我正在尝试通过单击按钮从消息扩展名打开我的母亲应用程序。我在扩展中使用了以下代码:
@IBAction func open(_ sender: UIButton) {
let url = URL(string: "swiftexamples://")
self.extensionContext?.open(url!, completionHandler: {(succes) in })
}
当我的母亲应用程序运行并进入后台时,一切正常,但是当我要打开关闭的应用程序时,它崩溃。没有崩溃日志,我对此iOS app crashes when first opened by URL Scheme有类似的情况。
我很确定我必须向应用程序委托中添加一些内容。
我找到了函数application(_:open:options :)。问题是,我现在不怎么实现“选项”部分。我写了这个:
func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
// Override point for customization after application launch.
let url = launchOptions?[UIApplicationLaunchOptionsKey.url] as? NSURL
let sourceApp = launchOptions?[UIApplicationLaunchOptionsKey.sourceApplication] as? String
let annotation = launchOptions?[UIApplicationLaunchOptionsKey.annotation] as? AnyObject
self.application(application, open: url, options:[sourceApp: String, annotation:AnyObject] )
return true
}
我也听说过通用链接,它们具有与url方案相似的功能。是否可以使用通用链接来实现我想要的功能?
最佳答案
您可以尝试此解决方案。它为我工作:
func openUrl(url: URL?) {
let selector = sel_registerName("openURL:")
var responder = self as UIResponder?
while let r = responder, !r.responds(to: selector) {
responder = r.next
}
_ = responder?.perform(selector, with: url)
}
func canOpenUrl(url: URL?) -> Bool {
let selector = sel_registerName("canOpenURL:")
var responder = self as UIResponder?
while let r = responder, !r.responds(to: selector) {
responder = r.next
}
return (responder!.perform(selector, with: url) != nil)
}
如https://stackoverflow.com/a/44694703/2064473中所建议
关于ios - 通过Swift 3中的邮件扩展程序打开应用,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/42737059/