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Spring mvc + maven: cannot access images

I'm using a standard maven archetype of webapp, so I put my images folder under src/main/resources directory and in the war package it goes under WEB-INF/classes/ directory.
My project tree is in the following image
So the images dir is in the classpath and i load it via getResource
final String path ="/resources/images/disegno.svg";
URL imagePath = this.getClass().getClassLoader().getResource(path);
but i get
Could not complete request
java.lang.NullPointerException
when declaring path
Also in .jsp, I try to access the image in this way:
<img src="<%=request.getContextPath() %>/WEB-INF/classes/images/disegno.svg"/>
but no image is displayed, even if it links to
http://localhost:8080/svgTest/WEB-INF/classes/images/disegno.svg
I cannot understand how maven works in this situation, and cannot find a complete guide, because everywhere all looks so easy, so I don't know which error i'm doing...

You're making 3 mistakes.
First: ClassLoader.getResource() expects a path that does NOT start with a /.
Second: what is under src/main/resources in the source project goes at the root of the classpath. There is no resources package, as your screenshot of WEB-INF/classes shows. So your path should be
this.getClass().getClassLoader().getResource("images/disegno.svg")
Third: by design, everything undex WEB-INF is not accessible from the browser. Everything out of WEB-INF is. So if you want the browser to be able to download your image, it should be in the src/main/webapp folder, not in src/main/resources. Of cource, once it's there, you won't be able to load it with the ClassLoader anymore, since it won't be in the classpath anymore. If you need to load it in addition to make it directly available to the browser, then use ServletContext.getResource().

Resources folder in a Java EE application

I'm developing a simple mail sender as Java EE application.
The project structure is shown as follows:
To properly setup email contents, I need to read the *.vm files placed inside the resource folder, that I supposed to have as path classpath:/templates/mail/*.vm (as with Spring)... But my supposition is wrong!
Which is the right path to use?
Should I have to use the META-INF folder? Is this solution more
java-ee-compliant? In that case, where have I to put the META-INF folder inside my project structure?
Update:
I packaged the project as .war, then I putted the files in:
/src/main/webapp/WEB-INF/classes/templates/mail/
Then:
org.apache.velocity.Template t = myVelocityEngine.getTemplate("classpath:/templates/mail/account_to_confirm.vm",
"UTF-8");
Nonetheless, the app returns an error at runtime:
Unable to find resource 'classpath:/templates/mail/account_to_confirm.vm'
What am I doing wrong?
Just to better understand:
Supposing that I'd like to deploy this app as jar (removing the servlet class, of course): in that case, should I have to edit the folder layout in order to still use the same path into the source code?

I think the problem is due to the prefix classpath:: where did you find that you have to use it?
You might find useful understanding how to initialize VelocityEngine reading Loading velocity template inside a jar file and how Configuring Resource Loaders in Velocity.

If you can, use Classloader.getResourceAsStream("templates/mail/*.vm"); or similar getResourceAsURL method.
If not, take a look at where files from resources are placed inside WAR. In your case, the file should be in /WEB-INF/classes/templates/mail .

Resource loading in Java not working as it should

This is the well known problem of loading resources from a jar file. This is not the first time I've tried to do this, but now it doesn't work the way I expect it to.
Normally I try to load the Resources with this.getClass.getResource("foo.png"), or getResourceAsStream()and it works. Now however it does not. The Resource is always null.
If I let System.out.println(this.getClass.getResource("")) print me the path (from eclipse) it shows /path/to/eclipseproject/package/structure/. Running this from a jar it just shows rsrc:package/structure
If I recall correctly this should print the path to the jar. Furthermore I thought this would print the package structure in both cases. Am I doing something wrong?

Here is the thing...
When Extracting the file from the Jar use:
this.getClass.getResource("/foo.png")
When running from a runnable Jar use, to reference an external file in the Jar folder path:
this.getClass.getResource("foo.png")
// When running this from Eclipse, it would refer to files in project root!
I have a code in the lower level determining where I'm running from to determine the correct path.

Doe this get the path you need?
this.getClass().getClassLoader().getResource("<your class name>.class").getPath();
See also this question for more on this issue.

Unless you prepend the path to the resources with '/', Class.getResource() will search for the resource in class package. E.g.: tld.domain.Foo.class.getResource("Bar.txt") will search for tld/domain/Bar.txt

Check the URLClassLoader for all the gory details, but it really depends on whether you are trying to access a ressource in the jar,
using a class loaded inside the same jar, in this case your file 'root' is the root of the jar
using a class loaded outside the jar (your eclipse case) where the root is your 'working directory'
To access resources inside a jar from outside, you should use something like
URL url = new URL( "jar", "", "file:" + jar.getCanonicalPath( ) + "!/" + localPathResource );
url.openStream(...)

This answer provides an explanation of how to load class resources from JAR files, even when the class is not in the JAR file and not in the Class-Path specified in the JAR file's manifest. There are also links to code.

Java (maven web app), getting full file path for file in resources folder?

I'm working with a project that is setup using the standard Maven directory structure so I have a folder called "resources" and within this I have made a folder called "fonts" and then put a file in it. I need to pass in the full String file path (of a file that is located, within my project structure, at resources/fonts/somefont.ttf) to an object I am using, from a 3rd party library, as below, I have searched on this for a while but have become a bit confused as to the proper way to do this. I have tried as below but it isn't able to find it. I looked at using ResourceBundle but that seemed to involve making an actual File object when I just need the path to pass into a method like the one below (don't have the actual method call in front of me so just giving an example from my memory):
FontFactory.somemethod("resources/fonts/somefont.ttf");
I had thought there was a way, with a project with standard Maven directory structure to get a file from the resource folder without having to use the full relative path from the class / package. Any advice on this is greatly appreciated.
I don't want to use a hard-coded path since different developers who work on the project have different setups and I want to include this as part of the project so that they get it directly when they checkout the project source.
This is for a web application (Struts 1.3 app) and when I look into the exploded WAR file (which I am running the project off of through Tomcat), the file is at:
<Exploded war dir>/resources/fonts/somefont.ttf

Code:
import java.io.File;
import org.springframework.core.io.*;
public String getFontFilePath(String classpathRelativePath) {
Resource rsrc = new ClassPathResource(classpathRelativePath);
return rsrc.getFile().getAbsolutePath();
}
In your case, classpathRelativePath would be something like "/resources/fonts/somefont.ttf".

You can use the below mentioned to get the path of the file:
String fileName = "/filename.extension"; //use forward slash to recognize your file
String path = this.getClass().getResource(fileName).toString();
use/pass the path to your methods.

If your resources directory is in the root of your war, that means resources/fonts/somefont.ttf would be a "virtual path" where that file is available. You can get the "real path"--the absolute file system path--from the ServletContext. Note (in the docs) that this only works if the WAR is exploded. If your container runs the app from the war file without expanding it, this method won't work.

You can look up the answer to the question on similar lines which I had
Loading XML Files during Maven Test run
The answer given by BobG should work. Though you need to keep in mind that path for the resource file is relative to path of the current class. Both resources and java source files are in classpath

Absolute Path of Project's folder in Java

Lots of confusion in this topic. Several Questions have been asked. Things still seem unclear.
ClassLoader, Absolute File Paths etc etc
Suppose I have a project directory structure as,
MyProject--
--dist
--lib
--src
--test
I have a resource say "txtfile.txt" in "lib/txt" directory. I want to access it in a system independent way. I need the absolute path of the project.
So I can code the path as abspath+"/lib/Dictionary/txtfile.txt"
Suppose I do this
java.io.File file = new java.io.File(""); //Dummy file
String abspath=file.getAbsolutePath();
I get the current working directory which is not necessarily project root.
Suppose I execute the final 'prj.jar' from the 'dist' folder which also contains "lib/txt/txtfile.txt" directory structure and resource,It should work here too. I should absolute path of dist folder.
Hope the problem is clear.

You should really be using getResource() or getResourceAsStream() using your class loader for this sort of thing. In particular, these methods use your ClassLoader to determine the search context for resources within your project.
Specify something like getClass().getResource("lib/txtfile.txt") in order to pick up the text file.
To clarify: instead of thinking about how to get the path of the resource you ought to be thinking about getting the resource -- in this case a file in a directory somewhere (possibly inside your JAR). It's not necessary to know some absolute path in this case, only some URL to get at the file, and the ClassLoader will return this URL for you. If you want to open a stream to the file you can do this directly without messing around with a URL using getResourceAsStream.
The resources you're trying to access through the ClassLoader need to be on the Class-Path (configured in the Manifest of your JAR file). This is critical! The ClassLoader uses the Class-Path to find the resources, so if you don't provide enough context in the Class-Path it won't be able to find anything. If you add . the ClassLoader should resolve anything inside or outside of the JAR depending on how you refer to the resource, though you can certainly be more specific.
Referring to the resource prefixed with a . will cause the ClassLoader to also look for files outside of the JAR, while not prefixing the resource path with a period will direct the ClassLoader to look only inside the JAR file.
That means if you have some file inside the JAR in a directory lib with name foo.txt and you want to get the resource then you'd run getResource("lib/foo.txt");
If the same resource were outside the JAR you'd run getResource("./lib/foo.txt");

First, make sure the lib directory is in your classpath. You can do this by adding the command line parameter in your startup script:
$JAVA_HOME/bin/java -classpath .:lib com.example.MyMainClass
save this as MyProject/start.sh or any os dependent script.
Then you can access the textfile.txt (as rightly mentioned by Mark) as:
// if you want this as a File
URL res = getClass().getClassLoader().getResource("text/textfile.txt");
File f = new File(res.getFile());
// As InputStream
InputStream in = getClass().getClassLoader()
.getResourceAsStream("text/textfile.txt");

#Mark is correct. That is by far the simplest and most robust approach.
However, if you really have to have a File, then your best bet is to try the following:
turn the contents of the System property "java.class.path" into a list of pathnames,
identify the JAR pathname in the list based on its filename,
figure out what "../.." is relative to the JAR pathname to give you the "project" directory, and
build your target path relative to the project directory.
Another alternative is to embed the project directory name in a wrapper script and set it as a system property using a -D option. It is also possible to have a wrapper script figure out its own absolute pathname; e.g. using whence.