[英]Reading NSNotification userInfo in Swift
(I'm new to Swift and iOS development in general). (一般来说,我是Swift和iOS开发的新手)。
I'm porting some Objective-C code over to Swift and I don't know how to translate this: 我正在将一些Objective-C代码移植到Swift,但我不知道该如何翻译:
-(void) fooDidBar:(NSNotification *)notification
{
Foo* foo = [[notification userInfo] objectForKey:BarKey];
// do stuff with foo
}
So far I have this: 到目前为止,我有这个:
private func fooDidBar(notification: NSNotification) {
var foo: Foo = notification.userInfo.objectForKey(BarKey)
}
But I get a compiler error: 但是我得到一个编译器错误:
/Users/me/src-me/project/FooBarViewController.swift:53:57: Value of type '[NSObject : AnyObject]?'
/Users/me/src-me/project/FooBarViewController.swift:53:57:类型'[NSObject:AnyObject]的值?' has no member 'objectForKey'
没有成员'objectForKey'
As userInfo
is declared as an NSDictionary
in UIApplicationShortcutItem.h
: 由于
userInfo
在UIApplicationShortcutItem.h
被声明为NSDictionary
:
@property (nullable, nonatomic, copy) NSDictionary<NSString *, id <NSSecureCoding>> *userInfo;
...I thought I'd try this: ...我想尝试一下:
= notification.userInfo[BarKey]
but then I get this error: 但是然后我得到这个错误:
/Users/me/src-me/project/FooBarViewController.swift:53:65: Type '[NSObject : AnyObject]?'
/Users/me/src-me/project/FooBarViewController.swift:53:65:输入'[NSObject:AnyObject]?' has no subscript members
没有下标成员
Your idea to use subscripts was correct, however as you can see by the presence of ?
您使用下标的想法是正确的,但是,通过出现
?
可以看到?
, the userInfo
dictionary is Optional , meaning that it can be nil (in Objective-C, no distinction is made). ,
userInfo
字典是Optional ,意味着它可以为nil(在Objective-C中,没有区别)。 Furthermore, Swift's Dictionary is not as lenient with types as NSDictionary, so you'll need to use as?
此外,Swift的Dictionary在类型上不如NSDictionary宽大,因此您需要使用
as?
to ensure the value is a Foo
instance as you expect. 确保该值是您期望的
Foo
实例。
You can't directly subscript the optional dictionary, but if you use a ?
您不能直接下标可选词典,但是如果使用
?
for optional chaining , then you can access the value if the dictionary is non-nil. 对于可选链接 ,如果字典为非nil,则可以访问该值。 I would also recommend
if let
to access the final value if it's non-nil. 我还建议
if let
访问非零的最终值。 (Or you might choose to use guard let
with a fatalError
or return
in case the Foo is not present.) (或者您可以选择使用带有
fatalError
guard let
fatalError
或者在不存在Foo的情况下return
。)
if let foo = notification.userInfo?[BarKey] as? Foo {
// foo is non-nil (type `Foo`) in here
}
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