当用户点击应用程序中的任何链接时，如何阻止ios应用程序打开url？

class CustomApplication: UIApplication {
    override func openURL(_ url: URL, options: [UIApplication.OpenExternalURLOptionsKey : Any] = [:], completionHandler: ((Bool) -> Void)? = nil) {
        if url.host == "yourappdomain.com" {
            // 执行应用程序内部的操作
            // ...
        } else {
            // 阻止打开URL
            completionHandler?(false)
        }
    }
}

@UIApplicationMain
class AppDelegate: UIResponder, UIApplicationDelegate {

    var window: UIWindow?

    func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey: Any]?) -> Bool {
        
        // 创建自定义的UIApplication实例
        let customApplication = CustomApplication.shared

        // 替换默认的UIApplication实例
        object_setClass(UIApplication.shared, type(of: customApplication))
        
        return true
    }
    
    // ...
}