I need to read a file in my code. It physically resides here:
C:\eclipseWorkspace\ProjectA\src\com\company\somePackage\MyFile.txt
I've put it in a source package so that when I create a runnable jar file (Export->Runnable JAR file) it gets included in the jar. Originally I had it in the project root (and also tried a normal sub folder), but the export wasn't including it in the jar.
If in my code I do:
File myFile = new File("com\\company\\somePackage\\MyFile.txt");
the jar file correctly locates the file, but running locally (Run As->Java Main application) throws a file not found exception because it expects it to be:
File myFile = new File("src\\com\\company\\somePackage\\MyFile.txt");
But this fails in my jar file. So my question is, how do I make this concept work for both running locally and in my jar file?
Use ClassLoader.getResourceAsStream or Class.getResourceAsStream. The main difference between the two is that the ClassLoader version always uses an "absolute" path (within the jar file or whatever) whereas the Class version is relative to the class itself, unless you prefix the path with /.
So if you have a class com.company.somePackage.SomeClass and com.company.other.AnyClass (within the same classloader as the resource) you could use:
SomeClass.class.getResourceAsStream("MyFile.txt")
or
AnyClass.class.getClassLoader()
.getResourceAsStream("com/company/somePackage/MyFile.txt");
or
AnyClass.class.getResourceAsStream("/com/company/somePackage/MyFile.txt");
If I have placed i file in a jar file, it only worked if and only if I used
...getResourceAsStream("com/company/somePackage/MyFile.txt")
If I used a File object it never worked. I got also the FileNotFound exception. Now, I stay with the InputStream object.
Related
I have a project with a folder "src/main/resources" where inside there is the hibernate configuration file, I load it using this line of code
HibernateUtil.class.getResource("/hibernate.cgf.xml").getPath()
From inside the IDE it is working well, but when I create the jar it doesn't file the file.
How can I load it properly in the jar file too?
Thanks
Could you please try this:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
I cannot say for ceratin that this is the issue without knowing how exactly you use the path extracted by:
HibernateUtil.class.getResource("/hibernate.cgf.xml").getPath()
but I can tell you this:
Run from an IDE the above line of code will return:
/path/to/project/src/main/resources/hibernate.cgf.xml
which is a valid filesystem path. You can then use this path to, for example, create an instance of File class and then use that instance to read the file contents.
However the same line of code run from inside a jar file will return:
file:/path/to/jar/jar_name.jar!/hibernate.cgf.xml
which is not a valid filesystem path. If you create an instance of File class using this path and then try to read the contents of the file you'll get an exception: java.io.FileNotFoundExeption
To read the contents of the file from inside of a jar you should use method Class.getResourceAsStream(String), which will return an instance of class sun.net.www.protocol.jar.JarURLConnection.JarURLInputStream (or equivalent in non-Oracle or non-OpenJDK Java). You can then use this object to read the contents of the file. For example:
InputStream inputStream = HibernateUtil.class.getResourceAsStream("/hibernate.cgf.xml");
Scanner scanner = new Scanner(inputStream).useDelimiter("\\A");
String fileContents = scanner.hasNext() ? sscanner.next() : "";
Most likely, the file is absent from the jar you create. There's too little information in your question, but I will try a guess:
Your hibernate.cgf.xml resides in the same directory as the Java sourcefles, and you are using a build tool (be it IDE, maven, gradle or an ant script) that expects resources to be stored in a separate directory.
It's easy to check: try to unzip your jar and see if the file is there (use any tool, you can just change the extension from .jar to .zip). I think you will see the file is absent.
Then come back with a question: "how to pack my non-java resources into a jar, using XXX", where XXX will be the name of the techology you are using for building the jar.
Most probably the slash in "/hibernate.cgf.xml" is not needed, if the hibernate.cgf.xml is in the same package as you class HibernateUtil.
You can access the file actually also via the classloader using the full path. Yet you never add to it the first slash.
Here is some code demonstrating how you can access the file using different methods:
public static void main(String[] args) {
// Accessing via class
System.out.println(SimpleTests.class.getResource("hibernate.cgf.xml").getPath());
// Accessing via classloader from the current thread
String path = Thread.currentThread().getContextClassLoader()
.getResource("simple/hibernate.cgf.xml").getPath();
System.out.println(path);
// Accessing via classloader used by the current class
System.out.println(SimpleTests.class.getClassLoader().getResource("simple/hibernate.cgf.xml").getPath());
}
In the example above the package 'simple' should be replaced by the package where your hibernate.cgf.xml is. But you should never have the slash at the beginning of the package declaration.
I have a runnable jar file which is not able to access my resources which reside outside of the default src directory. Based on my understanding from What is the difference between Class.getResource() and ClassLoader.getResource(), I should be able to access root/res/img/img1.png (see folder setup below) by using the following getResourceFile function:
public class Foo {
private static final ClassLoader CLASS_LOADER = Foo.class.getClassLoader();
public static File getResourceFile(String relativePath) {
// Since I'm using getClassLoader, the path will resolve starting from
// the root of the classpath and it'll take an absolute resource name
// usage: getResourceFile("img/img1.png")
// result: Exception in thread "main" java.lang.NullPointerException
return new File(CLASS_LOADER.getResource(relativePath).getFile());
}
}
folder setup:
root/
src/
foo/
bar/
res/
img/
img1.png
audio/
audio1.wav
The problem arises when I try to execute the jar executable itself. However, the strange thing is that I was not able to replicate this through eclipse IDE which was actually able to resolve the path correctly. I have added the resource directory to the build path via (Project -> Properties -> Java Build Path -> Add Folder) so Java should be able to find the resource folder at runtime.
Is there something I'm missing in terms of generating the jar file? When unpacking the jar file everything seems to be in order with the img and audio directories being in the root (given the above initial folder setup):
foo/
/bar
img/
img1.png
audio/
audio1.wav
Files can only be used to represent actual files in your filesystem. And once you package your files into a JAR, the resource (img/img1.png) is not a file anymore, but an entry in the JAR file. As long as you use the folder structure from within Eclipse, the resources are individual files so everything is fine.
Try this:
System.out.println(CLASS_LOADER.getResource(relativePath));
It will print a URL, but it will not be a valid path to a file in your file system, but to an entry within the JAR file.
Usually, you will only want to read a resource. In that case, use getResourceAsStream() to open an InputStream.
I have a jav aproject which is build through ant. It write the class files to output/classes/com/... path. One of my java classes needs input stream read from a file that is in a folder one level above output folder. Looks like if copy the file to the package folder under outptu/classes, it seems to work. But I do not want to palce my config file in output folder as it will be cleaned when I do ant clean. I want it to find it look above the output folder, in config folder and load it.
public static final String CONFIG_FILE="/../../../../../../../Config.txt";
public static ConfigObj getConfigObj() throws IOException {
InputStream i=ConfigLoader.class.getResourceAsStream(CONFIG_FILE);
...
I want to know when I want to give raltivepath, what should it be relative to. I tried looking up , it says relative to classloader. What is classloader in this case? Is it output/classes/com....../config folder where my ConfigLoader.class lives?
The problem is that getResourceAsStream() will only load resources from the classpath. I guess you only have output/classes on your classpath, so you will never be able to load the config file via getResourceAsStream() if it's outside that directory. Use a File with an absolut path pointing to the file, or place it in your classpath.
I am trying to access a data file from a public class, both of which are located within a JAR file. However, when I execute the jar on a Hadoop cluster, the system throws a FileNotFoundException. The bottom line is: is it possible to access resources within a Jar when running an application on a cluster, or does the resource need to be copied to the HDFS individually, and for either of the above, how would you go about implementing it?
Thanks!
Yes, if the JAR is in the CLASSPATH you can call getResourceAsStream() using a class loader or servlet context to get a reference to an InputStream for that file.
You will NOT have access to the file path. You give a path relative to the CLASSPATH and the classloader finds the file. You can get the contents, but not the absolute location.
You shouldn't want the absolute location. What will you do with it?
You can't write to the file. You can't alter anything. If your app is packaged in a WAR file, you can't alter its contents.
Currently, in my eclipse project, I have a file that I write to. However, I have exported my project to a JAR file and writing to that directory no longer works. I know I need to treat this file as a classpath resource, but how do I do this with a BufferedWriter?
You shouldn't have to treat it as a classpath resource to write to a file. You would only have to do that if the file was in your JAR file, but you don't want to write to a file contained within your JAR file do you?
You should still be able to create and write to a file but it will probably be relative to the working directory - the directory you execute your JAR file from (unless you use an absolute path). In eclipse, configure the working directory from within the run configuration dialog.
You're probably working in Linux. Because, in Linux, when you start your application from a JAR, the working directory is set to your home folder (/home/yourname/). When you start it from Eclipse, the working directory is set to the project folder.
To make sure you really know the files you are using are located in the project folder, or the folder where your JAR is in, you can use this piece of code to know where the JAR is located, then use the File(File parent, String name) constructor to create your files:
// Find out where the JAR is:
String path = YourClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath();
path = path.substring(0, path.lastIndexOf('/')+1);
// Create the project-folder-file:
File root = new File(path);
And, from now on, you can create all your File's like this:
File myFile = new File(root, "config.xml");
Of course, root has to be in your scope.
Such resources (when altered) are best stored in a sub-directory of user.home. It is a reproducible path that the user should have write access to. You might use the package name of the main class as a basis for the sub-directory. E.G.
our.com.Main -> ${user.home}/our/com/