ios - 在swift 3中从消息扩展打开应用程序
<p><p>我正在尝试通过单击按钮从消息扩展中打开我的母亲应用程序。我在我的扩展中使用了这个代码:</p>
<pre><code> @IBAction func open(_ sender: UIButton) {
let url = URL(string: "swiftexamples://")
self.extensionContext?.open(url!, completionHandler: {(succes) in })
}
</code></pre>
<p>当我的母亲应用程序正在运行并在后台运行时，一切正常，但是当我想打开已关闭的应用程序时，它崩溃了。没有崩溃日志，我有类似的情况 <a href="https://stackoverflow.com/questions/25814202/ios-app-crashes-when-first-opened-by-url-scheme" rel="noreferrer noopener nofollow">iOS app crashes when first opened by URL Scheme</a> .
我很确定我必须向应用程序委托(delegate)添加一些内容。
我找到了函数应用程序(_:open:options:)。问题是，我现在不知道如何实现“选项”部分。我写了这个:</p>
<pre><code> func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: ?) -> Bool {
// Override point for customization after application launch.
let url = launchOptions? as? NSURL
let sourceApp = launchOptions? as? String
let annotation = launchOptions? as? AnyObject
self.application(application, open: url, options: )
return true
}
</code></pre>
<p>我还听说过通用链接，它们具有与 url 方案相似的能力。是否可以使用通用链接来实现我想要的？ </p></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>您可以尝试此解决方案。它对我有用:</p>
<pre class="lang-swift prettyprint-override"><code>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)
}
</code></pre>
<p>正如 <a href="https://stackoverflow.com/a/44694703/2064473" rel="noreferrer noopener nofollow">https://stackoverflow.com/a/44694703/2064473</a> 中的建议</p></p>
<p style="font-size: 20px;">关于ios - 在swift 3中从消息扩展打开应用程序，我们在Stack Overflow上找到一个类似的问题：
<a href="https://stackoverflow.com/questions/42737059/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/42737059/
</a>
</p>
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