API。时间过去了 - API 已更改。就是这样。

第二季度：

尝试如下...

var validatedPassword: AnyPublisher<String?, Never> {
    return Publishers.CombineLatest($password, $passwordAgain)
        .map { password, passwordRepeat in
            guard password == passwordRepeat, password.count > 8 else { return nil }
            return password
        }
        .map { ($0 ?? "") == "password1" ? nil : $0 }
        .eraseToAnyPublisher()
}