I just want to read a file into my program. The file is located one directory above the working directory at "../f.fsh". So the following code runs correctly when I run it in the IDE
String name="../f.fsh";
InputStream is = getClass().getResourceAsStream(name);
InputStreamReader isreader=new InputStreamReader(is);//CRASHES HERE WITH NULL POINTER EXCEPTION
BufferedReader br = new BufferedReader(isreader);
but when I create a JAR file that has f.fsh zipped inside of it and run it, it crashes when creating the InputStreamReader, because the InputStream is null.
I've read a bunch of answers to questions about input streams and JAR files, and what I got out of it is that I should be using relative paths, but I am already doing that. From what I understand getResourceAsStream() can find files relative to the root of the project, that is what I want. Why does it not work in the JAR? What is going wrong, how can I fix it?
Does it have to do with the classpath? I thought that was only for including files external to the jar being run.
I have also tried, but still fail, when putting a slash in:
InputStream is = getClass().getResourceAsStream("\\"+name);
I looked at: How to get a path to a resource in a Java JAR file andfound that contents of a JAR may not necesarily be accesible as a file. So I tried it with copying the file relative to the jar (one directory up from the jar), and that still fails. In any case I'd like to leave my files in the jar and be able to read them there. I don't know what's going wrong.
You can't use .. with Class.getResourceAsStream().
To load a resource f.fsh in the same package as the class, use SomeClass.class.getResourceAsStream("f.fsh")
To load a resource f.fsh in a sub-package foo.bar of the package of the class, use SomeClass.class.getResourceAsStream("foo/bar/f.fsh")
To load a resource f.fsh in any package com.company.foo.bar, use SomeClass.class.getResourceAsStream("/com/company/foo/bar/f.fsh")
This is described in the javadoc of the getResource() method, although it lacks examples.
If .. works in Class.getResourceAsStream() while running from Eclipse, it's a bug in Eclipse. Eclipse and other IDEs implement custom class loaders to fetch resources from the project at runtime. It looks like the class loader implementation in Eclipse isn't performing all the necessary validations on input to getResourceAsStream() method. In this case the bug is in your favor, but you will still need to rethink how you structure your resources for your code to work in all cases.
it's mandatory that the name of the file is CASE SENSITIVE
it's mandatory to refresh (F5) the project explorer if the file is moved or copied outside Exclipse
Related
I've created a file inside a project package using this code:
File xmlFile = new File("src/com/company/project/xml/tags.xml");
I am able to read the file while running from eclipse. However, after creating .jar, I'm unable to read the file. So I want to put absolute path while reading the file from the project package. How it can be done? Help and suggestions are appreciated.
In most cases, IDE's will include no Java files in the resulting Jar. Most IDE's will also include the src directory in the classpath when you run/debug the program from within them.
As a general rule of thumb, never include src in any path, src will simply not exist once the program is built.
Instead you need to make use of Class#getResource or Class#getResourceAsStream, depending on your needs. You should remember, you should never treat an "embedded" resource as a File, as in most cases it won't be, it'll be a stream of bytes in a zip file.
Something like...
URL xmlFile = getClass().getResource("/com/company/project/xml/tags.xml");
will return a URL reference to the resource. Remember, if you need a InputStream, you'll have to Class#getResourceAsStream.
If you want the resource to be writable, then you will need to find a different location to store it, as embedded resources are read only
Try with getClass().getResource()
new File(getClass().getResource("src/com/company/project/xml/tags.xml").toURI());
I have a code where i create Java Actions and try to associate Icons with them. One snapshot of code is
FileOpenCommand fileOpen = new FileOpenCommand(this);
fileOpen.putValue("ImageOnly", false);
fileOpen.putValue(Action.NAME, "Open");
fileOpen.putValue(Action.SMALL_ICON, new ImageIcon(getClass().getResource("../resources/File-Open-icon24x24.png")));
fileOpen.putValue(Action.SHORT_DESCRIPTION, "Opens the existing file.");
fileOpen.putValue("Group", "File");
fileOpen.putValue(Action.ACCELERATOR_KEY, KeyStroke.getKeyStroke(KeyEvent.VK_O, ActionEvent.CTRL_MASK));
this.commands.put("FileOpen", fileOpen);
The accent is on the line where I try to set the Action.SMALL_ICON property to the action. This works when executed in NetBeans environment either in debug or release mode. But when I've tried to execute jar file from the command line, it fails with exception.
Any idea? Anything to do with classpath? Resources folder is put as the package inside the main package.
Thanks in Advance!
I'm not entirely sure what exception is being thrown in your case, although assuming it is a NullPointerException, IOException, or IllegalArgumentException deriving from
fileOpen.putValue(Action.SMALL_ICON, new ImageIcon(getClass().getResource("../resources/File-Open-icon24x24.png")));
Your issue should be resolved simply by adding getClassLoader() between the getClass() and getResource(), like so:
fileOpen.putValue(Action.SMALL_ICON, new ImageIcon(getClass().getClassLoader().getResource("../resources/File-Open-icon24x24.png")));
Additionally, you must be exact in your filenames, specifically the extension. In this case, you are accessing File-Open-icon24x24.png, which will work perfectly fine regardless of whether the actual file is extended by png or PNG within Netbeans, but once exported the extension case matters.
Lastly, if neither of those changes resolve your problem, I would check your filepath, as there is most likely a logical error somewhere down the line.
When using embedded resources in Netbeans, you should have a resources folder containing additional folder or whatever data you need, which you seem to have, but this folder should be located inside the Netbeans project's src folder. getClass().getResource() returns the directory at the top of the package line, meaning if your class package is com.example.code, then the compiler will look for files/folder on the same level as com. Opening the Netbeans src folder you should see the initial com folder. Your resource folder should be placed directly next to that folder, as then it will be properly embedded in the jar file export.
In your code your path is ../resources/File-Open-icon24x24.png, which confuses me as to why you begin with ... I cannot see your folder structure so I cannot give a precise answer on this note, but you may be accessing the wrong location, although I feel like you are not as you said your project runs correctly within Netbeans. However, your resource files may not be correctly encoding into the jar file due to placement as mentioned. To test what your jar file actually contains, make a copy of it (for safety reasons) and change the file extension from jar to zip. You can then look through its contents in Windows Explorer, and see its directory structure. Another debugging trick for folder structures is to create a text file at the URL you are trying to access to see where it is placed.
I really need your help to solve my own problem. Now, I'm dealing with small code app. In that project folder contain some resource files (*.xlsx, *.png,...). I placed them in current folder with code file. I just wonder that when I run my code in netbean ide, it just worked find.
After I build code project, I get a jar file in "dist" directory. I run it. It open normally since app used JFrame as user interface. However, when I execute some function of that app, it showed me the error log. Here is the error message:
java.io.FileNotFoundException:
src\sample.xlsx (The system cannot find the path specified)
What's the matter out there?
Here is some pieces of my code:
copyFile(new File("src\\sample.xlsx"),
new File(txtout.getText()+"\\sample.xlsx"));
Node: copyFile function is used for copy file from source to dest.
Here is my project folder structure in Netbean IDE:
Project Name
Source Pakage(src)
myClass.java, sample.xlsx, etc
First, never reference src directly, the directory will not exist once the program is built. Second, you can not access resources which have been embedded within in the application context via a File reference, they simply no longer exist on the file system.
Instead, you need to use Class#getResource or Class#getResourceAsStream
URL url = getClass().getResource("/sample.xlsx");
InputStream is = getClass().getResourceAsStream("/sample.xlsx");
// Don't forget to manage your streams appropriately...
Well you can create a folder named resources under the src folder put your resources in it and use them in your code by using getResourceAsStream() and getResource() methods that can access the embedded resources.Clean and Build will compile the code and embed the contents of the resources folder into the application’s .jar file.
Ways of Accessing resources :
String pathToImage = "resources/images/filling.png";
InputStream stream= ClassName.class.getResourceAsStream(pathToImage );
String pathToImage = "resources/images/filling.png";
InputStream stream= ClassName.class.getResource(pathToImage );
please refer the link information
I had problems while finding the path of file(s) in Netbeans..
Problem is already solved (checked answer).
Today I noticed another problem: When project is finished,
I have to execute the generated .jar to launch the program, but it doesn't work because an error occurs: NullPointer (where to load a file) when accessing/openning jar outside Netbeans.
Is it possible to open a file with the class file in Java/Netbeans which works in Netbeans and even in any directory?
I've found already some threads about my problem in site but none was helpful.
Code:
File file = new File(URLDecoder.decode(this.getClass().getResource("file.xml").getFile(), "UTF-8"));
The problem you have is that File only refer to files on the filesystem, not files in jars.
If you want a more generic locator, use a URL which is what getResource provides. However, usually you don't need to know the location of the file, you just need its contents, in which case you can use getResourceAsInputStream()
This all assumes your class path is configured correctly.
Yes, you should be able to load a file anywhere on your file system that the java process has access to. You just need to have the path explicitly set in your getResource call.
For example:
File file = new File(URLDecoder.decode(this.getClass().getResource("C:\\foo\\bar\\file.xml").getFile(), "UTF-8"));
The project that I am currently working on utilizes an old application contained within a .jar file. One of the responsibilities of this application is that it updates the database when changes to the configuration files are made. Every time I try to run this file (which is a simple Ant Task extension), I get an exception thrown early in the process. I decompiled the Java file responsible, and found that the exception being thrown happens here. I do not know what the issue is, as "hibernate.cfg.xml" is contained within the same .jar file as the .class throwing the exception.
ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream in = loader.getResourceAsStream("hibernate.cfg.xml");
if (in == null) {
throw new RuntimeException("Couldn't find built in hibernate config");
}
If anyone has any ideas, even pointing me in the right direction, I would be grateful.
Of course, any solution will have to be external, as the client already has this build of the program in production use.
Are you loading it from the root dir and you need "/hibernate.cfg.xml" instead of just hibernate.cfg.xml?
getResourceAsStream() expects the file to be in the same package as the Class that was the origin of the ClassLoader. Is your file in the right package?
Doh. Didn't read the question fully. Please ignore this.
try
InputStream in = YourClass.class..getResourceAsStream("hibernate.cfg.xml");
this will work if the class is in the same jar as the cfg file.
edit:
if you can't rebuild the application, you will need to add the jar to the bootstrap classloader. this is very ugly.
you can do it by running the jvm with (play with the exact arguments, you may need to add rt.jar from your jre to it as well).
-Xbootclasspath your.jar
your problem is that the code is using the classloader that loaded the Thread class, which is most likely the bootstrap classloader. and that you are now in a more complex environment (app server?) that loads your jar using a different classloader. so the bootstrap classloader can't find your resource.
It is possible that the classloader cannot open files contained in the jar file. Try one of two things 1) try extracting the jar file and running it from the file system, or 2) if the jar manifest has a ClassPath entry, try extracting the hibernate.cfg.xml into a directory in the classpath and see if it runs.
Apparently the hibernate.cfg.xml file isn't located in the source root of the JAR file, but it is instead placed inside a package. You'll need to specify the package path in the resource name as well.