问题
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 dont understand why such exceptions arise..
回答1:
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 that protocol, Java refuses to create a URL to save you from failure when you try to call openConnection.
Solution
You should probably be using the URI class if you don't plan on opening a connection of those protocols in Java.
回答2:
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
回答3:
You're getting that error because java doesn't have a standard protocol handler for telnet.
回答4:
The simple answer is that it only does recognize certain protocols, and the remainder of the infinity of protocols is not recognized.
来源：https://stackoverflow.com/questions/2406518/why-does-javas-url-class-not-recognize-certain-protocols