3

我正在使用以下代码打开 URL SFSafariViewController

import UIKit
import SafariServices

class ViewController: UIViewController, UITextViewDelegate {

    override func viewDidLoad() {
        super.viewDidLoad()
        openBrowser(url: "https://www.example.com")
    }
    fileprivate func openBrowser(url: String) {
        let svc = SFSafariViewController(url: NSURL(string: url)! as URL, entersReaderIfAvailable: false)
        svc.delegate = self
        present(svc, animated: true, completion: nil)
    }
    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
    }
}

extension ViewController: SFSafariViewControllerDelegate {
    func safariViewController(_ controller: SFSafariViewController, didCompleteInitialLoad didLoadSuccessfully: Bool) {
    }
    func safariViewController(_ controller: SFSafariViewController, activityItemsFor URL: URL, title: String?) -> [UIActivity] {
        return []
    }
    func safariViewControllerDidFinish(_ controller: SFSafariViewController) {
    }
}

这工作得很好。

但是,如果该网站包含一个链接target="_blank"并且我单击它,则不会发生任何事情。

我知道它SFSafariViewController不支持多个窗口或选项卡。

但我认为应该有可能忽略target="_blank"并打开链接,就像没有target="_blank".

有什么办法可以做到这一点?

4

0 回答 0