swift scheme 跳转外部应用

url地址为 weixin://dl/business 需要通过浏览器打开跳转到微信

一开始使用safari打开链接 报url异常

func showSafariVC(for url: String){
	guard let url = URL(string: url) else {
	    return
	}
	let safariVC = SFSafariViewController(url: url)
	self.present(safariVC, animated: true, completion: nil)
}

“The Seecified URL has an unsupported scheme.Only HTTP and HTTPS URLs are supported.”

解决方案:
应使用 UIApplication.shared.open方法打开链接

    guard let url = URL(string:s ?? "") else {
                                return
                            }
                        if #available(iOS 10,*) {
                              UIApplication.shared.open(url, options: [:], completionHandler: nil)
                          } else {
                              UIApplication.shared.openURL(url)
                           }