Closed ghost closed 5 years ago
I suppose this is more of an issue with openurl
I second that, the generated URL is invalid and the host in this case is in the path url component. The only way to fix it and get a valid url is to convert it to string, append a schema (http/https) and then generate a new URL object. I think ActiveLabel should do that, prior to passing the url to the block.
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when using this library and parsing strings that contain URLs, if they are constructed without a host the convenience URL is not to RFC 1808 spec and fails to have a valid host.
Example: label.text = "www.something.com" ... label.handleURLTap { url in UIApplication.shared.openURL(url) }
The URL is invalid and not openable because of the missing host.