I have a method that gets called at the start of my program, and it works fine when I run the jar, but when I run the jnlp file it crashes.
public void newImg(String i)
{
try
{
File file = new File(i);
img = ImageIO.read(file);
}
}
img is a bufferedImage, by the way.
Odds are the path you're giving to the file, or permissions, don't match. Put the File ctor into a try block, catch the exception, and find out what it is.
It should look something like this:
try {
File file = new File(i);
img = ImageIO.read(file);
} catch (Exception ex) {
// You probably want to open the java console, or use a logger
// as a JNLP may send stderr someplace weird.
Systemm.err.println("Exception was: ", ex.toString());
}
The code you have doesn't do anything with the exception.
You might want to look at the exceptions thread in the Java Tutorial.
update
See my comments. I just tried something and confirmed what I was thinking -- code with a try {} and no catch or finally won't even compile. if this is really the code you think you're working with, you've probably been loading an old class file; this one hasn't compiled.
$ cat Foo.java
public class Foo {
public void tryit() {
try {
File f = new File(null);
}
}
}
$ javac Foo.java
Foo.java:3: 'try' without 'catch' or 'finally'
try {
^
1 error
$
Maybe your not loading your image properly. Don't use the relative location of the file. This will be different for each OS. Your image in the JAR you should be loaded correctly like this:
URL url = this.getClass().getResource("image.jpg");
Image img = Toolkit.getDefaultToolkit().getImage(url);
This will load a file called image.jpg that is located in the same location as the class. You can also use things like File.pathSeparator if its in another location.
Use one of these two methods to load it as a resource:
http://java.sun.com/javase/6/docs/api/java/lang/Class.html#getResource(java.lang.String) http://java.sun.com/javase/6/docs/api/java/lang/Class.html#getResourceAsStream(java.lang.String)
Make sure you have the correct file name/path.
Make sure you have file access to the file system.
Related
tl;dr I'm more used to writing command-line scripts that can just output based on the current working directory, so I'm unsure what directory to use for output files in a program that will be launched from a JAR.
Program Description:
My program builds an HTML file from data given to it from the rest of the program, and then is supposed to write it to a file that we'll call "Output.html" for simplicity.
Relevant Code:
public void outputHTML()
{
String output = buildHTML();
// Expanded to explain my confusion better
String fileDirectory = ""; // ???
String fileName = "Output.html";
String fullPath = fileDirectory + "\\" + fileName;
try (BufferedWriter writer = new BufferedWriter(new FileWriter(fullPath)))
{
writer.write(output);
writer.close();
} catch (IOException e)
{
System.out.println("File not found.");
e.printStackTrace();
}
}
Problem
I don't know what to put the file directory as. Usually I run my programs from the command line and use ".\\Output.txt" as my output path, but I don't know where to put it if it's being run from a JAR.
The desired file structure is as follows:
Encompassing Folder
Program.jar
output
Output.html
Or alternatively (not sure if this makes it easier to understand or harder):
main\
main\Program.jar
main\output\
main\output\Output.html
Everything I can find on SE only relates to reading files that are both immutable and internal, but I'm trying to output a non-static file to a location outside of my jar.
Can anyone help with this? Thanks!
Misc Details
I'm using Eclipse without Gradle currently, because I don't know what Gradle is and new things are scary. If this particular problem would be easier to solve with Gradle, let me know and I'll look up more about it.
EDIT:
Added syntax highlighting to code block.
Formatted everything a bit better
Changed title to be more descriptive
You can use an absolute path: e.g. fileDirectory = "\\project\\test\\main\\output";
using normal slash should also work even on Windows ("/project/test/main/output")
Or use a relative path - this will start from the current working directory (user directory), the one where the JVM was started in - e.g. fileDirectory = "main\\output";
private void copyFile() throws IOException {
Path destination;
String currentWorkingDir = System.getProperty("user.dir");
File fileToCopy = component.getArchiveServerFile();
if (path.contains(File.separator)) {
destination = Paths.get(path);
} else {
destination = Paths.get(currentWorkingDir + File.separator + path);
}
if (!Files.exists(destination)) {
try {
Files.createDirectories(destination);
} catch (IOException ioe) {
ioe.printStackTrace();
}
}
FileUtils.copyFileToDirectory(fileToCopy, new File(destination.toString()));
}
}
Basically what I'm trying to do here is copying a file in some location using the path provided in the class's constructor. The logic is like this:
If the path has file separator, I consider it a full path and copy the file at the end.
If the path doesn't have file separator, I copy the file in the working directory from which the .exe file was launched.
So far, only the first option works (the full path). For some reason, the working directory option is not working and I can't figure out why.
UPDATE: If I just change the following line:
String currentWorkingDir = System.getProperty("user.dir");
to
String currentWorkingDir = System.getProperty("user.home");
It works. So I'm guessing the problem is coming from user.dir? Maybe at runtime, the folder is already being used and as a result, it can't copy the file into it?
The weird thing is, I don't have any exceptions or error, but nothing happens as well.
UPDATE 2: I think the problem here is that I'm trying to copy a file which is embedded in the application (.exe file) that I'm executing during runtime, and java can't copy it while the current working directory is being used by the application.
UPDATE 3:
Since this copy method is used in an external library, I had to come up with another way (other than logs) to see the content of system property user.dir. So I wrote I little program to create a file and write in it the value return by the property.
To my surprise, the path is not where my application was launched. It was in:
C:\Users\jj\AppData\Local\Temp\2\e4j1263.tmp_dir1602852411
Which is weird because I launched the program from :
C:\Users\jj\workspace\installer\product\target\
Any idea why I'm getting this unexpected value for user.dir?
I'm using a custom method to get pictures from the resources/ folder. The hardcoded path works well when programming during production (src/main/resources/). However when delivering, I would need to make this path relative to the .jar root. So I made this.
public static Image getImageFromFile(String file)
{
Image image = null;
try
{
String path = FileUtils.class.getClassLoader().getResource(file).toExternalForm();
System.out.println(path);
File pathToFile = new File(path);
image = ImageIO.read(pathToFile);
}
catch (IOException ex) {ex.printStackTrace();}
return image;
}
file:/C:/Users/Hugo/Desktop/Hugo/Java%20Workspace/ClashBot/bin/main/icons/level-label.png
javax.imageio.IIOException: Can't read input file!
at javax.imageio.ImageIO.read(Unknown Source)
at com.lycoon.clashbot.utils.FileUtils.getImageFromFile(FileUtils.java:55)
The printed path is valid and points to the corresponding picture. However, the program raises an IOException.
Why can't it find the file?
You're jumping through way too many hoops. It's quite simple:
FileUtils.class.getResource("path.png");
// -OR-
try (var in = FileUtils.class.getResourceAsStream("path.png")) {
// in is an inputstream.
}
is all you need. Note that this means the path.png file is searched for in the exact same place (and even same 'subdir') as where FileUtils lives. So if you have, say, a file on C:\Projects\Hugo\MyApp\myapp.jar, and if you were to unzip that, inside you'd find com/foo/pkg/FileUtils.class, then the string path.png would look in that jar, and for com/foo/pkg/path.png. In other words, AnyClass.class.getResource("AnyClass.class") will let a class find its own class file. If you want to go from the 'root' of the jar, add a slash, i.e. FileUtils.class.getResource("/path.png") looks in the same jar, and for /path.png inside that jar.
getResource returns a URL. getResourceAsStream returns a stream (which you need to close; use try-with-resources as I did). Just about every resource-using API out there will take one of these two as input. For example, ImageIO does so; it even takes a URL so you can use either one:
var image = ImageIO.read(FileUtils.class.getResource("imgName + ".png"));
Yes. It's a one-liner. This will load images straight from within a jar file!
You could try to use a slightly different call like this:
java.net.URL fileUrl = Thread.currentThread().getContextClassLoader().getResource(file);
String filePath = URLDecoder.decode(fileUrl.getPath(), "UTF-8");
image = ImageIO.read(filePath);
This is the first time I've implemented sounds but I can't figure out where to actually place the sounds to play them. I am using Eclipse as my IDE and I've put my sounds in a folder called sounds.
The following code is what I've used to create one of the audioclip objects:
private final String background = "." + slash + "sounds" + slash + "background.wav";
main(....){
try {
backgroundClip = Applet.newAudioClip(new File(background).toURI().toURL());
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
I don't hear anything. When I remove the try/catch I get an error saying that it is unable to find the file. I placed my sound folder in both /src and /bin but neither can find it. where do I put it?
The base for your project is the level above the src folder. So using . will put you at your project folder.
Basically,
. = project_root
./src = default_package
./src/packagename = inside the package named "packagename"
./sounds/background.wav = a .wav file in the sounds folder, in the project_root
Using your current path, you need to put your .wav file in the sounds folder in the project_root.
The path will end up being project_root/sounds/background.wav.
See this thread to understand why application resources should be obtained by URL obtained from getResource(), rather than a File converted to an URL.
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