Java

29-Dec-11 16:56

I have begun work on a 2d game. Everything was going as planned in netbeans, until I tried to make the .jar for a friend to view my progress. I am getting a URI is not hierarchial exception.

My issue is , for a game object, I am using a folder structure to keep sprite strips rather than having one large sprite. This results in :

Media/
CharacterName/
AnimationName/
image.extension

the programming object just holds it's folder as a string, and I pass the getResource() URL to an object to fill the map of images. My error comes from this code:

Java

dir = new File(s.toURI());

I take the directory, and call listFiles and pass the file names found to the sprite loader. Here is a code snippet:

Java

dir = new File(s.toURI());
 File[] chld = dir.listFiles();
        //get a list of files in the image/character folder
        for(File f:chld)
        {
            //get a list of the files for each dir
            File[] grandChild = f.listFiles();
            for(File t:grandChild)
            {
                 String fname = t.getAbsolutePath();
                 System.out.println(fname);
                 String temp = fname;
                 temp = temp.substring(temp.lastIndexOf("/") + 1,temp.lastIndexOf("."));
                 String animName = temp.replaceAll("\\d*$", "");
                 int numPics = 0;
                 Pattern p = Pattern.compile("[0-9]+");
                 Matcher m = p.matcher(temp);
                 while(m.find()){
                     numPics = Integer.parseInt(m.group());
                 }
                 System.out.println("animation name: " + animName);
                 System.out.println("file name: " + fname);
                 System.out.println("number of pictures: " + numPics);
                Animations.put(animName, sl.loadStripImageArray(fname, numPics));
            }
        }

Excuse the poor code for now, I'm still working on it. (like the variable named t).

sl is the sprite loader, and Animations is a hash map. This works fine until I package the project. I don't want to write a bunch of convoluted code that I have to switch back and forth between netbeans and the jar. I'd like to have something that "just works".

I have considered application data, and I can do that if necessary. I'd like it to work as a package if possible though, and not have to deal with external folders.

loctrice30-Dec-11 7:06

30-Dec-11 7:06

After some collaboration on stack overflow, I have decided to use a meta data file. This will work for my current purposes.

Eventually I will move to an application data folder.

jschell30-Dec-11 8:51

30-Dec-11 8:51

URIs and files are not the same.

Your posted code is solely file based.

I believe you can create a URI, with care, that will resolve itself, as a Resource (research it) regardless of whether it is file or jar based.

loctrice30-Dec-11 9:06

30-Dec-11 9:06

Thank you for taking the time to answer.

I know that uri's and files are not the same thing. I also know the code I posted was file based, that's what I was looking for help with. I needed a way to do the same type of thing using the jar resource. The thing is, I know where the animations are stored because the sprite object will have his folder (i.e. the dino will have Media/dino/{n}nodes/image.extension)

The thing I have found out is that package resources are not the way to go for this method. Getting the resource itself is not the problem, finding it in a generic fashion is. Finding one resource from a jar is trivial, however having multiples with only the root resource identified is not so easy.

The way to do it simply is to include a jar in the project, and use the jar access classes to do the discovery. I did not want to go this route. I wanted something simple so I could access them the same way through every object, and not have to include a jar in my project.

jschell wrote:
Resource (research it)

I do not usually participate in posting on forums. I do not post without sufficient research first, and usually I find enough to understand what I'm doing without ever needing to post a question. I can get single resources from a jar, however this was not the issue I was having. These resources also need opened as streams, which means I still can't get the effect the posted code was doing. Again, the intent was to find all the sub resources based on this uri, not to load the uri as a resource.

For now I will use a meta file with the resource information in it, sot hat I can load it as a reasource from inside the class generically after parsing the meta file. Later, I will be switching to an external folder structure holding my application data.

jschell31-Dec-11 13:29

31-Dec-11 13:29

loctrice wrote:
Finding one resource from a jar is trivial, however having multiples with only the root resource identified is not so easy.

I believe it is if I understand the question. If you have multiple jars then you should have multiple directories and thus you will have different resource names.

loctrice wrote:

I do not usually participate in posting on forums. I do not post without sufficient research first, a

Not a problem in the sense that it doesn't matter to me whether people (including you) have researched it first or not. Other respondents might have that problem but not me.

I said that because I wasn't going to go look up the methodology myself.

loctrice wrote:
<layer>These resources also need opened as streams,

You can do that with a Resource, see ClassLoader.getResourceAsStream()

loctrice1-Jan-12 4:29

1-Jan-12 4:29

I don't think I was clear enough when I origionaly asked the question. It's not loading the resource I was looking for. It was finding the resource. Specifically the listfiles (because finding a single resource from the class loader is trivial) Loading a known resource is not a problem, finding it was. Just using the resource loader wasn't going to work, because the path I was looking in may have {n} sub paths.

I found a way to do this with the jar files, but it's not what I want. I would have to use a jar package inside my project. This would be problematic at this time, which is why I posted. I wanted to know if there was a way around it.

What I have decided to do is create a meta file in each root and load that. I can find all the subs in the jar from there. This leaves my code alone , still using the getresource. The answer was finding a way to find out which sub resources this particular object had.

jschell2-Jan-12 10:39

2-Jan-12 10:39

loctrice wrote:
I don't think I was clear enough when I origionaly asked the question. It's not loading the resource I was looking for. It was finding the resource. Specifically the listfiles (because finding a single resource from the class loader is trivial) Loading a known resource is not a problem, finding it was. Just using the resource loader wasn't going to work, because the path I was looking in may have {n} sub paths.

I don't understand the question then.

To "find" a resource you must must either use a 'name' of some sort or look in the resource itself.

If you are using a name, then the 'path' to the resource is then just part of the name. So instead of using "thing.x" you use "resource/all/bright/thing.x".

If you do that then the method I suggested will find it.

loctrice3-Jan-12 7:36

3-Jan-12 7:36

resource/{unknown number of unknown names}/thing.x

Your method only works if you know the variable part of that equation.

I was trying to find a way to get all of the variable part, and I believe you were helping me figure out what to do with them after I know them.

For instance, I have a game object that animates via sprite. Here is that object structure

ObjectName/{animationname}/file.png
ObjectName/{other_animation_name}/file.png
ObjectName/{Yet_another_animation_name/file.png
ObjectName/{some_media_type}/file.extension

I was expecting to get all the paths under ObjectName/

Loading these after I find them is not an issue at all. Finding them in a file/folder structure is not an issue either. Finding them inside a jar is what I was looking for. -*Finding*- them, not loading them. I don't know when I run what all is in there.

jschell3-Jan-12 8:33

3-Jan-12 8:33

Then it requires two methodologies. One that sequences through the file system. Another that sequences through the jar via the zip api.

No other choice.

You could probably trigger off of a top level class as two which is used.

Although just to be clear I don't see how that matters to running a game. One could use it for some sort of listing of images but not one that is tied to game functionality.

loctrice3-Jan-12 9:30

3-Jan-12 9:30

jschell wrote:
No other choice.

Except for a meta file, which is in my first reply.

ObjectName/Meta.ext

Which will allow me to find dependant resources and load them.

jschell wrote:
I don't see how that matters to running a game

That's alright, it's not required for the context. I was asking how to find the information I was needing. The manner in which I was looking for them was the subject. I could have the same problem looking for text files unrelated in any way to a game.

jschell wrote:
I don't see how that matters to running a game

I can preload all the images, sounds, character information, etc, etc when the game loads. Because of the people I have doing media with me, and the way I organized it, this will not only be easier to load, but easier to save when I make the toolsets for building. It will also follow the same save/load format no matter what object it is, including my tile maps. This will allow me have consistency in my toolsets, file formats, etc.

We are just interested in doing things our own way. It's not something going for sale, or for any other reason then our own fun and/or learning. If it were for something else, we'd use prebuilt engine software and/or another language.

At any rate, thank you for the conversation. I'm sorry I was not clear in my question and I hope I didn't waste too much of your time.

jschell3-Jan-12 10:36

3-Jan-12 10:36

loctrice wrote:
I can preload all the images, sounds

You could use a single root then.

loctrice wrote:
Except for a meta file, which is in my first reply.

Keep in mind that you can generate that rather than maintaining it manually.

loctrice3-Jan-12 11:00

3-Jan-12 11:00

Yeah, I am planning on having a toolset that will generate the meta file for me.

jschell wrote:
You could use a single root then.

I couldn't figure this out without the meta file either. I'd still have an unkown name and number of files, which leads me back to how to find them. I can do it with the meta file, but I still want it to be easily viewed if we unpack it (with logical directory structures).

In the end, I'm going to end up with an application data folder. I think , in that folder, I'm still going to pack the stuff in jar files, so I'll still use the meta file in that case. Right now I'm just packing it into my project. The "end goal" was to be able to move it without having to hard code stuff.

UnOrdered SET

saadullah Bhutto21-Dec-11 4:39

saadullah Bhutto

21-Dec-11 4:39

hello this is Saad

there is a Question revolving in my Mind that why a Set is UnOrdered
? ?
Smile | :)

Luc Pattyn21-Dec-11 5:21

Luc Pattyn

21-Dec-11 5:21

by definition.
sometimes an order is irrelevant to a collection, then use a set. It is like a mathematical set (remember Venn diagrams?).
without any order, adding and removing items can be very fast operations.

Smile | :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

TorstenH.21-Dec-11 20:50

TorstenH.

21-Dec-11 20:50

as Luc says, it's so by definition.

the Interface Set[^] is an extension to the interface Collection. And a Collection is unsorted - so it's most likely an inherited behavior.

regards Torsten
When I'm not working

jschell26-Dec-11 8:54

26-Dec-11 8:54

As a guess, most collections do not need to be ordered.
Thus ordering is something that is best done by adding it to a collection when needed.

Also if collections started out ordered then attempting to make them unordered, for performance reasons, would probably be hard if possible at all.

So better if everything started out unordered.