How can I find path of a xml (static myXml.xml) file that is embedded into jar? Obviously not by absolute path but I am facing same kind of problem with relative paths. I cannot get it relative to home folder as Java returns different home folder depending upon from where I am calling the accessing Java class. For instance, from:
command prompt
App server
Eclipse launcher
Eclipse remote debugger etc
Is there someway that my accessing class (packed in same jar) may access embedded xml regardless of where jar file exists and who is trying to access it?
What you need to do is use the class loader to load the file:
InputStream stream = this.getClass().getClassLoader().
getResourceAsStream("myXml.xml");
The above code assumes that the file is at the top level of your jar.
Have you tried the getResource and getResourceAsStream methods in the Class class? Usually those are what I have to resort to in these situations.
Hope this helps.
You can use ClassLoader#getResource(..) to get InputStream of a file from the classpath:
Object.class.getClassLoader().getResource().openStream()
Also there are some other methods in ClassLoader which could be useful in your case.
Related
I'm trying to export a .JAR to be used as library to other projects. The problem is that I need to use relative paths when referencing files inside this library, but the only solutions I found were using absolute paths like:
private static final String FILE = new File("").getAbsolutePath().concat("/src/bla/file.txt");
Obviously whenever I try to run this line of code as an exported library I'll get something like DRIVE/project/src/bla/file.txt which is not correct since this .JAR can be anywhere inside DRIVE/projects like DRIVE/projects/lib/myLib.jar.
In Nodejs we had easy functions to retrieve relative paths according to the runtime location. How can I reference files in such a way that it will capture the "runtime path" so that I can safely reference them and the path will be dynamically solved?
For those who are so eager to mark this question as duplicate, please read with attention first. I'm NOT asking how to READ files from resources!
To use the "file.txt" present in the classpath,we need to make sure the "file.txt" is present in the directory represented by classpath.
Assume you have all the class files generated in a directory named "/home/abcuser/target".
For simplicity we will place the file.txt in the target directory root level.
The main class is say TestFileAccess.class(the class with the main method)
To execute the main class present in the target directory you can use the below command
java -cp /home/abcuser/target TestFileAccess
Now, the classpath in this case is /home/abcuser/target
To access the resources on classpath,you can go with two ways.
ClassLoader.getSystemResource and ClassLoader.getSystemResourceAsStream methods.
Class.getResource and Class.getResourceAsStream
The main difference between the ClassLoader and Class versions of the methods is in the way that relative paths are interpreted.
The Class methods resolve a relative path in the "directory" that corresponds to the classes package.
The ClassLoader methods treat relative paths as if they were absolute; i.e. the resolve them in the "root directory" of the classpath
Using ClassLoader you can use the below snippet
InputStream inputStream = ClassLoader.getSystemResourceAsStream("file.txt");
To explicitly reference a resource as a classpath file you can add the resource path to the classpath while executing the java code.
Let's say your resource "file.txt" is in /home/abcuser/resources.
You can add the the resource path to the classpath during the java execution start as shown below
java -cp "/home/abcuser/target:/home/abcuser/resources" TestFileAccess
I have a maven project and want to read file in it form its class path. The code that i am using is
InputStream is = getClass().getResourceAsStream ("filename.json");
But every time i am getting null inputstreams. I am not sure why ?
The file is places under /src/main/resources. The same folder which contains log4j.xml and it is being picked up decently.
Please note, I am trying to run this file from Eclipse i.e., run or debug mode. No vm arguments or whatsoever.
The Class.getResourceAsStream(String) method looks for the given resource within the same namespace (i.e. package) that the given class is in unless you give it an absolute path (see the API documentation); If it can't find the resource on the classpath in this namespace, it returns null. Since your class is likely inside e.g. com.myproject.resourcemanagement, your resource file has to analogously be under src/main/resources/com/myproject/resourcemanagement, similary to how your class source files are organised (under src/main/java/com/myproject/resourcemanagement).
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
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"));
when i am loading some data into my java program, i usually use FileInputStream. however i deploy the program as a jar file and webstart, so i have to use getRessource() or getRessourceAsStream() to load the data directly from the jar file.
now it is quite annoying to always switch this code between development and deployment?
is there a way autmate this? i.e. is there a way to know if the code is run from a jar or not?
when i try to load it withoug jar like this:
InputStream is = this.getClass().getResourceAsStream("file.txt");
the returned inputstream is simply null, although the file is definitely in the root directory of the application.
thanks!
Why do you use FileInputStream during development? Why not just use getResourceAsStream from the very start? So long as you place your files in an appropriate place in your classpath, you shouldn't have any problems. It can still be a file in the local filesystem rather than in a jar file.
It's helpful to develop with the final deployment environment in mind.
EDIT: If you want something in the root directory of your classpath, you should either use:
InputStream x = getClass().getResourceAsStream("/file.txt");
or
InputStream x = getClass().getClassLoader().getResourceAsStream("file.txt");
Basically Class.getResourceAsStream will resolve relative resources to the package containing the class; ClassLoader.getResourceAsStream resolves everything relative to the "root" package.
You could read your data always as a ressource. You only have to add the path where the data lies to your classpath.
If your data stays in WEB-INF/somewhere/mydata.txt inside your jar file, you will access it with:
getClass().getResourceAsStream( "/WEB-INF/somewhere/mydata.txt" )
Now, if you create a development directory /devel/WEB-INF/somewhere/mydata.txt and put /devel to your classpath, your code will work in development and production.
EDIT after explanation in question:
In your case this.getClass().getResourceAsStream( "mydata.txt" ) the resource is taken from the same position where the classfile of this is taken from. If you want to keep this, then you have to create a directory /devel/<path of package>/mydata.txt and again add /devel to your classpath.
How about setting a system property in your dev environment, via the -D switch? e.g. java -D:mypropertyname=mypropertyvalue
You could set the property in ant scripts in your dev environment, other environments don't get the property:
e.g.
public static boolean isDevEnvironment(){ return System.getProperty("mypropertyname")!=null;}
You might find a better way to hack it from one of the existing System Properties
If a file is considered part of your deployed application (as opposed to be part of the installation specific files) and can be located through the classpath then consider simply always using getResourceAsStream since it works regardless of the actual deployment scheme as long as it is in the classpath.
You might also find the information available from the JVM relevant (if allowed by the security manager):
// Get the location of this class
Class cls = this.getClass();
ProtectionDomain pDomain = cls.getProtectionDomain();
CodeSource cSource = pDomain.getCodeSource();
URL loc = cSource.getLocation(); // file:/c:/almanac14/examples/
http://www.exampledepot.com/egs/java.lang/ClassOrigin.html?l=rel
There shouldn't be any difference between development vs deployment, IHMO.
Classloader.getResource or getResourceAsStream works well, you can read resources and even write them.You can write your own Protocol handles and access everything as an URL/URI, which allows you to read and write resources and also allows proper identification of who actually provide the resource.
The only problem is if an URLStreamHandlerFactory is already registered(in a J2EE application the container could install a factory and you don't have any way to go over and install your own) and you cannot use your handlers "anywhere".
Knowing that, it is preferred to implement your own "resources". At that time when I need it I couldn't find something like that so I had to implement my own ResourceManager. For me it looks more intuitive to access a resource like
Resource layout = ResourceManager.resolve("view://layout/main.jsp")
instead of
URL layout = Classloader.getResource("some_package/view/layout/main.jsp")