URL u=new URL("telnet://route-server.exodus.net");
This line is generating :
java.net.MalformedURLException: unknown protocol: telnet
And I encounter similar problems with other URLs that begin with "news://"
These are URLs extracted from ODP, so I don't understand why such exceptions arise..
A protocol handler is a Java class that implements the communications protocol for accessing the URL resource. For example, given an http URL, Java prepares to use the HTTP protocol handler to retrieve documents from the specified server.
Constructors. The java. net. URL class represents a URL and has a complete set of methods to manipulate URL in Java.
The URL class provides several methods that let you query URL objects. You can get the protocol, authority, host name, port number, path, query, filename, and reference from a URL using these accessor methods: getProtocol.
Java throws a MalformedURLException
because it couldn't find a URLStreamHandler
for that protocol. Check the javadocs of the constructors for the details.
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 that protocol, Java refuses to create a URL
to save you from failure when you try to call openConnection
.
You should probably be using the URI
class if you don't plan on opening a connection of those protocols in Java.
Sounds like there's no registered handler for the protocol "telnet" in your application. Since the URL class can be used to open a InputStream to URL it needs to have a registered handler for the protocol to do this work if you're to be allowed to create an object using it.
For details on how to add handlers see: http://docs.oracle.com/javase/7/docs/api/java/net/URLStreamHandlerFactory.html
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