Every time I run the exported .jar file, that contains a JFrame with an image as its icon, the file doesn't run, unless I extract the file. In the compiler it is running. I dont want to make a launcher that saves both, the resources package and the jar file, in a directory.
"Why my Jar doesn't run unless I extract files?"
This seems to be the behavior of using File to your resources. Take for example
File file = new File("resources/image.png");
Image image = ImagIO.read(file);
And you project structure (Note the resources should actually be in the src, so that it builds into the jar automatically - unless you configure it differently. But for the sake of this argument, let's say you do confgigure it where resources is built to the jar)
C:\
Project
resources\image.png
Some examination:
Run from IDE - WORKS! Why? Using File looks for files on the file system. Using a relative path, the search will begin from the "working directory", which in the case of the IDE in generally the project root. So "resources/image.png" is a valid path, relative to ProjectRoot
Build jar, say it ends up in a dist dir in the project. This is what it looks like
ProjectRoot
dist
ProjectRoot.jar
Now for the sake of this argument (and is actually the correct way), let's try and print the URL of the resource in out program, so that when you run the jar, it prints out the URL of the file
URL url = Test.class.getResource("/resources/image.png");
System.out.println(url.toString());
When we run the jar C:\ProjectRoot\dist> java -jar ProjectRoot.jar We will see the print out C:\ProjectRoot\dist\ProjectRoot.jar!\resources\image.png. You can obviously see even though the current working directory is the location of the jar, the paths no longer match, with the added jar ProjectRoot.jar! location.
So why does it work when we extract it. Well when you extract it, then the path is correct
C:\ProjectRoot
dist
resources/image.png // from extracted jar
ProjectRoot.jar
When you run from the C:\ProjectRoot\dist >, the resource dir is where is should be.
Ok enough with the explanation.
For this reason, when you want to read embedded resources, they should be read from an URL as Andrew Thompson mentioned. This url should be relative to the class calling it, or the class loader. Here are a couple different ways:
As shown already
URL url = getClass().getResource("/resources/image.png");
Notice the /. This will bring us to the root of the classpath, where the resources dir will be. URL can be passed to many constructors, like ImageIcon(URL) or `ImageI.read(URL)
You can use:
InputStream is = getClass().getResourceAsStream("/resources/image.png");
Which will use an URL under the hood. You can use InputStream with many constructors also.
There's also ways to use the class loader, which will start at the root, so you don't need the /
URL url = getClass().getClassLoader().getResource("resources/image.png");
So there are a few ways you can go about it. But in general, reading File with hard coded string paths is never a good idea, when using embedded resources. It's possible to obtain the path dynamically so you can use File, but you will still need to use one of the aforementioned techniques, which unless you really need a File would be pointless, as you can do what you need with the InputStream or URL
To make a long story short
This would work
ProjectRoot
src\resources\image.png
URL url = getClass().getResource("/resources/image.png");
Image image = ImageIO.read(url);
Related
I have a maven project with these standard directory structures:
src/main/java
src/main/java/pdf/Pdf.java
src/test/resources
src/test/resources/files/x.pdf
In my Pdf.java,
File file = new File("../../../test/resources/files/x.pdf");
Why does it report "No such file or dirctory"? The relative path should work. Right?
Relative paths work relative to the current working directory. Maven does not set it, so it is inherited from whatever value it had in the Java process your code is executing in.
The only reliable way is to figure it out in your own code. Depending on how you do things, there are several ways to do so. See How to get the real path of Java application at runtime? for suggestions. You are most likely looking at this.getClass().getProtectionDomain().getCodeSource().getLocation() and then you know where the class file is and can navigate relative to that.
Why does it report "No such file or dirctory"? The relative path should work. Right?
wrong.
Your classes are compiled to $PROJECT_ROOT/target/classes
and your resources are copied to the same folder keeping their relative paths below src/main/resources.
The file will be located relative to the classpath of which the root is $PROJECT_ROOT/target/classes. Therefore you have to write in your Pdf.java:
File file = new File("/files/x.pdf");
Your relative path will be evaluated from the projects current working directory which is $PROJECT_ROOT (AFAIR).
But it does not matter because you want that to work in your final application and not only in your build environment. Therefore you should access the file with getClass().getResource("/path/to/file/within/classpath") which searches the file in the class path of which the root is $PROJECT_ROOT/target/classes.
No the way you are referencing the files is according to your file system. Java knows about the classpath not the file system if you want to reference something like that you have to use the fully qualified name of the file.
Also I do not know if File constructor works with the classpath since it's an abstraction to manage the file system it will depend where the application is run from. Say it is run from the target directory at the same level as source in that case you have to go one directory up and then on src then test the resources the files and finally in x.pdf.
Since you are using a resources folder I think you want the file to be on the classpath and then you can load a resource with:
InputStream in = this.getClass().getClassLoader()
.getResourceAsStream("<path in classpath>");
Then you can create a FileInputStream or something to wrap around the file. Otherwise use the fully qualiefied name and put it somewere like /home/{user}/files/x.pdf.
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 have a project that uses the normal Maven structure, e.g.
module
\ src
\ main
- java
- resources
\ test
- java
- resources
etc. Under test/resources, I'd like to keep a set of test input files for a parser I'm writing, then run all files in the directory through the test suite. As written now, the test code works from the command line, but fails when run through the Eclipse JUnit plugin:
File file = new File("src/test/resources");
file.list();
(I'm actually using a FilenameFilter, but I'm trying to simplify.)
The problem, after poking through the unit test with a debugger, turns out to be that the File I'm constructing points to /path/to/workspace/myproj/src/test/resources, whereas the actual files reside in /path/to/workspace/myproj/modulename/src/test/resources (it's a Maven multi-module project). Apparently, this isn't a problem when running mvn test from the command line.
I guess my question is two-fold: one, am I doing this wrong? I see a lot of people using the class loader to discover resources, as in this question, but I don't want all the resources of a particular type, just one directory under test/resources. Two, if this isn't a terrible idea in the first place, do I have a configuration error (e.g. it "should" work)? Is it Eclipse's fault, a Maven problem, or what?
One trick would be to place a file in resources with a known name, get the URI of this file through the classloader, then construct a File from this URI, then get the parent, and list() the contents of that directory. Kind of a hack, but it should work.
So here's what the code should look like, but place a file called MY_TEST_FILE (or whatever) in test/src/resources
URL myTestURL = ClassLoader.getSystemResource("MY_TEST_FILE");
File myFile = new File(myTestURL.toURI());
File myTestDir = myFile.getParentFile();
Then you have access to the directory you're looking for.
That said, I'd be surprised if there's not a more 'maven-y' way to do it..
Just for completeness, wanted to point out the way to get this without having to grab the current instance of the ClassLoader, using ClassLoader#getSystemResource. This example does the work without having to place a file at the top.
//Obtains the folder of /src/test/resources
URL url = ClassLoader.getSystemResource("");
File folder = new File(url.toURI());
//List contents...
Try this?
1)put test data files into the same package structure as you test classes. That is, if you have a test class named Apple in src/test/java/com/fruits, you test data file will be in src/resources/java/com/fruits.
2) When the files are compiled both the class and the data file should be in target/test-classes/com/fruits. If this is the case, in you code, you can obtain the file this way "this.getClass().getResourceAsStream("myFile")"
put desired resource into /src/test/resources/lipsum.pdf
find it's full path using
String fileName = ClassLoader.getSystemResource("lipsum.pdf").getFile();