在SFSafariViewController中处理弹出窗口/选项卡

import SafariServices

class ViewController: UIViewController, SFSafariViewControllerDelegate {
    func openSafariViewController() {
        let url = URL(string: "https://example.com")!
        let safariViewController = SFSafariViewController(url: url)
        safariViewController.delegate = self
        present(safariViewController, animated: true, completion: nil)
    }
    
    func safariViewController(_ controller: SFSafariViewController, activityItemsFor URL: URL, title: String?, completionHandler: @escaping ([UIActivity]?) -> Void) {
        if controller.dismissButtonStyle == .done {
            // 用户点击了一个链接，执行默认操作
            completionHandler([])
        } else {
            // 用户点击了一个弹出窗口/选项卡的链接，执行自定义操作
            let customActivity = CustomActivity()
            completionHandler([customActivity])
        }
    }
}

class CustomActivity: UIActivity {
    override var activityType: UIActivity.ActivityType? {
        return UIActivity.ActivityType("com.example.customactivity")
    }
    
    override var activityTitle: String? {
        return "Custom Activity"
    }
    
    override var activityImage: UIImage? {
        return UIImage(named: "custom_activity_icon")
    }
    
    override func canPerform(withActivityItems activityItems: [Any]) -> Bool {
        return true
    }
    
    override func prepare(withActivityItems activityItems: [Any]) {
        // 准备自定义操作所需的数据
    }
    
    override func perform() {
        // 执行自定义操作
        activityDidFinish(true)
    }
}