Tags: array, case, creating, directory, exception, file, files, gifs, java, listing, malformed, points, programming, protocol, url

Need help with "Malformed URL: No Protocol" exception

On Java Studio » Java Programming

1,377 words with 2 Comments; publish: Tue, 05 Feb 2008 21:20:00 GMT; (15078.13, « »)

I'm creating a File that points to a directory. Using File.list() I'm getting an array listing all the files (in this case, .gifs) in the directory. I'm trying to create a URL pointing to each one of these .gifs, but I get a "Malformed URL Exception: No protocol" for each one. My guess is that the Java URL class doesn't know what to do with the gif. Does anyone know any way around this?

try

{

String iconRootDir = "./Icons";

File iconRoot = new File(iconRootDir);

String[] fileList = iconRoot.list();

QLChoiceButton tempButton = null;

for (int i = 0; i < fileList.length; i++)

{

tempButton = new QLChoiceButton(this, new URL(iconRootDir+"/"+fileList[i])); //exception thrown here

buttonScrollPane.add(tempButton);

} //end for

} //end try

"There's nothing more dangerous than a resourceful idiot." --Dilbert

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  • 2 Comments
    • Bit of a guess, but try having a protocl of "file://" on the front.

      You'll also need a "/" on the front of the file path. so you'll get:

      file:///C:/blah/blah.doc in windows.

      Bayard

      bayard.java-program.developerfaqs.com.generationjava.com

      Brainbench MVP for Java

      http://www.brainbench.com

      #1; Sat, 10 Nov 2007 02:45:00 GMT
    • That did it (I feel dumb! :) ). Thanks!

      "There's nothing more dangerous than a resourceful idiot." --Dilbert

      #2; Sat, 10 Nov 2007 02:46:00 GMT