I have a folder called Etc which has an image I want to use in another file in the same folder, Example.java. So I have Etc\image.png and Etc\Example.java. I've tried using "Etc/image.png" as the image path, but that didn't work. How should I go about this?
Also, suppose the files were in different packages, how would I do it?
My main .java classes are in a package called Main, for the record.
EDIT:
I used this:
ClassLoader.getSystemClassLoader().getResource("Etc\image.png");
You can use Class.getResource(), which uses the class loader to obtain a URL to the resource. For example:
import java.net.URL;
import javax.swing.ImageIcon;
public class Example {
public ImageIcon getImage() {
URL url = Example.class.getResource( "image.png" );
if( url != null ) {
return new ImageIcon( url );
}
return null; // TODO: Better error handling
}
}
The important part is Example.class.getResource( "image.png" ) - the image path is specified relative to the named class; in this case, it's in the same directory as the class file. You could also use this line in any other class, leaving the reference to Example.class intact.
First things first.
Is Etc a package? In other words at the top of your Example file do you have a
package Etc;
???
Usually package names are lower case, which is why I ask.
Second, although you can use relative paths to access resources, I would recommend always using absolute paths.
So
URL url = Example.class.getResource("/Etc/image.png");
if Etc is a package, otherwise
URL url = Example.class.getResource("/image.png");
if it is not.
For that to work, you have to have the directory where Etc is in the classpath. If it is inside the jar, I don't remember if . works as a classpath, if not, add Etc to the classpath, and reference the image without the classpath, or put Etc in a sub directory. and put that sub directory in the class path.
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)
For some reason I keep getting an NPE in a gradle javafx project.
My folder structure is very basic. I have a package with my java files in the main/java folder. I also have my resources in the main/resources folder. When I try to load image.png it gives me an NPE.
public static Image createImage(Object context, String url) {
Image m = null;
InputStream str = null;
URL _url = context.getClass().getResource(url);
try {
m = new Image(_url.getPath());
} catch (NullPointerException e) {
e.printStackTrace();
}
return m;
}
This is a helper class.
From the Scene I call: Image image = Helper.createImage(this, "image.png");
The absolute path to the image would be main/resources/images/image.png.
I checked every tutorial on the internet but I couldn't find any solution for this. I also tried it with the path to the image as parameter and also with an InputStream but it never worked.
Resources
The Class#getResource(String) and related API are used for locating resources relative to the class path and/or module path. When using Class to get a resource you can pass an absolute name or a relative name. An absolute name will locate the resource relative to the root of the class path/module path; an absolute name starts with a /. A relative name will locate the resource relative to the location of the Class; a relative name does not start with a leading /.
In a typical Maven/Gradle project structure, the src/main/java and src/main/resources are roots of the class path/module path. This means all resource names are relative to those directories. It's slightly more complicated than that because the files under those directories are moved to the target/build directory and it's that location that's put on the class path/module path, but for all intents and purposes consider the source directories as the root. There's a reason a get-resource API exists in the first place, to provide an application-location-independent way of obtaining resources.
Issues in Your Code
From your question I gather your project structure looks something like:
<project-dir>
|--src/
|--main/
|--java/
|--resources/
|--images/
|--image.png
And you're calling your method with an Object and a resource name of image.png. The problem here is that, since you're passing a relative name, the resource is located relative to the Class of the passed Object (i.e. context). I doubt your class is located in a package named images which means the resource will not be found relative to said class. You need to pass an absolute name: /images/image.png.
The other problem is your use of URL#getPath(). The URL you obtain from Class#getResource(String) will, if the resource were to be found, look something like this:
file:/path/to/gradle/project/build/resources/main/images/image.png
But the result of URL#getPath() will give you:
/path/to/gradle/project/build/resources/main/images/image.png
This causes a problem due to the way Image works. From the documentation:
All URLs supported by URL can be passed to the constructor. If the passed string is not a valid URL, but a path instead, the Image is searched on the classpath in that case.
Notice the second sentence. If the passed URL does not have a scheme then it's interpreted as a resource name and the Image will locate the image file relative to the classpath. In other words, since you're passing the value of URL#getPath() to the Image constructor it searches for the resource image.png in the package path.to.gradle.project.build.resources.main.images. That package does not exist. You should be passing the URL as-is to the Image constructor via URL#toString() or URL#toExternalForm().
Solution
If you're going to use the URL returned by Class#getResource(String) to load the Image then no matter what you need to use URL#toString() or URL#toExternalForm() instead of URL#getPath().
public static Image createImage(Object context, String resourceName) {
URL _url = context.getClass().getResource(resourceName);
return new Image(_url.toExternalForm());
}
Then you have at least two options:
Pass the absolute resource name (i.e. "/images/image.png") to your #createImage(Object,String) method since the image.png resource is not in the same package as the passed Object (i.e. context).
Move the resource to the same package as the class of the passed in Object (i.e. context). For instance, if the context object's class is com.foo.MyObject then place the resource under src/main/resources/com/foo and it will be in the same package as MyObject. This will allow you to continue passing the relative resource name.
Of course, as noted by the documentation of Image you can pass a scheme-less URL and it's interpreted as a resource name. In other words, you could do:
Image image = new Image("images/image.png");
And that should work. A note of caution, however: When using modules the above will only work if the resource-containing package is opens unconditionally or if the module itself is open.
Try using the path /images/image.png.
The resources always get referenced from the class root, in your case src/main/resources, so from there going to /images/image.png should be the correct path.
this is how I am passing the images in my application. ivSerialAssignmentLogo is a FXML element (ImageView).
ivSerialAssignmentLogo.setImage(new Image(getClass().getResourceAsStream("/img/serialAssignment.svg")));
In your case, you could use something like that
public static Image createImage(Object context, String url) {
Image m = null;
InputStream str = null;
URL _url = context.getClass().getResource("/images/" + url);
try {
m = new Image(_url.getPath());
} catch (NullPointerException e) {
e.printStackTrace();
}
return m;
}
I'm trying to find a way to get the absolute path for the resources folder in my junit test.
My doubt is not get a file in the resources folder (I now how to do this) but get the path to the resources folder.
Per example, if my file in my resources folder have this structure:
/opt/project/view/api/target/classes/
/opt/project/view/api/target/classes/file_in_resources.txt
I would like to get the location /opt/project/view/api/target/classes/ without use something ugly like:
URL location = this.getClass().getResource("/file_in_resources.txt");
String path = location.getPath();
String rightPath = path.substring(0, path.lastIndexOf("/"));
The idea is have a method like this.getClass().getResourcesPath(), but this kind of method not exists.
How to do?
There could be a long philosophical discussion about the pro´s and con´s, if one should "have this question". But if you bump into it too like me, than this could be a way to go, if you want to do it with a current Java style (assumption: you´re file_in_resources.txt resides inside the appropriate and Unittest standard folder src/test/resources):
private String readFileFromResourcesAsString(String fileName) throws IOException {
Path filePath = Paths.get("./src/test/resources/yourCustomPathHere/" + fileName);
return Files.lines(filePath).collect(Collectors.joining());
}
If you really want to stick to the folder view/api/target/classes/, you could use that Paths.get-statement with a java.net.URI instead (like Chetan Jadhav CD already mentioned):
Paths.get(this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI());
This solution makes use of the more recent Java-APIs around NIO.2 (see oracle tutorials here) like
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
and some Java 8 stream API part to read the Files content into a String (you could change this part as you like :) ).
The
Paths.get(".")
is key here - it gives you the current working directory transparently whatever OS (Windows, Linux, MacOS) you´re using.
If you´re a Spring-User, the whole thing is even more easy! See so answer here.
If your test class is located within the same resources folder, you could try:
URL location = TestClass.class.getProtectionDomain().getCodeSource().getLocation();
String path = location.getPath();
String rightPath = path.substring(0, path.lastIndexOf("/"));
If the resources folder isn't the same, you could navigate to it based on its relative location taking the location of the Test Class (obtained using the code above) as a reference point.
String adjustedPath = rightPath+"/resources";
It looks Spring ClassUtils provides such method.
http://docs.spring.io/spring/docs/current/javadoc-api/org/springframework/util/ClassUtils.html#classPackageAsResourcePath-java.lang.Class-
If you are in the context of a unit test, then it is no problem referencing a file directly, as you know where it is and it is not going anywhere. You can get the file url, then use Path to navigate upwards in the folder structure until you have the correct folder.
final URL location = this.getClass().getResource("/file_in_resources.txt");
final Path path = Paths.get(location.getPath());
final String folder = path.getParent() //getParent().getParent()...
.toString();
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)
I have this issue of accessing a file in one of the parent directories.
To explain, consider the following dir structure:-
C:/Workspace/Appl/src/org/abc/bm/TestFile.xml
C:/Workspace/Appl/src/org/abc/bm/tests/CheckTest.java
In the CheckTest.java I want to create a File instance for the TestFile.xml
public class Check {
public void checkMethod() {
File f = new File({filePath value I want to determine}, "TestFile.xml");
}
}
I tried a few things with getAbsolutePath() and the getParent() etc but was getting a bit complicated and frankly I think I messed it up.
The reason I don't want to use "C:/Workspace/Appl/src/org/abc/bm" while creating the File instance is because the C:/Workspace/Appl is not fixed and in all circumstances will be different at runtime and basically I don't want to hard-code.
What could be the easiest and cleaner way to achieve this ?
Thank you.
You should load it from Classpath in this case.
In your CheckTest.java, try
FileInputStream fileIs = new FileInputStream(CheckTest.class.getClassLoader().getResourceAsStream("org/abc/bm/TestFile.xml");
Use System.getProperty to get the base dir or you set the base.dir during application launch
java -Dbase.dir=c:\User\pkg
System.getProperty("base.dir");
and use
System.getProperty("file.separator");
What could be the easiest and cleaner way to achieve this ?
For accessing static resources use:
URL urlToResource = this.getClasS().getResource("path/to/the.resource");
If the resource is expected to change, write it to a sub-directory of user.home, where it is easy to locate later.
First of all, you can't get a reference to the source file path on runtime.
But, you can access the resrources included at your classpath (where you complied .class files will be).
Normally, your compiler will copy the xml file included at your srouce directory into the build directory, so at last, you could end up having something like this:
C:/Workspace/Appl/classes/org/abc/bm/TestFile.xml
C:/Workspace/Appl/classes/org/abc/bm/tests/CheckTest.class
Then, with your classpath pointing to the compiled classes root dir, you get the resources from this directory, using the ClassLoader.getResource method (or the equivalent Class.getResource() method).
public class Check {
public void checkMethod() {
java.net.URL fileURL=this.getClass().getResource("/org/abc/bm/tests/TestFile.xml");
File f=new File( fileURL.toURI());
}
}
One could do this:
String pathOfTheCurrentClass = this.getClass().getResource(".").getPath();
File file = new File(pathOfTheCurrentClass + "/..", "Testfile.xml");
or
String pathOfTheCurrentClass = this.getClass().getResource(".").getPath();
File filePath = new File(pathOfTheCurrentClass);
File file = new File(filePath.getParent(), "Testfile.xml");
But as Tomas Naros points out this gives you the file located in the build path.
Did you try
URL some=Test.class.getClass().getClassLoader().getResource("org/abc/bm/TestFile.xml");
File file = new File(some.getFile());