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Spring not looking in the Config subfolder

I am new to Springboot and I am having trouble with externalizing my properties files.
I have multiple ".properties" files that I have kept in a subdirectory "config/". I have removed context-placeholders from my project and have included the comma-separated properties files while executing the jar.
eg. java -jar myjar.jar --spring.config.location=file:////config/PROP1.properties, file:///config/PROP2.properties -debug
I have few questions
Why are the files in the config directory not being read even after explicitly mentioning where to look?
I have my own dependencies in the project that have same-named properties files packed in its jar. Is that creating any sort of problem when SpringBoot tries to read the files from the config folder while executing my project jar?
Update
Now I am keeping only a single properties file suppose ABC.properties outside the jar in the same directory . I am using the name "ABC" instead of "application". I am using the below command
java -Dserver.log.dir=/path/to/log/dir -jar myjar.jar --server.port=9090 --spring.config.name=ABC --prop1=val1
I have overriden a property in my external property file but I don't see the overriden value being used when I run the application. My new questions are
Is there anything wrong with the command-line?
I have the following line in xml bean configuration
<context:property-placeholder location="ABC.properties" />. Is this causing any sort of problems to detect and use the external properties?
If the above is true and I have to remove the line how will SpringBoot understand from where the property values are to be imported?
(Not related directly to the question) Is there a order that I need to follow while giving command line arguments?

Spring boot has explicit indicate how to write this external configuration. See doc ref here: https://docs.spring.io/spring-boot/docs/2.5.2/reference/htmlsingle/#features.external-config.files
Basically, you need to specify the location like this:
--spring.config.location=classpath:/somefolder/somefile.properties
--spring.config.location=file:./somefolder/somefile.properties
From your command line, it seems that you are missing one dash, and also using a wrong format of file schema.

Q1: Why are the files in the config directory not being read even after explicitly mentioning where to look?
A1: If your config folder is beside your jar file, your command should be like
java -jar myjar.jar --spring.config.location=file:./config/PROP1.properties
meanwhile, you can also use the full path to target your config file.
Q2: If I don't mention the properties files names explicitly as an argument then they won't be picked up even when they are in the config directory
A2: No. See this doc, Spring Boot will try to find config files from the four places:
A /config subdir of the current directory.
The current directory
A classpath /config package
The classpath root
If spring.config.location is not set, files in these folders named application.properties will be treated as valid config file. You can change the accepted file name by setting property spring.config.name.
Q3 About Config File Priority
A3 As described in the doc mentioned earlier, if same name properties appear in different config files, locations higher in the list override lower items.
Another tip, it will be better to remove config files inside the jar file, so you can get a full view of configuration just in one place (the externalized config folder);

how to read config.properties j2ee

I need to read in config.properties for some configuration.
I am open to any way of doing it, with requirements.
I want it outside my war file so that it can be changed easily, and does not require a rebuild.
I have searched everywhere but cannot find how to do it, I'm sure this is java 101, but I cant figure it.
I have tried classloader but that seems to only load resources from inside the war, and I cannot find how to get the location I loaded the war into my server from to read it from there.
I also cannot find a way to pass in an argument via command line parameter's as its a soap endpoint, which I can access anywhere in my code?
I saw this Where to place and how to read configuration resource files in servlet based application?
and i want to use the file system approach but I don't want a hard coded path for the config file.
I just want something simple and easy, I know there is something but I just cannot find it.

use this this is helpful your question . i think.
In one method abcd
public static String abcd(String one) {
properties = new Properties();
properties.load(<classNmae>.class.getClassLoader().getResourceAsStream("AppResources.properties"));
return properties.getProperty(one);
call this code
String fileLocation = abcd("internal property file");
Properties properties = new Properties();
FileInputStream fis = new FileInputStream(fileLocation );
properties.load(fis);
fis.close();
acd = (String)properties.get("acd");
Note:
AppResources.properties have external file location D:/aaa.properties file
then in second method you read properties of external file

I want it outside my war file so that it can be changed easily, and does not require a rebuild.
May not be how you want it, but just to answer this, we have a similar scenario where our web applications are deployed in web servers located in locations /apps/servers/webserver-1, /apps/servers/webserver-2. And we have properties placed in some other locations like /apps/my-web-app1/app.properties and /apps/my-web-app2/app.properties.
Now good thing about this structure is that if I need to update any property, I just do the edit in the relative property and restart my web server. A downside of this is that I have to pass in paths to these properties file as system arguments to my web-server startup scripts (in my case, these are the catalina.sh files, yes I am using tomcats).
So my catalina.sh has line somewhere lying around something like
export JAVA_ARGS `-Dpath.to.properties.of.my-web-app2=/apps/my-web-app2/app.properties` ....
To read these properties, I have a Property Utility function that gets called by the StartupServlets of each application. The purpose of this function is to simply open up this file by reading the system property path.to.properties.of.my-web-app2 and puts these properties in something like a cache (a HashMap in my case) from where I can access them easily throughout the application.

Unable to find property file

I have a number of Java property files which used to be included in the standard resources directory and the following property holder of Spring works fine.
<context:property-placeholder location="classpath:/project/myproperty.properties/>
For some reason, I need to move them to another location so that they are not included as part of the JAR. The deployment script will handle them and copy them to a specific location called config.
I added the new location to the classpath argument of my java command, however, the above statement always fails, complaining no property file is found. My command is a follows
java -classpath "C:\Deployment\config" ............
And the property file does exist under:
C:\Deployment\config\project\myproperty.properties
Can someone let me know what I did wrong?

You can try to do it like this:
<context:property-placeholder location="file:${my.config.location}\project\myproperty.properties"/>
And to add the property in the environment properties:
java -Dmy.config.location="C:\Deployment\config" ...
Hope this helps.

Storing configuration file of an application in OS independent path

I am writing an application (basically eclipse plugin), so there are few combo-box, drop-downs etc, which I am getting values for them dynamically via XML file. My doubt is which is the best way to store these files in a particular directory so that it can be read in both Windows as well as Linux.
Initially I tried to create a config path under {eclipse.home.location} like:
String finalPath = System.getProperty("eclipse.home.location") +"/myAppConfig";
and store all of my plugin's configuration there (not only configuration but few helper jars which I programatically read in my plugin). But there is a probabilty that Eclipse installation maybe in shared location and user may not have write access to it.
I also tried to store it in a program files directory using:
System.getenv("ProgramFiles");
But this fails under non-windows environments. So my question is can anyone shed a light on this so that I can store in some common directory where it is valid for both windows and linux?
Kindly let me know if my wordings are confusing. Or is it possible to store my config files under plugins directory and get the path like this /plugins/myConfigDir ?

Try using the getStateLocation() method in Plugin.
That will give you an IPath that points to a user specific workspace location.

Creating a standalone package in java and take some settings from the configuration file

Anyone plz let us know what to do when we have some configuration file which is basically xml.I want to for example give the path to save the image(for my java program) in a folder from some config file (xml in my case).In that case where should the config file be kept.Rt now every thing is converted to jar file when i create a java standalone package.But i want to give some setting from xml file.What to do in that case.How is it possible.This article only provides to create a single jar file for java project but talks nothing about the configuration settings that u can provide from some external source.
Regards
Sagar

I'm not sure I fully understand your question, but if it is where to put the XML file with configuration information, you can place your xml file in the same directory as your jar file, and then pass the XML file name and path into the Jar on the command line when calling the Jar. If you're running it in Windows, this is often done using a shortcut. Then you can get the full path string for the Jar from the main method's String[] arg array that accepts the command parameters.

Sagar,
The fact your java program is a standalone package (.jar file) has no bearing on where your configuration file is stored. Your java package is a program and that program can read any file from the file system that it so desires; it does not have to be part of the code inside the IDE i.e. you don't have to write it when you write the program. What you do need is some way, when you start the program, to find and read said configuration file.
Depending on how you expect the program to be configured, you might put that file in a number of locations. For example, /etc/yourimageprogram/config.xml or c:\program files\yourimageprogram\config.xml or perhaps c:\users\Sagar\Application Settings\yourimageprogram\config.xml. Which you choose of those options really depends on what the use case is and that I can't help with.
However, there are some main points to reading any file:
Does it exist?
Are we allowed to open it for reading?
Are we allowed to open it for writing? Might want to know if we want to update the config?
In Java, typically, you would test this with:
File configfile = new File("C:\test.xml");
if ( configfile.exists() && configfile.canRead() )
{
// read the file
}
else
{
// decide what to do if no config exists.
// might be first run of app.
}
The next stage is to parse the file. There are a number of parsers available for XML including sax and org.w3c.dom. What you need to do is to use these to extract the information you require and store that in a class. Probably a singleton class as you're unlikely to have multiple configuration instances per instance of the program.
I suggest you read about XML Parsers and File Handling under Java. Also look at the File object. See all your options for file io in java. These should give you some indication of how to proceed.