Issue Java throws a MalformedURLException because it couldn’t find a URLStreamHandler for that protocol. Check the javadocs of the constructors for the details. Summary Since the URL class has an openConnection method, the URL class checks to make sure that Java knows how to open a connection of the correct protocol. Without a URLStreamHandler for … Read more
You need to encode your parameter’s values before concatenating them to URL. Backslash \ is special character which have to be escaped as %5C Escaping example: String paramValue = “param\\with\\backslash”; String yourURLStr = “http://host.com?param=” + java.net.URLEncoder.encode(paramValue, “UTF-8”); java.net.URL url = new java.net.URL(yourURLStr); The result is http://host.com?param=param%5Cwith%5Cbackslash which is properly formatted url string.
Use the validators package: >>> import validators >>> validators.url(“http://google.com”) True >>> validators.url(“http://google”) ValidationFailure(func=url, args={‘value’: ‘http://google’, ‘require_tld’: True}) >>> if not validators.url(“http://google”): … print “not valid” … not valid >>> Install it from PyPI with pip (pip install validators).