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Thread: If a file extension is known, but not the name, how do you find it?

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    Question If a file extension is known, but not the name, how do you find it?

    I'm working on a project that I started, and an attempting my first shot at networking so to speak. Lets say my program looks for "updates" by reading in a few lines from a textfile on a webserver. I can do that fine. Now lets say it wants to look for updates to plugins. If the base path to the plugin's update is known and the extention of the information file, how would I search for that file? I know that on some web servers, when you go up one in a directory, it lists the contents of that directory. Here's the code that i'm using:
    File[] listOfFiles = new File(currentURL.getPath()).listFiles(); 
    			for (int i = 0; i < listOfFiles.length; i++){
    			   if (listOfFiles[i].isFile()){
    			       if (listOfFiles[i].getName().endsWith(".info") || listOfFiles[i].getName().endsWith(".INFO"))
    			       {
    			          hasFoundInfoFile = true;
    			          InfoFile = listOfFiles[i].getAbsoluteFile();
    			       }
    			       if (hasFoundInfoFile)
    			    	   break;
    			   }
    			}
    but i'm getting null pointer exceptions on this line:
    for (int i = 0; i < listOfFiles.length; i++){

    I checked the path, and that exists, i even made sure that a file of the extension .info exists in that directory. What am I doing wrong?


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    Super Moderator Sean4u's Avatar
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    Default Re: If a file extension is known, but not the name, how do you find it?

    Have you looked in the API docs for java.io.File? There's an explanation in there of why you might be getting a null return value from listFiles(). You're doing the right kind of thing with your code, given what you say you want to achieve. While you're in the API docs for java.io.File, have a look at the directory listing methods that take a FileFilter or FilenameFilter argument.

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    Default Re: If a file extension is known, but not the name, how do you find it?

    Ok, so java doesn't treat it as a directory. Is there a way around this without mirroring the whole directory on the local filesystem?

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    Super Moderator Norm's Avatar
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    Default Re: If a file extension is known, but not the name, how do you find it?

    my program looks for "updates" by reading in a few lines from a textfile on a webserver.
    Can you explain where the code you show is executing?
    What does a server have to do with looking at files in a folder on your computer?
    The listFiles() method has an overload that takes a Filter you can use to chose what files to include.

    File[] listOfFiles = new File(currentURL.getPath()).listFiles();
    I'd do this in two steps:
    File theDir = new File(currentURL.getPath());
    // Here test  theDir 
    File[] listOfFiles = theDir.listFiles();

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    Default Re: If a file extension is known, but not the name, how do you find it?

    Basically, My program pulls information from plugin developers "repositories" that they host on their web servers. In the base path that they give out to users is a file PLUGIN_NAME.repo. At runtime I do not know the exact name of PLUGIN_NAME, but only that it exists in the URL given and it has the extension of .repo. Does this help?

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    Super Moderator Norm's Avatar
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    Default Re: If a file extension is known, but not the name, how do you find it?

    Can you explain where the code you show is executing? On the server or on your PC or ???

    What does a server have to do with looking at files in a folder on your PC?

    If the files are on a server and your code is executing on your PC, is there a way for you to ask the server for a list of files it has locally on its computer? What protocol are you using?

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    Default Re: If a file extension is known, but not the name, how do you find it?

    The code executes client side. My program is opened by the end user on their computer. When given a web url entered by the user(ex: "http://example.com/Plugin/"), the program should try to search IANA &mdash; Example domains for *.repo. The first time it finds a match, it reads the contents in as a text file. The problem is that java doesn't appear to treat URLs as directories. I'm trying to avoid having to write server code, as it will be easier for the audience i'm targeting to just upload a folder with some files to a web server than to have them run a server applet 24/7

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    Super Moderator Norm's Avatar
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    Default Re: If a file extension is known, but not the name, how do you find it?

    The problem is that java doesn't appear to treat URLs as directories.
    Can you explain this a bit? What class are you using the URL with?
    It could depend on the protocol you are using in the URL.

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    Default Re: If a file extension is known, but not the name, how do you find it?

    Here's an example of the structure i'm using: "http://techwizmc.co.cc/Test/"
    In that url, there is a file called project.repo. Using standard directory listing methods, I get null pointer exceptions, I think because java doesn't treat this as a directory, as
    File path = new File("http://techwizmc.co.cc/Test/");
    if(!path.isDirectory()){
        System.out.println("error");
    }
    always executes.

  10. #10
    Super Moderator Norm's Avatar
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    Default Re: If a file extension is known, but not the name, how do you find it?

    As far as I know the File class looks on the local PC.
    Otherwise the server would need to cooperate using some protocol, like FTP. Not all servers support that protocol.

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    Default Re: If a file extension is known, but not the name, how do you find it?

    I was afraid of that. I think I might have to change the standard then, and make the name of the file match the name of the last directory. Thanks for the help though.

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    Default Re: If a file extension is known, but not the name, how do you find it?

    Norm is right - you can't do that. URL.getPath() returns the string that ends up in the middle part of the first line in a HTTP request. It's completely useless as a File path on the client side. If your webserver supports directory listings out-of-the-box, you could do what you want by requesting the directory listing page from the webserver and parsing the returned page. It would be grim. Browse archive.ubuntu.com/ for an example Apache directory listing.

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