I'm trying to figure out how to add a lot of images into my Java program without having to add every file name in the code.
So I want to take all the .png images in the correct folder and make them into an IconImage and then place them in an ArrayList.
What I've got right now is this.
private void fillList()
{
File[] files = new File("C:/Users/marre/Google Drive/Java/lek/imageplay/src/img").listFiles();
for (File f : files)
{
if (f.isFile())
{
ImageIcon tempIcon = new ImageIcon(getClass().getResource("/img/"+f.getName()));
List.add(tempIcon);
}
}
}
This works the way I want it to regarding getting the names of all images. But the file only reads from absolute file path.
What can I use instead to get the same result, but that works within .jar file (so it reads from src/img/ and not the whole c:/ bla bla..)?
I'm not sure if this will be helpful but I had a similar issue in attempting to obtain files within a jar file. In the case this can be useful to you, my approach was to obtain the URL for the files inside the jar using the ClassLoader:
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
URL url = classLoader.getResource("src/img/");
Check inside your jar to make sure of the directory structure for the images so you can specify the correct relative path.
Keep in mind, a jar is a packaged form, so this will most likely return a zipped file (and not a simple File object). You might have to dabble in ZipFile.
ZipFile folderInsideJar = new ZipFile(new File(url.getPath());
Enumeration<? extends ZipEntry> entries = folderInsideJar.entries();
while (entries.hasMoreElements()) {
// ideally this would be your image file, zipped
ZipEntry childFile = entries.nextElement();
// ... etc ...
}
For environments such as JBoss, etc, you might also get virtual files (vfsfile/vfszip). Point being, you can determine what you are dealing with by looking at the result URL.getProtocol().
It would also appear ImageIcon has a constructor that takes a URL. You might try giving this a shot to see if it handles the "resource inside jar" case before manually attempting to retrieve the zip file.
Related
I have written a project where some images are used for the application's appearance and some text files will get created and deleted along the process. I only used the absolute path of all used files in order to see how the project would work, and now that it is finished I want to send it to someone else. so what I'm asking for is that how I can link those files to the project so that the other person doesn't have to set those absolute paths relative to their computer. something like, turning the final jar file with necessary files into a zip file and then that the person extracts the zip file and imports jar file, when runs it, the program work without any problems.
by the way, I add the images using ImageIcon class.
I'm using eclipse.
For files that you just want to read, such as images used in your app's icons:
Ship them the same way you ship your class files: In your jar or jmod file.
Use YourClassName.class.getResource or .getResourceAsStream to read these. They are not files, any APIs that need a File object can't work. Don't use those APIs (they are bad) - good APIs take a URI, URL, or InputStream, which works fine with this.
Example:
package com.foo;
public class MyMainApp {
public void example() {
Image image = new Image(MyMainApp.class.getResource("img/send.png");
}
public void example2() throws IOException {
try (var raw = MyMainApp.class.getResourceAsStream("/data/countries.txt")) {
BufferedReader in = new BufferedReader(
new InputStreamReader(raw, StandardCharsets.UTF_8));
for (String line = in.readLine(); line != null; line = in.readLine()) {
// do something with each country
}
}
}
}
This class file will end up in your jar as /com/foo/MyMainApp.class. That same jar file should also contain /com/foo/img/send.png and /data/countries.txt. (Note how starting the string argument you pass to getResource(AsStream) can start with a slash or not, which controls whether it's relative to the location of the class or to the root of the jar. Your choice as to what you find nicer).
For files that your app will create / update:
This shouldn't be anywhere near where your jar file is. That's late 80s/silly windows thinking. Applications are (or should be!) in places that you that that app cannot write to. In general the installation directory of an application is a read-only affair, and most certainly should not be containing a user's documents. These should be in the 'user home' or possibly in e.g. `My Documents'.
Example:
public void save() throws IOException {
Path p = Paths.get(System.getProperty("user.home"), "navids-app.save");
// save to that file.
}
After using images for example on a Button, when I build the application creating the .jar file and execute only the file, the images are not there but would only show if I copy the images folder in the same directory as the jar file. Why is that and how can I resolve this if possible?
I am currently using the following code to set the icon/image:
JButton btn = new JButton("Text", "img/icon.png");
The fact that you can use the images when they are stored outside the jar, suggests that you are doing something of the kind:
File image = new File("directory/image.jpg");
InputStream is = new FileInputStream(image);
This reads a file from a directory on the file system, not from the classpath. Now, if you have packaged your image in a "directory" inside your Jar, you must load the image from the classpath.
InputStream is = getClass().getResourceAsStream("/directory/image.jpg");
(note the slash in the path)
Or
InputStream is = getClass().getClassLoader().getResourceAsStream("directory/image.jpg");
(note the absence of the slash in the path)
Your example, as it is now, should not compile. (The second argument of your JButton construtor is an Icon, not a String, java 8). So when you were getting the image from the file system, you were probably doing something else.
With your example, you need to read an image from the inputstream and convert it to an Icon:
try (InputStream is = getClass.getClassLoader().getResourcesAsStream("directory/image.jpg")) {
BufferedImage image = ImageIO.read(is);
return new JButton("Text", new ImageIcon(image));
} catch (IOException exc) {
throw new RuntimeException(exc);
}
That should use the image that is located in "directory" inside your jar. Of course, you need to include the image within your jar, or you will get a NullPointerException on the inputstream is.
I think you need to understand the
ClassLoader:
A typical strategy is to transform the name into a file name and then
read a "class file" of that name from a file system.
So with this you will be able to lead Resources of your project with getResource
public URL getResource(String name)
Finds the resource with the given name. A resource is some data
(images, audio, text, etc) that can be accessed by class code in a way
that is independent of the location of the code.
Right now I'm manually writing song file names into a String array, then sending the strings to InputStream is = this.getClass().getResourceAsStream(filename); to play the song.
I would rather loop through all the song names in my resources folder and load them into the array so that each time I add a song, I don't need to manually type it into the array.
All of my songs are located in resources folder:
Is there a java method or something to grab these names?
The files need to work when I export as .jar.
I thought I could use something like this?
InputStream is = this.getClass().getClassLoader().getResourceAsStream("resources/");
Is there a listFiles() or something for that?
Thanks
No, because the classpath is dynamic in Java. Any additional Jar could contribute to a package, a classloader can be as dynamic as imaginable and could even dynamically create classes or resources on demand.
If you have a specific JAR file and you know the path within that JAR file then you could simply treat that file as archive and access it via java.util.jar.JarFile which lets you list all entries in the file.
JarFile jar = new JarFile("res.jar");
Enumeration<JarEntry> entries = jar.entries();
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
System.out.println(entry.getName());
}
If you want to do this without knowing the JAR filename and path (e.g. by a class within the JAR itself) you need to get the filename dynamically as Evgeniy suggested; just use your own class name (including package; your screenshot looks like you are in default package, which is a bad idea).
You can try something like this
Enumeration<URL> e = ClassLoader.getSystemResources("org/apache/log4j");
while (e.hasMoreElements()) {
URL u = e.nextElement();
String file = u.getFile();
...
file:/D:/.repository/log4j/log4j/1.2.14/log4j-1.2.14.jar!/org/apache/log4j
extract jar path from here and use JarFile class to know org/apache/log4j contents.
If it is unpacked, you will get straight path and use File.listFIles
I am trying to access a directory inside my jar file. I want to go through every of the files inside the directory itself. I tried using the following:
File[] files = new File("ressources").listFiles();
for (File file : files) {
XMLParser parser = new XMLParser(file.getAbsolutePath());
// some work
}
If I test this, it works well. But once I put the contents into the jar, it doesn't because of several reasons. If I use this code, the URL always points outside the jar.
structure of my project :
src
controllers
models
class that containt traitement
views
ressources
See this:
How do I list the files inside a JAR file?
Basically, you just use a ZipInputStream to find a list of files (a .jar is the same as a .zip)
Once you know the names of the files, you can use getClass().getResource(String path) to get the URL to the file.
I presume this jar is on your classpath.
You can list all the files in a directory using the ClassLoader.
First you can get a list of the file names then you can get URLs from the ClassLoader for individual files:
public static void main(String[] args) throws Exception {
final String base = "/path/to/folder/inside/jar";
final List<URL> urls = new LinkedList<>();
try (final Scanner s = new Scanner(MyClass.class.getResourceAsStream(base))) {
while (s.hasNext()) {
urls.add(MyClass.class.getResource(base + "/" + s.nextLine()));
}
}
System.out.println(urls);
}
You can do whatever you want with the URL - either read and InputStream into memory or copy the InputStream into a File on your hard disc.
Note that this definitely works with the URLClassLoader which is the default, if you are using an applet or a custom ClassLoader then this approach may not work.
NB:
You have a typo - its resources not ressources.
You should use reverse domain name notation for your project, this is the convention.
Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?
Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}
Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.
When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).
If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.
The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.
If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.
say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left