I am trying to load an image to display on the screen (just to get an idea of how to do it).
The problem is that when the program tries to load "apple.png" (which is saved on my desktop), it cannot find the image - Where do image files need to be stored in order for them to be found? Here is my loading method:
private void loadImage() {
ImageIcon appleIcon = new ImageIcon("apple.png");
Image appleImage = appleIcon.getImage();
}
If you want to reach it from the desktop, you should use the complete path. The easiest way to handle resources would be to create a folder in you java project, which you can access via "folderName/fileName.example".
Using the full path is an option
C:\filefolder\file.jpg
To answer your question though
Wherever your java file is is where it's going to think "home is" make yourself a java workspace to put your java source files in and inside it create a folder to put any assets you want in that way you will simply be able to call "apple.jpg"
if you want use the file name only, the file path ("apple.png")is relative to the source folder. so you have to place it in there.
you could also use absolute the file path to your desktop (sthg like "C:\Users\your.name\Desktop")
From the javadoc:
Creates an ImageIcon from the specified file. The image will be preloaded by using MediaTracker to monitor the loading state of the image. The specified String can be a file name or a file path. When specifying a path, use the Internet-standard forward-slash ("/") as a separator. (The string is converted to an URL, so the forward-slash works on all systems.) For example, specify:
new ImageIcon("images/myImage.gif")
As #Shriram mentioned, when you only specify the filename (with extension), it will search for that file in the current directory.
Hint: There exists an constructor overload which takes an URL as argument.
Related
I am trying to load an image to use as an icon in my application. The appropriate method according to this tutorial is:
protected ImageIcon createImageIcon(String path, String description)
{
java.net.URL imgURL = getClass().getResource(path);
if (imgURL != null) {
return new ImageIcon(imgURL, description);
} else {
System.err.println("Couldn't find file: " + path);
return null;
}
}
So, I placed the location of the file, and passed it as a parameter to this function. This didn't work, i.e. imgURL was null. When I tried creating the ImageIcon by passing in the path explicitly:
ImageIcon icon = new ImageIcon(path,"My Icon Image");
It worked great! So the application can pick up the image from an explicitly defined path, but didn't pick up the image using getResources(). In both cases, the value of the path variable is the same. Why wouldn't it work? How are resources found by the class loader?
Thanks.
getClass().getResource(path) loads resources from the classpath, not from a filesystem path.
You can request a path in this format:
/package/path/to/the/resource.ext
Even the bytes for creating the classes in memory are found this way:
my.Class -> /my/Class.class
and getResource will give you a URL which can be used to retrieve an InputStream.
But... I'd recommend using directly getClass().getResourceAsStream(...) with the same argument, because it returns directly the InputStream and don't have to worry about creating a (probably complex) URL object that has to know how to create the InputStream.
In short: try using getResourceAsStream and some constructor of ImageIcon that uses an InputStream as an argument.
Classloaders
Be careful if your app has many classloaders. If you have a simple standalone application (no servers or complex things) you shouldn't worry. I don't think it's the case provided ImageIcon was capable of finding it.
Edit: classpath
getResource is—as mattb says—for loading resources from the classpath (from your .jar or classpath directory). If you are bundling an app it's nice to have altogether, so you could include the icon file inside the jar of your app and obtain it this way.
As a noobie I was confused by this until I realized that the so called "path" is the path relative to the MyClass.class file in the file system and not the MyClass.java file. My IDE copies the resources (like xx.jpg, xx.xml) to a directory local to the MyClass.class. For example, inside a pkg directory called "target/classes/pkg. The class-file location may be different for different IDE's and depending on how the build is structured for your application. You should first explore the file system and find the location of the MyClass.class file and the copied location of the associated resource you are seeking to extract. Then determine the path relative to the MyClass.class file and write that as a string value with "dots" and "slashes".
For example, here is how I make an app1.fxml file available to my javafx application where the relevant "MyClass.class" is implicitly "Main.class". The Main.java file is where this line of resource-calling code is contained. In my specific case the resources are copied to a location at the same level as the enclosing package folder. That is: /target/classes/pkg/Main.class and /target/classes/app1.fxml. So paraphrasing...the relative reference "../app1.fxml" is "start from Main.class, go up one directory level, now you can see the resource".
FXMLLoader loader = new FXMLLoader();
loader.setLocation(getClass().getResource("../app1.fxml"));
Note that in this relative-path string "../app1.fxml", the first two dots reference the directory enclosing Main.class and the single "." indicates a file extension to follow. After these details become second nature, you will forget why it was confusing.
getResource by example:
package szb.testGetResource;
public class TestGetResource {
private void testIt() {
System.out.println("test1: "+TestGetResource.class.getResource("test.css"));
System.out.println("test2: "+getClass().getResource("test.css"));
}
public static void main(String[] args) {
new TestGetResource().testIt();
}
}
output:
test1: file:/home/szb/projects/test/bin/szb/testGetResource/test.css
test2: file:/home/szb/projects/test/bin/szb/testGetResource/test.css
getResourceAsStream() look inside of your resource folder. So the fil shold be placed inside of the defined resource-folder
i.e if the file reside in /src/main/resources/properties --> then the path should be /properties/yourFilename.
getClass.getResourceAsStream(/properties/yourFilename)
I am creating a program that applies a filter to an image and afterwards gives the user the option of darkening the image or making it lighter. Before the image is filtered the user picks a save location for the new image. Is there anyway I can get the filepath of that location they choose to save it to? I am using JFileChooser to accomplish this.
JFileChooser#getSelectedFile returns a java.io.File of the selected file (or null if they didn't pick something).
Have a look at How to Use File Choosers for more details
You could use File#getParentFile to return just the path element (without the file name) of the File object if all you want is the path, otherwise you have an abstract representation of the File which the user selected...
If you're using JFileChooser you can use getSelectedFile() and getPath() or getAbsolutePath()
I've built a Java application that loads an image at runtime. The location of the image is fixed relative to the project.
I would like to be able to run the program from both within Eclipse and the command line and for it to load the image correctly. However, I can only do one or the other but not both. This seems like such a trivial thing to want to do but I can't find out how to do it.
The project is set up so that it creates a bin directory for the output and puts the image in a resources sub-folder. This is fine when running from the command line as I can write my code to look in that sub folder for the file.
But when I run the program from within eclipse the current working directory is different.
What am I missing?
TIA
Update - adding some code
This is what I had originally:
BufferedImage awtImage = ImageIO.read(new File(System.getProperty("user.dir") + "/resources/image-name.png"));
Following the advice in the comments I am trying to use getResourceAsStream but I don't know what to pass to the File constructor.
InputStream temp = MyClass.class.getResourceAsStream("resources/image-name.png");
BufferedImage awtImage = ImageIO.read(new File(???));
The resource is being found because temp is not null.
I think there's 2 solutions.
1) you specify an absolute path
2) your image is in the classpath so you could load it via :
YouClass.class.getResourceAsStream("YourImg.png");
The working directory, if that's really what you mean, is not a great place to load an image from. It appears that you have an image that you would distribute with your finished program so that the program could use it. In that case, I suggest that you use Class.getResourceAsStream(), and put the image in the directory with (or near) that class.
EDIT:
Here is code I used in one of my programs for a similar purpose:
ImageIcon expandedIcon = null;
// ...
expandedIcon = new ImageIcon(TreeIcon.class.getResource("images/Expanded.png"));
The ImageIcon class is part of Swing; I don't know if you're using that, but this should serve to show you the idea. The getResource() method takes a URL; again, you might need something a little different. But this shows the pathname relative to the path of the class on which the method is called, so if TreeIcon is in x/y/z/icons, the PNG file needs to be in x/y/z/icons/images, wherever that is on that computer.
TreeIcon is a class of mine, and its internals will not help you, so I'm not posting them. All it's doing here is providing a location for the PNG file I'm loading into an ImageIcon instance.
In addition to working on a disk with a directory structure, this also works in a jar file (which is a common way to distribute a java program or library). The jar file is just a zip file, and each file in the jar/zip file has its directory associated with it, so the image can be in the jar in the correct directory just as the java classes are in their directories.
getResourceAsStream() returns a stream; if you want to use that byte stream to load as an image, find a class that converts an stream to something your image class can use as a constructor or in a load method and hook them up. This is a common thing to have to figure out with Java i/o, unfortunately there is no cookbook way to do it across all images and situations, so we can't just tell you what it is.
EDIT 2:
As from the comment, try:
ImageIO.read(new File(MyClass.class.getResource("resources/image-name.png");
I set up my Eclipse projects like this.
The input directory is added to the classpath (JavaBuildPath in Eclipse).
Finally, you access the image and / or text files like this.
private BufferedImage getIconImage() {
try {
return ImageIO.read(getClass().getResourceAsStream(
"/StockMarket.png"));
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
I'm trying to make a Jframe class that has buttons with icons
public ImageIcon Flag = new ImageIcon(getClass().getResource("resources/Flag.png"));
Gives me a null pointer exception
"C:\Users\Khalidi\Documents\NetBeansProjects\MineSweeper\resources\Flag.png"
Is the full directory to reach the image I want
I created the folder resources to house the images I need
Where should i place the images? And what line of code should i be writing?
Thanks in advance
Try using the package path with the same method.
for example, if my image is in org.nisheeth.resources package, it should be written as follows.
public ImageIcon flag = new ImageIcon(Test.class.getResource("/org/nisheeth/resource/Flag.png"));
When you're loading your images using getClass().getResource(), the JVM in fact uses your CLASSPATH to load the resource (you should take a close look at Class#getResource(...)).
As a consequence, to have your resource available, you must push it to your project output folder (must be something like target, no ?).
As you asked for conventions: using the Maven build infrastructure, you would have
src/main/java - java packages
src/main/resources - resources like your images
src/main/resources/images/Flag.png - some resource (no convention)
In your created jar / .classes
/images/Flag.png
public ImageIcon Flag = new ImageIcon(getClass().getResource("/images/Flag.png"));
Here, Class.getResource() uses a relative path to the package of the class; therefore, using an absolute path as above ("/...") is easiest.
You can simply place a folder resources to the folder with compiled classes and make ImageIcon like this.
public ImageIcon Flag = new ImageIcon("resources/Flag.png");
Create a package say ImgPack and paste all your images to this package and access all the images using
ImageIcon ii = new ImageIcon(System.getProperty("user.dir")+"\\src\\ImgPack\\<Imagename.extention>");
When I create ImageIcon class objects I use the following code:
iconX = new ImageIcon (getClass().getResource("imageX.png"))
The above code works correctly either in an applet or a desktop app when the .png is in the same folder of the class.
The question is: how to avoid a NullPointerException when the .Png is in another folder? Or how load the image in the object ImageIcon when it is in a different location to the class?
I don't understand how this method works, if anyone can help me I appreciate it. Thanks!!
Take a look at this - Class#getResource(java.lang.String)
Please click the link above and read the docs and follow to understand what's going on.
It says -
If the name begins with a '/', then the absolute name of the resource is the portion of the name following the '/'.
and
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 '.'.
So, if this object (where you call getResource) is in package /pkg1 (/ meaning pkg1 is right under the root of the classpath) and you used "imageX.png" then the result would be pkg1/imageX.png which is correct because that's where the image is located at.
But, if we moved the resource (imageX.png) to some other package /pkg2 and you called the method same way then the result would still be pkg1/imageX.png but this time it would be incorrect because the resource is actually located in /pkg2. That's when you end up with NPE.
It's good to explicitly specify the full path of the resource starting from the root of the classpath. (e.g. "/pkg/imageX.png").
Hope this helps.
Simply supply the path to the resource.
So, if you put the image in "/resources/images" within your Jar, you would simply use
iconX = new ImageIcon(getClass().getResource("/resources/images/imageX.png"))
Essentially what you're saying is, class loader, please search your class path for the following resource.
If the image is internal (you want a location relative to your project, or perhaps packaged into your jar), do what mad programmer said:
iconX = new ImageIcon(getClass().getResource("/path/imageX.png"))
The path is relative, so path/ will be a folder in the same folder as your project (or packaged into your jar).
If you want an external image, simply hand ImageIcon constructor the path (ex. "C:/.../file.png"). This isn't recommended though, as it's better to use it as a resource.
For more info on the ImageIcon constructor, see here. for more info on loading class resources, see here (Javadoc links)