I'm tried several ways to zip a directory structure in a zip file with Java. Don't matter if I use ZipOutputStream or the Java NIO zip FileSystem, I just can't add empty folders to the zip file.
I tried with unix zip, and it works as expected, so I discarded a possibly zip-format issue.
I could also do a little workaround, adding an empty file inside the folders, but I don't really want to do that.
Is there a way to add empty folders in zip files using java API's?
EDIT: Based on answers and comments, this is pretty much the solution I got.
Thanks!
Java NIO makes this as easy as working with a normal file system.
public static void main(String[] args) throws Exception {
Path zipfile = Paths.get("C:\\Users\\me.user\\Downloads\\myfile.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(zipfile, null);) {
Path extFile = Paths.get("C:\\Users\\me.user\\Downloads\\countries.csv"); // from normal file system
Path directory = zipfs.getPath("/some/directory"); // from zip file system
Files.createDirectories(directory);
Files.copy(extFile, directory.resolve("zippedFile.csv"));
}
}
Given a myfile.zip file in the given directory, the newFileSystem call will detect the file type (.zip mostly gives it away in this case) and create a ZipFileSystem. Then you can just create paths (directories or files) in the zip file system and use the Java NIO Files api to create and copy files.
The above will create the directory structure /some/directory at the root of the zip file and that directory will contain the zipped file.
Related
We are programming a game, which shall be startable from a .jar file. First we created a Project in IntelliJ and loaded the Images from a ZIP with the following code:
ZipFile zf = null;
try {
zf = new ZipFile(zipPath);
Image Image = ImageIO.read(zf.getInputStream(zf.getEntry("Block/Air.png")));
} catch (IOException ignored) {}
Now the attempt without the ZIP (just from the .jar) is:
Image image=ImageIO.read(getClass().getResourceAsStream(path+ "Block/Air.png"));
It doesn't load any texture. Do you have a better way to do this in combination?
Edit:Seems not to be the Problem.
Since jars are zip files you could place them in the jar file and placing them the classpath.
Image image=ImageIO.read(getClass().getResourceAsStream(path+ "Block/Air.png");
Path must be a relative path from a source path root. E.g. I have a file in "src/main/resource/my/cool/game/" path is "/my/cool/game".
If you want to use a zip file, it must be outside of your jar file. To load the zip, you could use a relative file path, which is the same if you start your game from Intellij and from dekstop.
To change the working directory in intellij look here.
The best way would be to place the zip file alongside the jar so you can use "." as working directory to load the zip file.
Alternatively you could use a fixed directory, but then your game needs some sort of installation so it knows where to finde the zip file.
If you use ".", the zip file needs to be in the root of the project directory.
The Class.getResource and Class.getResourceAsStream methods take a URL (not a file path!) which is relative to the root of each classpath entry. For classpath entries which are .jar files, this means the path of a file packaged within the respective .jar file.
If your entire program is in one .jar file, the classpath consists of just one item: that .jar file. Therefore, there is only one classpath root, and the String you pass to getResourceAsStream is the URL of an entry within your .jar file. Do not include the path to the .jar file in that String.
If you are not sure what you should pass, examine your .jar file's contents. Every IDE (that I know of) provides a way to do this. You can also use any unzip utility to examine a .jar file, since every .jar file is actually a .zip file. (If you only have Windows, with no zip tools installed, make a copy of the .jar file and change the copy's extension to ".zip", then open it.)
Inside the .jar file are, of course, zip entries. The full path of the entry you want to load (without the path to the .jar file) is what you must pass to getResourceAsStream. getResourceAsStream accepts a URL, and URLs always use forward slashes (/) on all platforms, so do not use any backslashes. Also, the first character of the String must be /.
It is actually possible to specify a shorter path, depending on how your images are packaged in the .jar, but that is a separate topic. See the documentation for full details.
Side note: Never, ever write an empty catch block. Ever. That caught exception is by far the easiest way for you or anyone else to know when and why your program is not working. At the very least, put exc.printStackTrace(); in your catch block. More often, the correct course of action is to abort the program with something like throw new RuntimeException(exc);. After all, your program can't continue to function properly if it can't load that image, right?
Why do you need to store your images in a zip file? If you're doing it to reduce the file size, you gain absolutely nothing from zipping it first. JAR files are zipped files anyway (if you don't believe me, rename your .jar file to .zip, and try to open it). What you're basically doing is attempting to zip an already zipped file, which doesn't really do anything.
I would recommend you unzip your images and store them somewhere like < resources >/images
If you insist on leaving them zipped, you'll need to change it to something like this. Otherwise, it's looking for the zip file in the working directory (directory from which the jar was executed)
ZipFile zf = new ZipFile(getClass().getResourcesAsStream("path/to/zip"));
Disclaimer: I am not familiar with the ZipFile class, so I do not know if that constructor exists.
I am having problem to use JWNL wordnet in a Jar file.
JWNL uses RandomAccessFile to read wordnet dictionary files. In order to create a Jar file, wordnet dictionary files are put in resources/wordnet folder. As resources is in my Build Path, I have no problem to run the application I created in Eclipse. However, when I use another application to run the created jar file, I get the following error:
java.io.FileNotFoundException: resources/wordnet/data.noun (No such file or directory)
from the following code:
RandomAccess _file = new RandomAccessFile(path, _permissions);
I use the following code to check the current working directory:
URL location = PrincetonRandomAccessDictionaryFile.class.getProtectionDomain().getCodeSource().getLocation();
System.out.println(location.getFile());
It seems both situation have the same location: /project/bin/
How should I fix the problem? Thank you
The key information you seem to be missing is that Jar files are compressed, and you can't "seek" because of the compression (which is I believe the DEFLATE algorithm).
However, you could extract the file(s) into temp file(s) on start and then use that. Temp files would be removed on application exit, and are the best answer I can think of.
RandomAccessFile to read files in a Jar file
There are no files in a JAR file. There are JAR entries. You can't read them with FileInputStreams, RandomAccessFiles, or FileReaders.You need to use a JarInputStream or its friends.
I would like to get a list of file contained in a directory which is in a jar package.
I have an "images" folder, within it I have an Images class that should load all images from that directory.
In the past i used the MyClass.class.getResourceAsStream("filename"); to read files, but how do I read a directory?
This is what I tried:
System.out.println(Images.class.getResource("").getPath());
System.out.println(new File(Images.class.getResource("").getPath()).listFiles());
I tried with Images.class.getResource because I have to work with File and there isn't a constructor that accepts an InputStream.
The code produces
file:/home/k55/Java/MyApp/dist/Package.jar!/MyApp/images/
null
So it is finding the folder which I want to list files from, but it is not able to list files.
I've read on other forums that in fact you can't use this method for folders in a jar archive, so how can I accomplish this?
Update: if possible, i would like to read files without having to use the ZipInputStream
You can't do that easily.
What you need to do:
Get the path of the jar file.
Images.class.getResource("/something/that/exists").getPath()
Strip "!/something/that/exists".
Use Zip File System to browse the Jar file.
It's a little bit of hacking.
I have an application that creates a temporary mp3-file and puts it in a directory like C:\
File tempfile = File.createTempFile("something", ".mp3", new File("C:\\));
I'm able to read it by just using that same tempfile again.
Everything works fine in the Eclipse IDE.
But when I export my project for as a Runnable jar, my files are still being made correctly (I can play them with some normal music player like iTunes) but I can't seem to read them anymore in my application.
I found out that I need to use something like getClass().getResource("/relative/path/in/jar.mp3") for using resource files that are in the jar. But this doesn't seem to work if I want to select a file from a certain location in my file system like C:\something.mp3
Can somebody help me on this one?
It seems you dont have file name of the temp files . When you was running your program in eclipse that instance was creating a processing files, but after you made a runable you are not able to read those file that instance in eclipse created, You runable file can create its own temp file and can process them,
To make temp files globe put there (path + name ) entries in some db or property file
For example of you will create a temp file from the blow code
File tempfile = File.createTempFile("out", ".txt", new File("D:\\"));
FileWriter fstream = new FileWriter(tempfile);//write in file
out = new BufferedWriter(fstream);
the out will not be out.txt file it will be
out6654748541383250156.txt // it mean a randum number will be append with file
and you code in runable jar is no able to find these temp files
getClass().getResource() only reads resources that are on your classpath. The path that is passed to getResource() is, in fact, a path relative to any paths on your current classpath. This sounds a bit confusing, so I'll give an example:
If your classpath includes a directory C:\development\resources, you would be able to load any file under this directory using getResource(). For example, there is a file C:\development\resources\mp3\song.mp3. You could load this file by calling
getClass().getResource("mp3/song.mp3");
Bottom line: if you want to read files using getResource(), you will need those files to be on your classpath.
For loading from both privileged JARs and the file system, I have had to use two different mechanisms:
getClass().getClassLoader().getResource(path), and if that returns null,
new File(path).toURI().toURL();
You could turn this into a ResourceResolver strategy that uses the classpath method and one or more file methods (perhaps using different base paths).
I have a scanner that's trying to read a file named info.data in the src folder.I get Exception in thread "main" java.io.FileNotFoundException: info.data (The system cannot find the file specified). What's the address I should put in the scanner?
If the input file is always part of your application (i.e. you also put this into the .jar file later) you should use getResourceAsStream() in order to read its contents.
InputStream in = getClass().getResourceAsStream(filename);
Scanner scanner = new Scanner(in);
In netbeans, the src folder isn't the destination of the compiled classes, so if you are using a relative path, the location your program launches is not going to be the src folder.
That means you typically should "extend" your build to copy a non-source file into the build path if you want it to operate in the manner you imply. Many files already copy over to the build path (like properties files), but if you are including a data file that doesn't have a rule for being place in the build path, you need to add the rule yourself.
Try putting the path to it.
File f = new File("C:\\path\\src\\info.data");