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)

}

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