Swift Open link in Safari

        let url = URL(string: "https://www.baidu.com")
        //根据iOS系统版本，分别处理
        if #available(iOS 10, *) {
            UIApplication.shared.open(url!, options: [:],
                                      completionHandler: {(success) in
                                        print(success)
            })
        } else {
            UIApplication.shared.openURL(url!)
        }

Reproduced in: https: //www.jianshu.com/p/20fcc749c7d2