如何使用其命名空间（或框架名称）将Swift类型名称作为字符串

Is there any way to get a Swift type name as a string with its namespace (or framework name)?

有没有办法将Swift类型名称作为带有命名空间（或框架名称）的字符串？

For example, if Foo.framework has a class named Bar, I would like to get a string something like "Foo.Bar".

例如，如果Foo.framework有一个名为Bar的类，我想得到类似“Foo.Bar”的字符串。

The followings just return the class name "Bar".

以下只返回类名“Bar”。

let barName1 = String(Bar.self)       // returns "Bar"
let barName2 = "\(Bar.self)"          // returns "Bar"
let barName3 = "\(Bar().dynamicType)" // returns "Bar"

I would like to also get the framework name "Foo" as a namespace.

我还想将框架名称“Foo”作为命名空间。

2 个解决方案

#1

Use String(reflecting:):

使用String（反映:)：

struct Bar { }

let barName = String(reflecting: Bar.self) 
print(barName) // <Module>.Bar

From the Xcode 7 Release Notes:

来自Xcode 7发行说明：

Type names and enum cases now print and convert to String without qualification by default. debugPrint or String(reflecting:) can still be used to get fully qualified names.

默认情况下，类型名称和枚举案例现在打印并转换为字符串，无需限定。 debugPrint或String（反射:)仍可用于获取完全限定名称。

#2

String(reflecting:) works right now, but could change in the future. So if you only need the fully-qualified name temporarily, you should be fine. However, if the name is something you're going to rely on over a longer period of time (spanning multiple releases of Swift) you shouldn't use it.

字符串（反映:)现在可以使用，但将来可能会改变。因此，如果您只需要临时的完全限定名称，那么您应该没问题。但是，如果名称是您将在更长的时间内（跨越Swift的多个版本）依赖的名称，则不应使用它。

The Swift community is trying to come to a consensus on a longer-term method for getting the information from a type.

Swift社区正试图就从类型中获取信息的长期方法达成共识。

From Joe Groff:

来自Joe Groff：

In 4.2, String(reflecting:) still shows a fully qualified name for local types, but the local type context is anonymized and printed in a way that's intended to be clear is non-stable, since no matter what reflection support we add in the future, it'd be a bad idea to incidentally rely on the presence or identity of local types in dynamic code.

在4.2中，字符串（反射:)仍然显示本地类型的完全限定名称，但本地类型上下文是匿名的，并且打算以明确的方式打印是不稳定的，因为无论我们添加什么反射支持未来，顺便依赖动态代码中本地类型的存在或身份是一个坏主意。

https://forums.swift.org/t/getting-the-more-fully-qualified-name-of-a-type/14375/9

#1