在Swift中读取UIApplicationLaunchOptionsURLKey

Question

只想阅读Swift中的启动选项.

这是我的旧obj-C代码,工作正常:

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{   
    NSURL *URL = [launchOptions valueForKey:UIApplicationLaunchOptionsURLKey];  
    if (URL)

这就是我认为Swift代码应该是这样的:

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool
{
    if let options = launchOptions
    {
        let URL: NSURL = options(UIApplicationLaunchOptionsURLKey) as String

但它给出了错误:

'(NSString!) - > $ T2'与'[NSObject:AnyObject]'不同

出现错误？但难以正确地投射它并且找不到如何做的链接.

Answer 1

斯威夫特3:

在Swift 3中,launchOptions是一个类型的字典[UIApplicationLaunchOptionsKey: Any]?,所以你可以像这样访问值:

launchOptions?[UIApplicationLaunchOptionsKey.url]

由于键类型是UIApplicationLaunchOptionsKey,您可以enum简单地将类型缩写为.url:

launchOptions?[.url]

与该键关联的值是a URL,而不是a String.此外,键可能不存在于字典中,因此您应该使用条件转换as?而不是正常转换.

在Swift中,您想要:

func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
    if let url = launchOptions?[.url] as? URL {
        // If we get here, we know launchOptions is not nil, we know
        // key .url was in the launchOptions dictionary, and we know
        // that the type of the launchOptions was correctly identified
        // as URL.  At this point, url has the type URL and is ready to use.
    }

斯威夫特2:

在你的代码中,launchOptions是一个类型的字典[NSObject: AnyObject]?,所以你想要访问这样的值:

options?[UIApplicationLaunchOptionsURLKey]

与该键关联的值是a NSURL,而不是a String.此外,键可能不存在于字典中,因此您应该使用条件转换as?而不是正常转换.

在Swift中,您想要:

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool
{
    if let url = launchOptions?[UIApplicationLaunchOptionsURLKey] as? NSURL {
        // If we get here, we know launchOptions is not nil, we know
        // UIApplicationLaunchOptionsURLKey was in the launchOptions
        // dictionary, and we know that the type of the launchOptions
        // was correctly identified as NSURL.  At this point, url has
        // the type NSURL and is ready to use.
    }