Develop Reference

The system cannot find the path specified in Eclipse(Web project)

When I try to get a pdf file in my project,the compiler is always saying "java.io.FileNotFoundException: \WEB-INF\Blank_A4.pdf(The system cannot find the path specified)":
Error Message
And my code is:
Code snipe
And my project structure is:
Project Structure

Is this error actually a run-time error? I don't see why you would get an error when the project is built.
Assuming a run-time error, common problems could be:
The code starts with a "\" instead of a relative path
The pdf is not in the output project (is the pdf in the created war file?)
The pdf would be a resource, and you would have to read it as a resource

java.io.File and every File-related class in that package use locations on disk. "\WEB-INF\Blank_A4.pdf" is not a location on disk. Without knowing the context of the code you showed--what variables it has to work with--there's not much to suggest other than looking at the methods in the javax.servlet.ServletContext interface for how to open input streams if it's available to you at runtime.

How do I access the resources directory for a calling program in Java

I have a library which is used by a program. This library loads a special directory in the resources folder.
in the library I have the method
public class DataRegistry{
public static File getSpecialDirectory(){
String resourceName = Thread.currentThread().getContextClassLoader().getResource("data").getFile().replace("%20", " ");
File file = new File(resourceName);
return file;
}
}
in my program I have the main method
public static void main(String args[]){
System.out.println(Data.getSpecialDirectory());
}
When I execute the getSpecialDirectory() in a junit test within the data program, the resource is fetched and all is well.
When I execute the getSpecialDirectory() in the main method outside the data program (imported jar) I get the jars data directory and not the directory the program executing the thread is expecting.
I figured getting the parent class loader would have solved this issue... I believe I may have a fundamental issue in my understanding here.
For clarity:
(library)
Line 15 of this file: https://gist.github.com/AnthonyClink/11275442
(Usage)
Line 31 of this file:
https://gist.github.com/AnthonyClink/11275661
My poms may have something to do with it, so sharing them is probably important:\
(Library)
https://github.com/Clinkworks/Neptical/blob/master/pom.xml
(Usage)
https://github.com/Clinkworks/Neptical-Maven-Plugin/blob/master/pom.xml

Maven uses the target/classes and target/test-classes directories to compose the classpath used to run unit tests. The content of your src/main/resources and src/test/resources gets copied to these locations together with the compiled java sources (the .class files).
java.lang.ClassLoader.getResource returns a java.net.URL.
In the unit test situation, ClassLoader.getResource returns a file:// schemed URL, and your code works as expected.
When your code is executed when packaged in a jar file, ClassLoader.getResource no longer returns a file:// schemed URL. It can't because the resource is no longer a separate entity in a file system - it's buried inside the jar file. What you actually get is a jar:file:// schemed URL.
jar: URLs cannot be accessed through a [java.io.File] object.
If you only need to read the content of the resource you should use java.lang.ClassLoader.getResourceAsStream(...) and read the content from that.
Note that it is not possible to update the content of class loader resources.

Do you want the directory where the Java program was loaded from ? Can try
File f = new File(".").getCanconicalFile();
Or if you want all the places could try parsing the classpath from System.env or use the default class loader
If you want a directory relative to the jar from where your classes were loaded: Quoting from http://www.nakov.com/blog/2008/06/11/finding-the-directory-where-your-java-class-file-is-loaded-from/
URL classesRootDir = getClass().getProtectionDomain().getCodeSource().getLocation();
The above returns the base directory used when loading your .class
files. It does not contain the package. It contains the classpath
directory only. For example, in a console application the above
returns someting like this:
file:/C:/Documents%20and%20Settings/Administrator/workspace/TestPrj/bin/
In a Web application it returns someting like this:
file:/C:/Tomcat/webapps/MyApp/WEB-INF/classes/
FYI This might apply if the user id running program cannot read all folders: http://docs.oracle.com/javase/7/docs/api/java/lang/Class.html#getProtectionDomain%28%29
or pass a .class object of the calling class as a parameter and get its classloader

There is no guarantee that you have a directory for your resources as far as I know. Java provide methods to get a resource as a InputStream, but nothing that can easily get the File for a resource, although it is possible to get the URL for a given resource.
If you need a special working directory in your application, I recommend having that passed as a system property, a program parameter, a web application parameter, or something else that lets you know where your application files are. You can even populate this directory with a ZIP file that you include as a resource in your application, as the ZIP libraries in Java do accept input streams as parameters.

While user #tgkprog is right and his answer correct, I can hardly imagine that you really want what you seem to have asked for and what he correctly answered in (3). Please edit your question and explain
what the output should look like if resourceName would be printed to the console,
if you really want the JAR file's location (in my test case that would be something like file:/C:/Users/Alexander/.m2/repository/com/clinkworks/neptical/0.0.1-SNAPSHOT/data, which does not make sense because you do not want to read from or even write to the local Maven cache),
or if you maybe rather want to read a resource directly from the JAR file or any other location such as the current working directory etc.
Otherwise I am afraid there is no good way to answer the question.

Why does getResourceAsStream() work in the IDE but not the JAR?

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

Netbeans Built .jar doesn't work with class file inside

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"));

How to resolve path of a static file in Java?

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.