在PromiseKit 6中返回void [英] Returning void in PromiseKit 6

问题描述

这就是我使用PromiseKit 4.5

This is what I had working with PromiseKit 4.5

api.getUserFirstName().then { name -> Void in
  print(name)
}

getUserFirstName（）返回 Promsise< String> 。我更新到PromiseKit 6，现在抛出一个错误：
无法转换类型'（_） - >的值Void'到预期的参数类型'（_） - > _'

getUserFirstName() returns a Promsise<String>. I updated to PromiseKit 6 and this now throws an error: Cannot convert value of type '(_) -> Void' to expected argument type '(_) -> _'

此错误消息对我来说没什么意义。我该如何解决这个问题？

This error message makes little sense to me. How do I fix this?

编辑：所以这似乎解决了这个问题，但我对此发生的事情几乎一无所知：

So this seems to fix it, but I have little understanding as to what's happening with this:

api.getUserFirstName().compactMap { name in
  print(name)
}

then（）和 compactMap（）<之间有什么区别？ / code>？

What's the difference now between then() and compactMap()?

推荐答案

根据 PromiseKit 6.0指南 然后被拆分为然后，已完成和地图

• 然后被提供先前的承诺价值并要求您返回承诺。


• 已完成被提供前一个承诺值并返回一个Void承诺（这是链使用量的80％）


• map 被提供前一个承诺值，并要求您返回一个非承诺，即。一个值。

• then is fed the previous promise value and requires you return a promise.
• doneis fed the previous promise value and returns a Void promise (which is 80% of chain usage)
• map is fed the previous promise value and requires you return a non-promise, ie. a value.

为什么会这样？正如开发人员所说：

Why that was happend? As said developers:

使用PromiseKit我们的然后做了多件事，我们依赖在Swift上从上下文推断出正确的然后。但是多行然后 s它将无法执行此操作，而不是告诉您情况不明确，它会发明一些其他错误。通常可怕的无法将T转换为AnyPromise 。我们有一个故障排除指南来解决这个问题，但我相信只有工作的工具，当你花了4年时间等待Swift解决问题而Swift没有解决问题时，你会怎么做？我们选择在更高级别找到解决方案。

With PromiseKit our then did multiple things, and we relied on Swift to infer the correct then from context. However with multiple line thens it would fail to do this, and instead of telling you that the situation was ambiguous it would invent some other error. Often the dreaded cannot convert T to AnyPromise. We have a troubleshooting guide to combat this but I believe in tools that just work, and when you spend 4 years waiting for Swift to fix the issue and Swift doesn’t fix the issue, what do you do? We chose to find a solution at the higher level.

所以可能在你的情况下需要使用 done

So probably in your case needs to use done

func stackOverflowExample() {
    self.getUserFirstName().done { name -> Void in
        print(name)
    }
}

func getUserFirstName() -> Promise<String> {
    return .value("My User")
}

compactMap 让你在返回nil时得到错误传输。

compactMap lets you get error transmission when nil is returned.

firstly {
    URLSession.shared.dataTask(.promise, with: url)
}.compactMap {
    try JSONDecoder().decode(Foo.self, with: $0.data)
}.done {
    //…
}.catch {
    // though probably you should return without the `catch`
}

在发布指南

compactMap 已重命名为 flatMap 在此处查看讨论

这篇关于在PromiseKit 6中返回void的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！