Develop Reference

How can I output a random image when in a jar file?

The below code works when running from my editor but the image fails to load when compiled into a runnable jar file with eclipse.
public static BufferedImage getRandomImage() {
// returns a random image from the Images folder
Random rand = new Random();
URL res = Card.class.getResource("Images"); // located in /src/.../Images
File f = new File(res.getFile());
if (!f.exists()) {
return new BufferedImage(1, 1, BufferedImage.TYPE_INT_RGB);
}
File[] files = f.listFiles();
int random = rand.nextInt(files.length);
BufferedImage img = null;
try {
img = ImageIO.read(files[random]);
} catch (IOException e) {
e.printStackTrace();
}
return img;
}
Could someone please suggest how I can modify my code or editor to load the files when compiled.
I have read other methods of accessing files but since I need to select randomly from a folder, I need to use the File class.

There is no safe way to list resources at runtime.
(Some people may suggest approaches which work sometimes, but will not work all the time. Class.getResource is not guaranteed to provide a listing; ProtectionDomain.getCodeSource can return null.)
But you don’t need to. It’s your application; you already know what files you put into it.
The best way is to either hard-code the list of files, or include a simple text file that contains a list of the files.
As an example, assume you created (or generated) a file named image-files.txt in which each line contains the base name of an image file, and embedded that file in your application:
List<String> imageNames;
try (BufferedReader linesReader = new BufferedReader(
new InputStreamReader(
Card.class.getResourceAsStream("image-files.txt"),
StandardCharsets.UTF_8));
Stream<String> lines = linesReader.lines()) {
imageNames = lines.collect(Collectors.toList());
} catch (IOException e) {
throw new UncheckedIOException(e);
}
int random = rand.nextInt(imageNames.length());
String imageName = imageNames.get(random)));
BufferedImage img;
try {
img = ImageIO.read(Card.class.getResource(imageName));
} catch (IOException e) {
throw new UncheckedIOException(e);
}
return img;
Note: The getFile() method of URL does not return a valid filename. It only returns the path portion of a URL. There are many characters which would be illegal in URLs, so the path portion percent-escapes them. If you ignore this fact, the value returned by getFile() will eventually fail.
(The reason for the misleading method name is that the URL class was part of Java 1.0, and in the mid-1990s, all URLs actually referred to physical files.)
I need to use the File class
Each .jar entry is just a subsequence of compressed bytes within a single .jar file, so you will never be able to use File to read such an entry. Class.getResource and Class.getResourceAsStream are the only correct ways to read those entries.

The problem is that you are trying to access a URL of a resource as a file.
with this you can get all the images, and then you can do this:
List<String> arr = getResourceFiles("Images");
String imgPath = arr.get(rand.nextInt(arr.size()));
InputStream stream = Card.class.getResourceAsStream("Images/" + imgPath);
try {
img = ImageIO.read(stream);
} catch (IOException e) {
e.printStackTrace();
}
return img;

Why am I getting 'nullfiles' when trying to copy a file from inside a jar to disk?

I am attempting to copy a file from inside my JAR to disk, outside the JAR file. The files that I will need to copy are default configuration files for a large-scale accounting system and are needed on the computer file system.
I have searched StackOverflow, as well as other sites (found with Google) and have read around fifty answers, which I've tried all of them. The code below is the first that has not simply blown up (with NullPointerException or FileNotFoundException), but has actually attempted to get the resource located in the JAR file.
I have my JAR file set up as follows:
com.is2300.isis
MainClass.java (actual name is crazy long and I don't want to type it out right now)
com.is2300.isis.resources
Location of the resource file I would like to copy out to disk
com.is2300.isis.utils
Location of my class ResourceExporter (below - bottom) that has the file exporting methods.
My MainClass.main() entry-point function:
public static void main(String[] args) {
// Test our 'utils.ResourceExporter.exportResource(String resourceName)
//+ function.
// Set the start point of our substring to either 5 or 9, depending upon
//+ if we are debugging (in NetBeans) or not (executing the JAR).
if ( isDebugging ) {
startPoint = 5;
} else {
startPoint = 9;
}
// First, we'll try just the name of the resource file to export.
String rsName = "nwind.conf";
try {
System.out.println(ResourceExporter.exportResource(rsName,
MainClass.class, "/home/user/tmp", startPoint));
} catch (Exception ex) {
System.err.println(ex.getCause());
System.err.println(ex.getMessage());
ex.printStackTrace(System.err);
}
// Then, we'll try it with the absolute path.
rsName = "/com/is2300/isis/resources/nwind.conf";
try {
System.out.println(ResourceExporter.exportResource(rsName,
MainClass.class, "/home/user/tmp", startPoint));
} catch (Exception ex) {
System.err.println(ex.getCause());
System.err.println(ex.getMessage());
ex.printStackTrace(System.err);
}
// Then, we'll try it with the relative path.
rsName = "../resources/nwind.conf";
try {
System.out.println(ResourceExporter.exportResource(rsName,
MainClass.class, "/home/user/tmp", startPoint));
} catch (Exception ex) {
System.err.println(ex.getCause());
System.err.println(ex.getMessage());
ex.printStackTrace(System.err);
}
// Last, we'll try it using dots instead of slashes.
rsName = "com.is2300.isis.resources.nwind.conf";
try {
System.out.println(ResourceExporter.exportResource(rsName,
MainClass.class, "/home/user/tmp", startPoint));
} catch (Exception ex) {
System.err.println(ex.getCause());
System.err.println(ex.getMessage());
ex.printStackTrace(System.err);
}
}
My ResourceExporter.exportResource() method:
public static String exportResource(String resourceName, Class cls,
String outPath, int startPoint) throws Exception {
File files = new File(cls.getResource(
cls.getResource(cls.getSimpleName() +
".class").toString().substring(
startPoint, cls.getResource(
cls.getSimpleName() + ".class").toString().lastIndexOf("/")
+ 1)) + "files");
InputStream in = new FileInputStream(files);
FileOutputStream out = new FileOutputStream(outPath +
resourceName.substring(resourceName.lastIndexOf("/")));
int readBytes;
byte[] buffer = new byte[4096];
while ( (readBytes = in.read(buffer)) > 0 )
out.write(buffer, 0, readBytes);
in.close();
out.close();
return files.getAbsolutePath();
}
With what I'm doing in public static void main(String[] args), I would expect one of the calls to the ResourceExporter.exportResource() method to actually cause the file to be copied.
However, when I step through the exportResource() method, on each call after the line:
File files = new File(cls.getResource(
cls.getResource(cls.getSimpleName() +
".class").toString().substring(
startPoint, cls.getResource(
cls.getSimpleName() + ".class").toString().lastIndexOf("/")
+ 1)) + "files");
The variable files.getCanonicalPath() call shows /home/user/Projects/ProjectName/nullfiles and I do not understand why this is, nor what this is.

#JBNizet and #AndrewThompson:
Thank you both for your comments. Especially #JBNizet! You gave me a swift kick in the head that made me look closer at what I had written and immediately saw the issue. Thank you very much for that.
The fix was this: Instead of the convoluted thing I was doing:
File files = new File(cls.getResource(
cls.getResource(cls.getSimpleName() +
".class").toString().substring(
startPoint, cls.getResource(
cls.getSimpleName() + ".class").toString().lastIndexOf("/")
+ 1)) + "files");
InputStream in = new FileInputStream(files);
FileOutputStream out = new FileOutputStream(outPath +
resourceName.substring(resourceName.lastIndexOf("/")));
Which I had written about 10 years ago and don't remember what I was thinking, I was able to simplify it to this:
InputStream in = ClassLoader.getSystemClassLoader().getSystemResourceAsStream(resourceName);
FileOutputStream out = new FileOutputStream(outPath +
resourceName.substring(resourceName.lastIndexOf("/")));
Now, the "file" (as a nod to the semantic correction by #AndrewThompson) is being located within the JAR file.
However, that was not the only change that had to be made. Where I set up the variable that is passed into the parameter resourceName was also not correct. I had it set up like so:
String rsName = "/com/is2300/isis/resources/nwind.conf";
However, the leading slash (/) was not supposed to be there. As soon as I changed the above line to:
String rsName = "com/is2300/isis/resources/nwind.conf";
everything just worked the way I expected.
Again, thanks to those two who commented. I appreciate your thoughts and assistance in getting my brain engaged.

Moveing files with java

I'm trying to move files using this java code and it can locate the file but not move it, just deletes the directory I'm moving it to.
public void ch() throws Exception{
if (FC.showOpenDialog(null) == JFileChooser.APPROVE_OPTION){
java.io.File file = FC.getSelectedFile();
Scanner input = new Scanner(file);
System.out.println(file);
Path source = Paths.get(file + "");
Path target = Paths.get("C:\\Users\\Marcus\\Desktop\\2");
try {
Files.move(source, target, REPLACE_EXISTING);
} catch (IOException e){
System.out.println("Failed to move the file");
}
}else{
System.out.println("?");
}
}

Add the file name at the end of your destination path, like below:
You could move files with File.ranameTo() method, like this:
file.renameTo(new File("C:\\Users\\Marcus\\Desktop\\2\\"+file.getName()));
In your example:
public void ch() throws Exception{
if (FC.showOpenDialog(null) == JFileChooser.APPROVE_OPTION){
java.io.File file = FC.getSelectedFile();
try {
file.renameTo(new File("C:\\Users\\Marcus\\Desktop\\2\\"+file.getName()));
} catch (Exception e){
System.out.println("Failed to move the file");
}
}else{
System.out.println("?");
}
}

If you want to use Files.move(), your target path should probably be the full path of the target file, not the destination directory where you want to place it.
Path target = Paths.get("C:\\Users\\Marcus\\Desktop\\2\\" + source.getName());

You should use Files.copy() instead of Files.move().

I strongly recommend the use of a third party tool such as Apache Commons IO's FileUtil class for this type of operation.
For example: FileUtil.moveFileToDirectory
Using these types of utilities saves you from many problems you aren't even aware are lurking. Yes, there are limitations to these common utils, but the benefits usually outweigh them in simple cases.
Google Guava is also an option, but I've got less experience there.

Your code is close but there a couple of potential issues. Before I start, I should say that I'm using a Mac (hence the path change), so while this is working for me, there may be some underlying permission issue on your system I can't account for.
1) You aren't using the name of the file you want to move to. You're using the directory you want to move the file to. That's a fair assumption, but you need to make it the fully qualified path and file name.
2) You are creating a Scanner to the to file but not using it. This probably doesn't really matter, but it's best to eliminate unnecessary code.
3) You don't validate the path that was created by getting the Path instance returned from Files.move().
Here is my example code. I tested it and it worked fine. Again, I'm using a Mac, so take that into account.
public void moveFile(){
JFileChooser fc = new JFileChooser("/");
if (fc.showOpenDialog(null) == JFileChooser.APPROVE_OPTION){
File file = fc.getSelectedFile();
System.out.println("Chosen File: " + file.getAbsolutePath());
String newFileName = System.getProperty("user.home")+File.separator+file.getName();
System.out.println("Attempting to move chosen file to destination: " + newFileName);
Path target = Paths.get(newFileName);
try {
Path newPath = Files.move(file.toPath(), target, REPLACE_EXISTING);
System.out.println("Path returned from move: " + newPath);
} catch (IOException e){
// Checked exceptions are evil.
throw new IllegalStateException("Unable to move the file: " + file.getAbsolutePath(),e);
}
}
}
The output from one of the tests:
Chosen File: /Users/dombroco/temp/simpleDbToFileTest1.txt
Attempting to move chosen file to destination: /Users/dombroco/simpleDbToFileTest1.txt
Path returned from move: /Users/dombroco/simpleDbToFileTest1.txt

Importing Fonts from JAR file

I've attempted to create a custom font using the following method, however it throws an exception :
Stream closed
and nothing happens! How can I import a ttf file from my JAR and use it in Java 2D! I'v managed to get it to work with external files, but it just doesn't work with an InputStream!
public Font gameFont(String filename, float fontSize) {
Font myfont = null;
Font myfontReal = null;
try {
InputStream is = new BufferedInputStream(this.getClass().getResourceAsStream("com/or/dungeon/" + filename));
myfont = Font.createFont(Font.TRUETYPE_FONT, is);
myfontReal = myfont.deriveFont(fontSize);
is.close();
} catch (FontFormatException | IOException e) {
System.out.println(e.getMessage());
}
return myfontReal;
}

You are missing a leading slash. Without it, it is searching relative to the class making the call. Try:
this.getClass().getResourceAsStream("/com/or/dungeon/" + filename));
Alternatively, try:
this.getClass().getClassLoader().getResourceAsStream("com/or/dungeon/" + filename));

How to get a path to a resource in a Java JAR file

I am trying to get a path to a Resource but I have had no luck.
This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents:
ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));
If I do this:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());
The result is:
java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)
Is there a way to get a path to a resource file?

This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.
Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.

When loading a resource make sure you notice the difference between:
getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path
and
getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning
I guess, this confusion is causing most of problems when loading a resource.
Also, when you're loading an image it's easier to use getResourceAsStream():
BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));
When you really have to load a (non-image) file from a JAR archive, you might try this:
File file = null;
String resource = "/com/myorg/foo.xml";
URL res = getClass().getResource(resource);
if (res.getProtocol().equals("jar")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile("tempfile", ".tmp");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.close();
file.deleteOnExit();
} catch (IOException ex) {
Exceptions.printStackTrace(ex);
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if (file != null && !file.exists()) {
throw new RuntimeException("Error: File " + file + " not found!");
}

The one line answer is -
String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()
Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm().

I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)
try {
//Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
//Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
//Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.
//Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
try {
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
} catch (Exception e) { }
//Find the last / and cut it off at that location.
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
//If it starts with /, cut it off.
if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
//If it starts with file:/, cut that off, too.
if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
PROGRAM_DIRECTORY = ""; //Current working directory instead.
}

This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names.
File file = null;
String resource = "/view/Trial_main.html" ;
URL res = getClass().getResource(resource);
if (res.toString().startsWith("jar:")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile(new Date().getTime()+"", ".html");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
input.close();
file.deleteOnExit();
} catch (IOException ex) {
ex.printStackTrace();
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}

if netclient.p is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really file:/path/to/jarfile/bot.jar!/config/netclient.p.

You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:
src/main/resources/myfile.txt
If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.

In my case, I have used a URL object instead Path.
File
File file = new File("my_path");
URL url = file.toURI().toURL();
Resource in classpath using classloader
URL url = MyClass.class.getClassLoader().getResource("resource_name")
When I need to read the content, I can use the following code:
InputStream stream = url.openStream();
And you can access the content using an InputStream.

follow code!
/src/main/resources/file
streamToFile(getClass().getClassLoader().getResourceAsStream("file"))
public static File streamToFile(InputStream in) {
if (in == null) {
return null;
}
try {
File f = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
f.deleteOnExit();
FileOutputStream out = new FileOutputStream(f);
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
return f;
} catch (IOException e) {
LOGGER.error(e.getMessage(), e);
return null;
}
}

A File is an abstraction for a file in a filesystem, and the filesystems don't know anything about what are the contents of a JAR.
Try with an URI, I think there's a jar:// protocol that might be useful for your purpouses.

The following path worked for me: classpath:/path/to/resource/in/jar

private static final String FILE_LOCATION = "com/input/file/somefile.txt";
//Method Body
InputStream invalidCharacterInputStream = URLClassLoader.getSystemResourceAsStream(FILE_LOCATION);
Getting this from getSystemResourceAsStream is the best option. By getting the inputstream rather than the file or the URL, works in a JAR file and as stand alone.

It may be a little late but you may use my library KResourceLoader to get a resource from your jar:
File resource = getResource("file.txt")

Inside your ressources folder (java/main/resources) of your jar add your file (we assume that you have added an xml file named imports.xml), after that you inject ResourceLoader if you use spring like bellow
#Autowired
private ResourceLoader resourceLoader;
inside tour function write the bellow code in order to load file:
Resource resource = resourceLoader.getResource("classpath:imports.xml");
try{
File file;
file = resource.getFile();//will load the file
...
}catch(IOException e){e.printStackTrace();}

Maybe this method can be used for quick solution.
public class TestUtility
{
public static File getInternalResource(String relativePath)
{
File resourceFile = null;
URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
String codeLocation = location.toString();
try{
if (codeLocation.endsWith(".jar"){
//Call from jar
Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
resourceFile = path.toFile();
}else{
//Call from IDE
resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
}
}catch(URISyntaxException ex){
ex.printStackTrace();
}
return resourceFile;
}
}

When in a jar file, the resource is located absolutely in the package hierarchy (not file system hierarchy). So if you have class com.example.Sweet loading a resource named ./default.conf then the resource's name is specified as /com/example/default.conf.
But if it's in a jar then it's not a File ...

This Class function can get absolute file path by relative file path in .jar file.
public class Utility {
public static void main(String[] args) throws Exception {
Utility utility = new Utility();
String absolutePath = utility.getAbsolutePath("./absolute/path/to/file");
}
public String getAbsolutePath(String relativeFilePath) throws IOException {
URL url = this.getClass().getResource(relativeFilePath);
return url.getPath();
}
}
If target file is same directory with Utility.java, your input will be ./file.txt.
When you input only / on getAbsolutePath(), it returns /Users/user/PROJECT_NAME/target/classes/. This means you can select the file like this /com/example/file.txt.