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How can I dynamically use externalized properties file with Spring jar?

With this setup (from Eclipse using Windows10)
I was able to correctly start my SpringBoot application. This one worked too (same directory pattern):
Now I'm packaging my project as JAR and I want to use an external properties file. I had an teste32.yml file beside my JAR at the same directory (also tried to use it inside /config directory, as show here, but it didn't work either)
I want to dynamically use a properties file beside my JAR file everytime. Doesn't matter at which directory they are, I wanted to dynamically point to a properties file always at the same directory as the JAR is. I want to say to my client: "take this JAR and this file, put them wherever you want and run this command X and everything will be alright". I'm trying to discover command X but before I add some dynamic path, I'm trying with absolutes paths. I'm using this:
java -jar myJar.jar -Dspring.config.name=teste32 -Dspring.config.location=C:\workspace\myProject\target\
I manually copied teste32 inside target\ to test this. But this didn't work. This didn't work either (only spring.config.location variants):
-Dspring.config.location=file:C:\workspace\myProject\target\
-Dspring.config.location=classpath:/
-Dspring.config.location=file:C:/workspace/myProject/target/
I also tried with no spring.config.location, only name
So my questions are:
What does classpath: and file: mean? Until now I got the 2 correct setups by pure luck and I would like to understand when to use them.
When I have my project package as a JAR, what classpath becomes?
Finally, which combination is necessary to dynamically use a properties always at the same directory as the JAR?
UPDATE
Using --debug at the correct example got me this line at the very begging (Spring banner was still visible):
2018-09-25 15:45:14.480 DEBUG 11360 --- [ main] o.s.b.c.c.ConfigFileApplicationListener : Loaded config file 'file:src/main/resources/xirulei/teste32.yml' (file:src/main/resources/xirulei/teste32.yml)
But after moving myJar.jar and teste32.yml to a specific directory and running java -jar myJar.jar -Dspring.config.name=teste32 --debug (without spring.config.location, since teste32 is at the same directory as JAR), I simply didn't get any ConfigFileApplicationListener debug line.

a) java -jar myJar.jar -Dspring.config.name=teste32 -Dspring.config.location=C:\workspace\myProject\target
Did you check content of target dir? I'm pretty sure your cfg file is placed to target\classes\xirulei and it is why Spring cannot find it in target
b) When you place teste32.yml in the same directory as jar file then Spring must be able to find it (given this directory is working directory) without -Dspring.config.location (but you still need to provide -Dspring.config.name=teste32)
c) When you use -jar and do not provide additional class paths then classpath: points to the root of packages inside jar. Spring cannot find your file at classpath:/ because your file is at classpath:/xirulei/

Well, after all it was a simple mistake. As documentation says and as already pointed here, it should be
java -jar myproject.jar --spring.config.name=myproject
and not
java - jar myproject.jar -Dspring.config.name=myproject
As stated on question, only when using Eclipse -D(JVM argument) is necessary. When using bash/cmd, just --(program argument) is the correct option:

Unable to locate a file in java project running in a executable jar

Both class.getResource(FILE_NAME) and class.getClass().getClassLoader().getResource(FILE_NAME) run perfectly inside my eclipse but the same code getting failed to locate the file which is inside the jar file, when run as an executable jar in windows machine.
I have gone through all related links available for this problem (well, not exactly the same issue but 90% in sync), asked for solution but no reply came from any of those posts, so I'm posting my issue as a separate question hoping for help on this.
In total, 4 cases I have ran to resolve but none worked so far and I'm out of ideas now.
class.getClass().getClassLoader().getResource("/resources/readme.txt");
class.getResource("/resources/readme.txt");
class.getClass().getClassLoader().getResource("resources/readme.txt");
class.getResource("resources/readme.txt");
Ouf of all the above 4 cases, only 2 cases ran successfully in eclipse which are as mentioned below.
class.getResource("/resources/readme.txt");
class.getClass().getClassLoader().getResource("resources/readme.txt");
The other 2 cases just throwing me Exception in thread "main" java.lang.NullPointerException
Coming to the executable jar, all 4 cases are throwing me the Exception in thread "main" java.lang.NullPointerException.
So I have created a folder named resources where my jar is residing and placed my files inside this folder and ran the jar. Now the jar is running without any issues referring to the files inside the resources folder I created. So wherever I run this jar (windows, linux etc.,) I need to create a resources folder and place my files under the folder. Now the question is, can it be possible to make my jar refer the resources folder which is inside the jar itself?
Any help on this is much appreciated!

To get your txt file:
File yourFileIsHere = new File("resources/readme.txt");
Where put your file?
In the same location of your jar, example:
myapp/yourjar.jar
myapp/resources/readme.txt
If you want read file inside of your "src" folder:
InputStream yourInputStream = new YourClass().getClass().getClassLoader().getResourceAsStream("readme.txt");

If you are using Spring:
org.springframework.util.ResourceUtils.getFile("classpath:readme.txt")
Otherwise:
import com.google.common.io.Resources
byte[] byteSource = Resources.asByteSource(Resources.getResource("readme.txt")).read()

method class.getClass().getClassLoader().getResource() may take 3 prefixes: url:, classpath: and file: each prefix tells what is your base of search. If you want to search inside your jar use classpath: prefix. That tells your classloader to search everywhere within your classpath. Here is one example how to deal with it with Spring tools. Look also at ResourceLoader class in Spring

Heroku Unable To Find XML Config File

I uploaded a Spring application to Heroku but the application crashed with the following error:
java.io.FileNotFoundException: class path resource [com/myname/myapp/config/dao-context.xml
The file is definitely there, and it is in GIT, and the app runs successfully locally.
Any ideas what is happening here?

I suspect that when you are running locally, it is picking up the file on the classpath as a regular file on the filesystem (i.e. not inside of a JAR).
On Heroku, it is probably inside of a JAR file, which means it is not a regular file, and must be read as an input stream, which might look like this:
ClassLoader cl = this.getClass().getClassLoader();
InputStream inputStream = cl.getResourceAsStream("com/myname/myapp/config/dao-context.xml");
You can probably reproduce the problem locally by running the same command that's in your Procfile.
If this is not the case, then make sure the file exists on Heroku by running this command:
$ heroku run ls com/myname/myapp/config/dao-context.xml

For future visitors to this question, I overcame the problem by converting my DAO XML config file to the Java Config method, therefore Spring no longer required that XML file. This didn't directly solve the issue of being unable to find the XML file, but the added benefit is that I am now using the more modern and less verbose Java Config method.

Opening a xml file from eclipse and from a .jar file in java

Yesterday, I had a problem because I couldn't manage to open a xml file (it owuld give me a FileNotFoundException) located in the ressources folder of my .jar file, which I managed to open on eclipse using the following lines of code. You can see my old problem here. This was my code with the problem :
File xmlFile = new File("ressources/emitter.xml");
ConfigurableEmitter emitter = ParticleIO.loadEmitter(xmlFile);
Someone told me it that one way was to use getClassLoader().getRessourceAsStream method to open a xml file in a .jar file that was exported
InputStream i= this.getClass().getClassLoader().getResourceAsStream("ressources/emitter.xml");
ConfigurableEmitter emitter = ParticleIO.loadEmitter(i);
Unfortunately, that solution only works when I export my project into a .jar file, so if I want to go back debugging my program, I have to take the old code that would only works on eclipse.
My question is: is there any better way to do this without having to change my code if I want to export it or if I want to debug it?
Thank you
edit :
Thank you all, it works perfectly fine now
my problem was that I put my ressources folder like that :
+project
+src
+ressources
+emitter.xml

InputStream i= this.getClass().getClassLoader().getResourceAsStream("/ressources/emitter.xml");
The above should work in both cases (Note is is /resources/.... This is assuming say your directory structure is below:
MyProject
+src
+ressources
emitter.xml

Place the file alongside your source files, then you can use the getResourceAsStream() method in both cases. Don't forget to update the path (which should be the package name of your class, but with slashes instead of dots).

My question is: is there any better way to do this without having to
change my code if I want to export it or if I want to debug it?
Yes, use Maven. Maven will handle that and it hooks into Eclipse beautifully (NetBeans too!) What you do is place the resource in src/main/resources and then you can have Eclipse run the test goal of the Maven project or you can just run mvn test from the command line. Another advantage of using Maven here is that you can also have src/test/resources/emitter.xml which overrides the one in src/main with environment-specific test instructions and it won't affect your deployment.

InputStream i= getClass().getClassLoader().getResourceAsStream("ressources/emitter.xml");
or
InputStream i= getClass().getResourceAsStream("/ressources/emitter.xml");
(note the absolute positioning)
both work when the class is in the same jar, on the same class path.
In the jar the names must be case sensitive, but as the jar already works. Ensure that the ressources directory is on the class path too, or copied to the target directory.
As "ressources" is probably configured yourself (not named "resources" as in English), you probably need to add it to the build somehow.

Path resolution in eclipse package structure

This is a very simple question for many of you reading this, but it's quite new for me.
Here is a screenshot for my eclipse
When i run this program i get java.io.FileNotFoundException: queries.xml (The system cannot find the file specified) i tried ../../../queries.xml but that is also not working. I really don't understand when to use ../ because it means go 1 step back in dir, and in some cases it works, can anyone explain this? Also how can I refer to queries.xml here. Thanks
Note: I might even use this code on a linux box

I assume it is compiling your code into a build or classes folder, and running it from there...
Have you tried the traditional Java way for doing this:
def query = new XmlSlurper().parse( GroovySlurping.class.getResourceAsStream( '/queries.xml' ) )
Assuming the build step is copying the xml into the build folder, I believe that should work
I don't use Eclipse though, so can't be 100% sure...

Try
file = new File("src/org/ars/groovy/queries.xml");
To check the actual working directory of eclipse you can use
File f = new File(".");
System.out.println(f.getAbsolutePath());

You could try using a property file to store the path of the xml files.
This way you can place the xml files in any location, and simply change the property file.
This will not require a change/recompilation of code.
This would mean you will only need to hardcode the path of the property file.
If you prefer not hardcoding the path of the property file, then you could pass it as an argument during startup in your server setup file. (in tomcat web.xml). Every server will have an equivalent setup file where you could specify the path of the property file.
Alternatively you could specify the path of the xml in this same file, if you don't want to use property files.
This link will show you an example of reading from property files.
http://www.zparacha.com/how-to-read-properties-file-in-java/