Having researched this issue extensively here & elsewhere, without finding a working solution, I thought I'd ask...
I have a jar file (deployed on a RaspberryPi), with an internal structure like this:
myApp
MyClass
....
textFiles
foo.txt
....
I need 'MyClass' to read 'foo.txt'.
the general advice here & elsewhere is to use something like the following:
InputStream in = getClass().getResourceAsStream("../textFiles/foo.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
I have also read that the path to use (as the param for getResourceAsStream()) is the path to the target file, relative to the location of the class reading the file.(..?)
However, regardless of the path I use, I cannot get the above 2 lines to work. I always get an NPE thrown by the 2nd line.
I'm assuming that the NPE indicates that 'in' is null because 'foo.txt' has not been found.
any advice leading to a successful resolution, gratefully received.
cheers
Paul
Try removing the .. from the path, as per documentation:
If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'.
Otherwise, the absolute name is of the following form:
modified_package_name/name
Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').
So my immediate problem has been solved, but not to my satisfaction.
I relocated the contents of 'textFiles' (in the Jar), so all the text files were in the Jar's root. I then successfully managed to find & load the file using:
InputStream in = getClass().getResourceAsStream("/foo.txt");
However, I still don't undserstand why this worked. I don't see why "/textFiles/foo.txt" didn't work, when the files were in that subdir, off the Jar's root...?
Anyway...I have a new problem now, which I think is hardware related....it takes about 40 seconds to read the files!! I think I blame the old SD card in my Raspberry Pi.
But that's a problem for another day!
Thanks for your input.
Related
I found many similar questions on this topic in this forum, but none of the solutions of those questions working for me and this problem is really making me frustrated.
I have the following method which should play a wav file when I call it.
Directory of the wav file is: ProjectFolder/src/resources/Sounds/click.wav
public static void Click()
{
String clickSound = "/resources/Sounds/click.wav";
Media hit = new Media(new File(clickSound).toPath().toString());
MediaPlayer mediaPlayer = new MediaPlayer(hit);
mediaPlayer.play();
}
But it results in the following exception:
Exception in thread "AWT-EventQueue-0" java.lang.IllegalArgumentException: java.net.URISyntaxException: Illegal character in path at index 0: \resources\Sounds\click.wav
at javafx.scene.media.Media.<init>(Media.java:385)
When I set the value of the string clickSound "resources/Sounds/click.wav" instead of "/resources/Sounds/click.wav", the exception says illegal character in path at index 9.
So I am guessing that it is considering '/' character as an illegal character. I tried using '\' instead of '/', but the result was same.
I do not want to change the location of the wav file for certain reason. How can access I access that wav file from ProjectFolder/src/resources/Sounds/click.wav and play it without any exception?
Any kind of help is appreciated.
Thanks in advance.
As stated in the documentation:
The source must represent a valid URI and is immutable. Only HTTP, HTTPS, FILE, and JAR URLs are supported. If the provided URL is invalid then an exception will be thrown.
You are passing in the path to a file, instead of a URI.
You almost certainly don't want a file here anyway; for example, when you deploy your application it will typically be deployed as a jar file, and the media will be an entry in that jar file (so it won't be a file in its own right at all). On top of that, the resources folder is typically part of the source code structure, and for obvious reasons the source code is not usually available at runtime.
Assuming your resources are being deployed to the root of the classpath, which is the usual setup, you need something like
String clickSound = "/Sounds/click.wav";
Media hit = new Media(getClass().getResource(clickSound).toExternalForm());
You listed the file as in directory "ProjectFolder/src/resource/Sounds/click.wav" but when trying to access you have "resources" instead of "resource"
I need to get a resource image file in a java project. What I'm doing is:
URL url = TestGameTable.class.getClass().
getClassLoader().getResource("unibo.lsb.res/dice.jpg");
The directory structure is the following:
unibo/
lsb/
res/
dice.jpg
test/
..../ /* other packages */
The fact is that I always get as the file doesn't exist. I have tried many different paths, but I couldn't solve the issue.
Any hint?
TestGameTable.class.getResource("/unibo/lsb/res/dice.jpg");
leading slash to denote the root of the classpath
slashes instead of dots in the path
you can call getResource() directly on the class.
Instead of explicitly writing the class name you could use
this.getClass().getResource("/unibo/lsb/res/dice.jpg");
if you are calling from static method, use :
TestGameTable.class.getClassLoader().getResource("dice.jpg");
One thing to keep in mind is that the relevant path here is the path relative to the file system location of your class... in your case TestGameTable.class. It is not related to the location of the TestGameTable.java file.
I left a more detailed answer here... where is resource actually located
The code I am using to load the image is:
ImageIO.read(SpriteSheet.class.getResource(path));
The path being the path to the resource. But it would error with IllegalArgumentException. I wondered what might be causing and came to the conclusion that the resource should be added into the same path as the class.
Is it possible to load the image from another folder, like a res folder outside of the bin folder? (folder holding compiled classes)
EDIT:
So i messed around with a few things, and came to a solution. But now I have another problem. Here is my code
File sheet = new File(SpriteSheet.class.getProtectionDomain().getCodeSource().getLocation().getPath());
URI uri = sheet.toURI();
BufferedImage image = ImageIO.read(uri.toURL());
When I try to run it, it gives me an IIOException: Can't read Input File
This means that I can never actually get it work. I tried debugging by prining the URL to the console and this is the URL.
C:\Users\Amma\Abhijeet\Eclipse%20Workspace1\Test%20Game\bin
The %20 comes in the middle. Meaning that the file is and never can be acceesed. Is there anyway I can fix this?
Thanks.
Class.getResource will return null if the resource could not be found or the invoker doesn't have adequate privileges to get the resource.
All variants of ImageIO.read will throw an IllegalArgumentException if they receive a null input.
Take a look at the documentation of the getResource to understand how an absolute resource name is constructed from the given resource named and what are the rules for searching resources.
You can read images from any location as long as you have permissions to do so, the ImageIO.read method accepts a File, URL or InputStream so you have many option to do it.
I am trying to retrieve a jrxml file in a relative path using the following java code:
String jasperFileName = "/web/WEB-INF/reports/MemberOrderListReport.jrxml";
File report = new File(jasperFileName);
FileInputStream fis = new FileInputStream(report);
However, most probably I didn't succeed in defining the relative path and get an java.io.FileNotFoundException: error during the execution.
Since I am not so experienced in Java I/O operations, I didn't solve my problem. Any helps or ideas are welcomed.
You're trying to treat the jrxml file as an object on the file-system, but that's not applicable inside a web application.
You don't know how or where your application will be deployed, so you can't point a File at it.
Instead you want to use getResourceAsStream from the ServletContext. Something like:
String resourceName = "/WEB-INF/reports/MemberOrderListReport.jrxml"
InputStream is = getServletContext().getResourceAsStream(resourceName);
is what you're after.
You should place 'MemberOrderListReport.jrxml' in classpath, such as it being included in a jar placed in web-inf\lib or as a file in web-inf\classes.
The you can read the file using the following code:
InputStream is=YourClass.class.getClassLoader().getResourceAsStream("MemberOrderListReport.jrxml");
String jasperFileName = "/web/WEB-INF/reports/MemberOrderListReport.jrxml";
Simple. You don't have a /web/WEB-INF/reports/MemoberOrderListReport.jrxml file on your computer.
You are clearly executing in a web-app environment and expecting the system to automatically resolve that in the context of the web-app container. It doesn't. That's what getRealPath() and friends are for.
check that your relative base path is that one you think is:
File f = new File("test.txt");
System.out.println(f.getAbsoluteFile());
I've seen this kind of problem many times, and the answer is always the same...
The problem is the file path isn't what you think it is. To figure it out, simply add this line after creating the File:
System.out.println(report.getAbsolutePath());
Look at the output and you immediately see what the problem is.
Hallo,
i have following line of code:
InputStream passoloExportFileInputStream = getClass().getClassLoader().getResourceAsStream("/com/thinkplexx/lang/de/general.xml");
and i know that jar with com/thinkplexx/lang/de/general.xml is in classpath.
It worked under "previous environment", which is maven2 build.
Now, i evaluate maven3 and it doesn't work! I know, that if i change the code to be:
InputStream passoloExportFileInputStream = getClass().getClassLoader().getResourceAsStream("com/thinkplexx/lang/de/general.xml");
it works great (i just removed the first slash from the resource path).
Btw, i use Linux. First slash in path normally means "from the root directory", so if this logic is sound for java resource loading as well, first example should never have worked!?
Questions: is something wrong with the first code sample, i.e. with /com/ and not com/? Is it just bad code or it means something different?
thank you!
It depends on how you are getting the resource. When you use a ClassLoader as in:
InputStream stream= getClass().getClassLoader().getResourceAsStream("/com/thinkplexx/lang/de/general.xml");
The leading '/' is meaningless. So, the correct form is "com/thinkplexx/lang/de/general.xml".
If, instead you use a 'Class', as in:
InputStream stream= getClass().getResourceAsStream("/com/thinkplexx/lang/de/general.xml");
You get a different behavior. The Class.getResourceAsStream will consider classes without a leading '.' to be relative to the package containing the class. Resources specified with a leading '.' are absolute, or resolved relative to the root of the jar.
So, if this is a reference to com.example.SomeThing, then the expected behavior is:
getClass().getResourceAsStream("/a/b/c.xml") ==> a/b/c.xml
getClass().getResourceAsStream("a/b/c.xml") ==> com/example/a/b/c.xml
getClass().getClassLoader().getResourceAsStream("a/b/c.xml") ==> a/b/c.xml
getClass().getClassLoader().getResourceAsStream("/a/b/c.xml") ==> Incorrect
Maven2 was being lax and allowing the last form.