I'm coding this in java and put my sounds folder with the bgmusic.wav in there but it doesn't seem to still identify it. Am I doing something wrong?
Here is my Sound class:
import java.applet.Applet;
import java.applet.AudioClip;
import java.io.File;
import java.net.MalformedURLException;
public class Sound {
private AudioClip myClip;
public Sound(String fileName) {
try {
File file = new File(fileName);
if (file.exists()) {
myClip = (AudioClip) Applet.newAudioClip(file.toURI().toURL());
} else {
throw new RuntimeException("Sound: file not found: " + fileName);
}
} catch (MalformedURLException e) {
throw new RuntimeException("Sound: malformed URL: " + e);
}
}
public void play() {
myClip.play();
}
}
This is then ran as an Object in my other class to run the bgmusic.wav:
String fileName = "bgmusic.wav";
String soundDir = "." + File.separator + "sounds" + File.separator;
String filePath = soundDir + fileName;
Sound bgMusic = new Sound(filePath);
bgMusic.play();
If I'm not mistaken I think I have everything setup correctly but I get a Sound: file not found: ./sounds/bgmusic.wav which is returned from to throw exception in my Sound class. Since I have the sounds folder as a subdirectory of the main java folder passing ./sounds/bgmusic.wav should be correct right?
The new File(".") represents the base folder (the root folder) of your application.
So check your directory structure that where does the file appear with respect to the project base folder. else share your project structure.
If you are placing the file in the java classpath (The src folder) then you can load the sound file from class path (refer here)
I think you need to have the full path for File to recognize it.
Related
I'm new to Java, and I am facing this issue in Eclipse. Even after pointing it to the correct file, it shows a file not Found Error.
I am trying to compile code from a Java file using the Java Compiler API.
The code words fine in Visual Studio with setting everything in root, But gives this error in Eclipse with all these directories.
Also, why are there three different src folders in the image?
My project structure
package com.example.app;
import javax.tools.JavaCompiler;
import javax.tools.ToolProvider;
import java.io.File;
import java.io.IOException;
public class compilier {
public static void main(String[] args) throws IOException {
JavaCompiler compiler = ToolProvider.getSystemJavaCompiler();
int result = compiler.run(null, null, null, new File("com/example/app/Code.java").getAbsolutePath());
if (result == 0)
{
System.out.println("File Compiled");
}
try {
String package_dir = "/demo/src/main/java/com/example/app";
try{
ProcessBuilder builder = new ProcessBuilder("java", package_dir.concat("/Code"));
builder.redirectErrorStream(true);
File outfile = new File((package_dir.concat("/output.txt")));
builder.redirectOutput();
builder.start();
if (outfile.length() > 3000)
{
System.out.println("Exceeded buffer limit");
System.exit(1);
}
} catch(IOException e) {
e.printStackTrace();
}
} catch (Exception err) {
System.out.println("Error!");
err.printStackTrace();
}
}
}
Error Message
Your path looks wrong. The /demo directory would need to be in the root of your current drive.
Also, the output of a Maven build is found in the target directory. The Java class files are generated there, and the resource files are copied over from src/main/res hierarchy. The .Java files are lost. You could add a Maven task to copy the .Java files but this would be very nonstandard.
Finally you need to load resource files using the classpath. There are lots of examples on the Internet. Otherwise you may end up with a project that finds the file in Eclipse but not when deployed in a .jar or .war file.
Happy hunting.
Is it possible to access Assets inside the Java code in Play Framework? How?
We access assets from the scala HTML templates this way:
<img src="#routes.Assets.versioned("images/myimage.png")" width="800" />
But I could not find any documentation nor code example to do it from inside the Java code. I just found a controllers.Assets class but it is unclear how to use it. If this is the class that has to be used, should it maybe be injected?
I finally found a way to access the public folder even from a production mode application.
In order to be accessible/copied in the distributed version, public folder need to be mapped that way in build.sbt:
import NativePackagerHelper._
mappings in Universal ++= directory("public")
The files are then accessible in the public folder in the distributed app in production form the Java code:
private static final String PUBLIC_IMAGE_DIRECTORY_RELATIVE_PATH = "public/images/";
static File getImageAsset(String relativePath) throws ResourceNotFoundException {
final String path = PUBLIC_IMAGE_DIRECTORY_RELATIVE_PATH + relativePath;
final File file = new File(path);
if (!file.exists()) {
throw new ResourceNotFoundException(String.format("Asset %s not found", path));
}
return file;
}
This post put me on the right way to find the solution: https://groups.google.com/forum/#!topic/play-framework/sVDoEtAzP-U
The assets normally are in the "public" folder, and I don't know how you want to use your image so I have used ImageIO .
File file = new File("./public/images/nice.png");
boolean exists = file.exists();
String absolutePath = file.getAbsolutePath();
try {
ImageInputStream input = ImageIO.read(file); //Use it
} catch (IOException e) {
e.printStackTrace();
}
System.out.println("EX = "+exists+" - "+absolutePath);
I've been browsing the site since yesterday and I can't see to find anything that answers my question, so I decided to just ask.
I'm making a pretty basic java GUI, it's designed to be run alongside files that wont be included in the actual java package for compatibility and easier customization of those files, I doubt they could be included either way as they have their own .jars and other things.
So, the problem I'm having is that the GUI application is in the main folder and I need it to locate and open txt files a couple sub-folders deep, in notepad without requiring a full file path as I'll be giving this project out to some people when it's done.
Currently I've been using this to open the files, but will only work for files in the main folder and trying to edit in any file paths did not work.
private void jButton4ActionPerformed(java.awt.event.ActionEvent evt) {
Runtime rt=Runtime.getRuntime();
String file;
file = "READTHIS.txt";
try {
Process p=rt.exec("notepad " +file);
} catch (IOException ex) {
Logger.getLogger(NumberAdditionUI.class.getName()).log(Level.SEVERE, null, ex);
}
}
if someone knows a way to do this, that'd be great.
On another note, I'd like to include the file shown above (READTHIS.txt) inside the actual java package, where would I put the file and how should I direct java towards it?
I've been away from java for a long time so I've forgotten pretty much anything, so simpler explanations are greatly appreciated.
Thanks to anyone reading this and any help would be awesome.
Update 2
So I added to the ConfigBox.java source code and made jButton1 open home\doc\READTHIS.txt in Notepad. I created an executable jar and the execution of the jar, via java -jar Racercraft.jar, is shown in the image below. Just take the example of what I did in ConfigBox.java and apply it to NumberAdditionUI.java for each of its JButtons, making sure to change the filePath variable to the corresponding file name that you would like to open.
Note: The contents of the JTextArea in the image below were changed during testing, my code below does not change the contents of the JTextArea.
Directory structure:
\home
Rasercraft.jar
\docs
READTHIS.txt
Code:
// imports and other code left out
public class ConfigBox extends javax.swing.JFrame {
// curDir will hold the absolute path to 'home\'
String curDir; // add this line
/**
* Creates new form ConfigBox
*/
public ConfigBox()
{
// this is where curDir gets set to the absolute path of 'home/'
curDir = new File("").getAbsolutePath(); // add this line
initComponents();
}
/*
* irrelevant code
*/
private void jButton1ActionPerformed(java.awt.event.ActionEvent evt) {//GEN-FIRST:event_jButton1ActionPerformed
Runtime rt = Runtime.getRuntime();
// filePath is set to 'home\docs\READTHIS.txt'
String filePath = curDir + "\\docs\\READTHIS.txt"; // add this line
try {
Process p = rt.exec("notepad " + filePath); // add filePath
} catch (IOException ex) {
Logger.getLogger(NumberAdditionUI.class.getName()).log(Level.SEVERE, null, ex);
}
// TODO add your handling code here:
}//GEN-LAST:event_jButton1ActionPerformed
/*
* irrelevant code
*/
Update
This is the quick and dirty approach, if you would like me to add a more elegant solution just let me know. Notice that the file names and their relative paths are hard-coded as an array of strings.
Image of the folder hierarchy:
Code:
Note - This will only work on Windows.
import java.io.File;
import java.io.IOException;
public class Driver {
public static void main(String[] args) {
final String[] FILE_NAMES = {"\\files\\READTHIS.txt",
"\\files\\sub-files\\Help.txt",
"\\files\\sub-files\\Config.txt"
};
Runtime rt = Runtime.getRuntime();
// get the absolute path of the directory
File cwd = new File(new File("").getAbsolutePath());
// iterate over the hard-coded file names opening each in notepad
for(String file : FILE_NAMES) {
try {
Process p = rt.exec("notepad " + cwd.getAbsolutePath() + file);
} catch (IOException ex) {
// Logger.getLogger(NumberAdditionUI.class.getName())
// .log(Level.SEVERE, null, ex);
}
}
}
}
Alternative Approach
You could use the javax.swing.JFileChooser class to open a dialog that allows the user to select the location of the file they would like to open in Notepad.
I just coded this quick example using the relevant pieces from your code:
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.io.File;
import java.io.IOException;
import javax.swing.JButton;
import javax.swing.JFileChooser;
import javax.swing.JFrame;
import javax.swing.SwingUtilities;
public class Driver extends JFrame implements ActionListener {
JFileChooser fileChooser; // the file chooser
JButton openButton; // button used to open the file chooser
File file; // used to get the absolute path of the file
public Driver() {
this.fileChooser = new JFileChooser();
this.openButton = new JButton("Open");
this.openButton.addActionListener(this);
// add openButton to the JFrame
this.add(openButton);
// pack and display the JFrame
this.pack();
this.setVisible(true);
}
public void actionPerformed(ActionEvent e) {
// handle open button action.
if (e.getSource() == openButton) {
int returnVal = fileChooser.showOpenDialog(Driver.this);
if (returnVal == JFileChooser.APPROVE_OPTION) {
// from your code
Runtime rt = Runtime.getRuntime();
try {
File file = fileChooser.getSelectedFile();
String fileAbsPath = file.getAbsolutePath();
Process p = rt.exec("notepad " + fileAbsPath);
} catch (IOException ex) {
// Logger.getLogger(NumberAdditionUI.class.getName())
// .log(Level.SEVERE, null, ex);
}
} else {
System.exit(1);
}
}
}
public static void main(String args[]) {
SwingUtilities.invokeLater(new Runnable() {
public void run() {
Driver driver = new Driver();
}
});
}
}
I've also included a link to some helpful information about the FileChooser API, provided by Oracle: How to use File Choosers. If you need any help figuring out the code just let me know, via a comment, and I'll try my best to help.
As for including READTHIS.txt inside the actual java package, take a gander at these other StackOverflow questions:
Getting file from same package?
Reading a text file from a specific package?
How to include text files with executable jar?
Creating runnable jar with external files included?
Including a text file inside a jar file and reading it?
Usually to rename a file I use:
File oldFile = new File("file path");
oldFile.renameTo(new File("file path with new name"));
But what if the file I want to rename is inside a .jar executable is there a way to rename it from there?
No, you can't do that unless you extract the JAR file, rename the file and repackage it.
You can copy a jar, one entry at a time, renaming the entry you want to change. This might be more efficient than unpacking, renaming and repacking.
You cannot rename a class file without changing all the references to that name. Without recompiling all the code, you can use a library like ObjectWebs ASM to inspect the byte code and change references to that class. If the class is referenced in a String, you may want to change the string as well.
Yes, you may rename a file within a jar. For example you may use JarEntryFilter
like this:
...
import org.springframework.boot.loader.jar.JarEntryData;
import org.springframework.boot.loader.jar.JarEntryFilter;
import org.springframework.boot.loader.jar.JarFile;
import org.springframework.boot.loader.tools.JarWriter;
import org.springframework.boot.loader.util.AsciiBytes;
...
JarWriter writer = new JarWriter(destination);
try {
JarFile filteredJarFile = sourceJar.getFilteredJarFile(new JarEntryFilter() {
#Override
public AsciiBytes apply(AsciiBytes name, JarEntryData entryData) {
String string = name.toString();
String exp = "^a.*";
if (string.matches(exp)) {
string = string.replaceFirst(exp, "replaced");
return new AsciiBytes(string);
}
return name;
}
});
writer.writeEntries(filteredJarFile);
} finally {
try {
writer.close();
} catch (Exception ex) {
ex.printStackTrace();
}
}
suppose I put a file a.txt in package com.xyz and the try access it like following. Will it work?
Hi All,
import com.xyz.*;
public class Hello
{
File f = new File("a.txt");
...
}
It is not working for me. Is there any workaround?
Use Class.getResource() or Class.getResourceAsStream(). see for example the Sun demo source at http://jc.unternet.net/src/java/com/sun/WatermarkDemo/WatermarkDemo.java
I will offer the same answer as jcomeau_ictx, but a lot shorter (around 30 lines in one file as opposed to >380 in 1 source file of 5), ..and with a screenshot. ;)
import javax.swing.*;
import java.net.URL;
class GetResource {
GetResource() {
Class cl = this.getClass();
final URL url = cl.getResource( cl.getName() + ".java" );
SwingUtilities.invokeLater(new Runnable() {
public void run() {
JEditorPane ep = new JEditorPane();
try {
ep.setPage(url);
JScrollPane sp = new JScrollPane(ep);
sp.setPreferredSize(new java.awt.Dimension(400,196));
JOptionPane.showMessageDialog(null, sp);
} catch(Exception e) {
e.printStackTrace();
JOptionPane.showMessageDialog(
null,
e.getMessage() + " See trace for details.");
}
}
});
}
public static void main(String[] args) {
new GetResource();
}
}
Based on your responses to the comments above. If you are looking for a work around, just specify the path to the .txt file on the file system. Putting it in a package does not help.
new File ("a.txt")
looks for a file on the the file system and not within a package.
Please also read the javadocs on File:
http://download.oracle.com/javase/6/docs/api/java/io/File.html
However I do not see the rationale in putting the file inside a package unless you would want to use it as a resource. In which case #jcomeau_ictx has the right solution
It's depend on your class path of java from where you can run this class. If both are in same place then it will work. Then no need to define path in file. But the file was not in the classpath dir then must be define path of that file otherwise file not found.