Develop Reference

Best Practices on refering to resource files in Java code by relative path?

I have a repository which looks something like this:
/
/resources/resource1.txt
/resources/resource2.txt
/code/
/code/C/...
/code/java/pom.xml
/code/java/src/main/XYZ.java
/code/java/src/main/ABC.java
In other words, there is a maven project rooted in /code/java/, and there are resource files located at /resources/.
Now, I want to refer to the resource files in my code. The code will be something like the files in /code/java/src/main/XYZ.java. How should I refer to this file?
I do not want to hard code the filepath from my local machine. For example, I do not want something like /Users/agnishom/projectX/resources/resource1.txt. This will make it difficult for other people who clone my repository to make it work.
I would like to be able to name a file by their relative path, possibly relative to the pom.xml file.
What is the best way to do this? Moving the resource files into the /code/java/ is not an option because they are shared by other projects which are not part of the java

In the build section of your pom.xml, you can specify the resources directory relative to the project's base directory (/pom.xml), as you were suggesting:
<build>
...
<resources>
<resource>
<targetPath>.</targetPath>
<directory>${basedir}/../../resources</directory>
</resource>
</resources>
</build>
targetPath is the path to where included resources should be placed, relative to the jarfile/resulting build.
directory is the directory to include resources from.

How to update environment value in property file using pom.xml

I have configuration.properties in configs/Configuration.properties where I have url path = http://finacial/dev3/api
http:// financial/{env}/api
I want to load the environment value from the pom.xml to replace dev 3 according to the environment property set in pom.xml.

You could use the filter option of the maven resources plugin, it does exactly that.
Variables can be included in your resources. These variables, denoted by the ${...} delimiters, can come from the system properties, your project properties, from your filter resources and from the command line.
E.g. if you have a property url in your pom
<properties>
<url>abc</ulr>
</properties>
and your .properties file contains a value ${url} and you include the file in the resource section of the pom, then ${url} will be replaced in the output file in the build target location by abc upon issuing mvn resources:resources
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>

Where exactly do I put project-defaults.yml?

I'm working on a Wildfly Swarm project (using Wildfly Swarm version 2017.8.1, Maven 3.5.0, OpenJDK 1.8.0_141) where users will often upload files WAY bigger than Undertow's default 10485760 bytes (10MB) max-post-size setting.
I see from other questions that what I need to change is the key swarm.undertow.servers.default-server.http-listeners.default.max-post-size.
To try this setting out, I put it on the command line when I start the jar:
java -jar project-swarm.jar -Dswarm.undertow.servers.default-server.http-listeners.default.max-post-size=4000000000
This worked. The following also works if I want to run it using the Wildfly Swarm maven plugin:
mvn wildfly-swarm:run -Dswarm.undertow.servers.default-server.http-listeners.default.max-post-size=4000000000
Using either of the above two commands I can upload files bigger than 10485760 bytes (10MB). The next step is to make this the default so that I don't have to pass it in via command line each time. The official Wildfly Swarm configuration documentation said that I should put the undertow setting in a project-defaults.yml file. This is what that looks like for me:
swarm:
undertow:
servers:
default-server:
http-listeners:
default:
max-post-size: 4000000000
The thing is, none of that documentation says WHERE the file should ultimately end up. Should it end up as part of the war, under WEB-INF/classes, or the root of the war, or somewhere else? From the wildfly-swarm examples repo on github, usually the project-defaults.yml file is insrc/main/resources`, which is where I had it originally. However, it didn't work for me. Any attempts to upload a file bigger than 10485760 bytes (10MB) fail.
I found out that anything in a maven project that is put into src/main/resources gets put into the war file's /WEB-INF/classes (and the war file itself is wrapped by the swarm jar). So, I updated my maven pom so that the project-defaults.yml would be stored in the base directory of the war file itself. This STILL didn't work.
Next, I tried putting project-defaults.yml in the swarm jar itself, rather than the enclosed war. Yet again, I could not break past the 10485760 bytes (10MB) max-post-size default.
Finally, I tried using the command line option to refer to an external yml file, like so:
java -jar project-swarm.jar -s../src/main/root-resources/project-defaults.yml
This loaded the file, but alas I was STILL STUCK at the default max-post-size of 10485760 bytes (10MB).
At this point, I'm stumped. My project-defaults.yml file might be wrong or I'm not putting it in the right location. How do I fix this so I can upload files bigger than 10485760 bytes (10MB) and specify it via the project-defaults.yml file?

For those searching for an answer to this question, you have two alternatives for placing project-defaults.yml in your project.
Use Default Location
src/main/resources/project-defaults.yml
Use Maven Resource Declaration
Place the file somewhere else, for example: src/main/java/project-defaults.yml
Then, declare it as a resource using maven inside of the build section like so:
<build>
<resources>
<resource>
<directory>src/main/java</directory>
<includes>
<include>project-defaults.yml</include>
</includes>
</resource>
</resources>
</build>

Java path to file in src/main/resources

I have a maven module with the following directory structure and I want to get the path to abc.access in abcManager.java.
src
|-main
|-java
|-org.abc.xyz
|-init
|-abcManager.java
|-resources
|-abc.access
I tried it using abcManager.class.getResource("abc.access") but it gives a null.
I went through the below questions but the solutions didn't work.
How to get the path of src/test/resources directory in JUnit?
Can't get path to resource

The problem was that I have not included the resource in the maven build in pom file. When I included the following it worked with abcManager.class.getResource("/abc.access")
<build>
<resources>
<resource>
<directory>src/main/resources</directory>
<includes>
<include>abc.access</include>
</includes>
</resource>
</resources>
</build>

According to the javadoc of URL java.lang.Class.getResource(String name) :
Before delegation, an absolute resource name is constructed from the
given resource name using this algorithm:
If the name begins with a '/' ('\u002f'), then the absolute name of
the resource is the portion of the name following the '/'.
Otherwise,
the absolute name is of the following form:
modified_package_name/name
Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').
When you invoke :
abcManager.class.getResource("abc.access")
You don't prefix the path with a "/" character . So, you use the second way.
It means that the resource should be located in
the org.abc.xyz.init package (or folder) but it is not the case since the resource is located in the root of the resources folder.
In Maven, resources is the base directory for resources.
So you can get the resource by invoking the first way:
abcManager.class.getResource("/abc.access")
or you can also simply do it :
getClass().getResource("/abc.access")

try this :
String filename = "abc.access";
InputStream in = getClass().getClassLoader().getResourceAsStream(filename);

Following should work.
InputStream resourceAsStream =
abcManager.class.getClassLoader().getResourceAsStream("abc.access");
provided you are working inside an IDE (Ex. Eclipse).
If you are trying this from commandline, you need to explicitely setup the classpath.

class.getResource is local to the class, in this case package org.abc.xyz.init. You are trying to read src/main/resources/org/ABC/xyz/init/abc.access. Alternatively to the class loader which always loads from the classpath root you can also do class.getResource("/abc.access")

Resources and config loading in maven project

I'm using Maven for desktop build application. I read about Maven standard directory layout and I have this project structure for now:
App
|-- pom.xml
`-- src
|-- main
|-- java
| `-- java classes
|-- resources
| `-- images
| `-- app images
`--config
`--config.xml
I want to find a way to load my resources and config files. I read some articles and topics and found this (simplified example from my code):
//class for loading config
public class Preferences {
public Preferences() {
InputStream image = Preferences.class.getResourceAsStream("images/image.png");
InputStream config = Preferences.class.getResourceAsStream("config.xml");
}
}
But image and config variables contains null. I was trying different variants of loading (from root folder, with this.getClass() instead of Preferences.class, and others), but it's always null. I really don't understand this resource loading system and I didn't find any good documentation about it. It would be nice, if somebody gives a good explanation about this mechanism (or just give a link on docs).
So, the main question is: How can I load my resources and config files?

You can use getResourceAsStream() method of java.lang.Class as you have done, but you have to add / before the path.
This question is tricky.
1. Two methods with same name
First of all, exist two methods of same name and same signature in these two classes:
java.lang.Class
java.lang.ClassLoader
They have the same name: getResource(String) (and getResourceAsStream(String) is alike).
2. They accept params of different format
Then, the param of them has different format:
The method java.lang.Class.getResouce<asStream>() accepts path with and without the leading /, resulting in different resources searching strategies. If a path has no /, Java will search the resource in the package/folder where the .class file resides. If it has /, Java will begin the searching from classpath root.
The method java.lang.ClassLoader.getResource<asStream>() accepts only path without /, because it always search from classpath. In a classpath based path, / is not a valid character. *
*: As this answer states: this.getClass().getClassLoader().getResource("...") and NullPointerException
How to add a folder to classpath? In Eclipse we resolve to the context menu of a project: "Build path" - "Configure build path..." and add some folder to build path.
3. When it comes to Maven
At last, if a project is a Maven project, by default src/main/resources is in the classpath, so we can use
Class.getResource("/path-to-your-res");
or,
ClassLoader.getResource("path-to-your-res");
, to load anything under src/main/resources.
If we want to add another resources folder, as you have mentioned, it is done in pom.xml. And they are added into classpath as well, done by Maven. No extra config is needed.
4. Example
For example, if your config.ini is under src/main/resources/settings, myAvatar.gif under src/main/images, you can do:
In pom.xml:
<build>
<resources>
<resource>
<directory>src/main/images</directory>
</resource>
</resources>
</build>
In code:
URL urlConfig = MyClass.class.getResource("/settings/config.ini"); //by default "src/main/resources/" is in classpath and no config needs to be changed.
InputStream inputAvatar = MyClass.class.getResourceAsStream("/myAvatar.gif"); //with changes in pom.xml now "src/main/images" is counted as resource folder, and added to classpath. So we use it directly.
We must use / above.
Or, with ClassLoader:
URL urlConfig = MyClass.class.getClassLoader().getResource("settings/config.ini"); //no leading "/"!!!
InputStream inputAvatar = MyClass.class.getClassLoader().getResourceAsStream("myAvatar.gif"); //no leading "/"!!!

I think I found the solution. As Juned Ahsan and mR_fr0g write, I need to use ClassLoader class, instead of this.getClass().getResource(). But, it works only for resource folder. But maven allows to add other folders as resource folders. I was just needed to add this section to pom.xml:
<build>
<resources>
<resource>
<directory>src/main/config</directory>
</resource>
</resources>
</build>
And working java code is:
InputStream image = this.getClass().getClassLoader().getResourceAsStream("images/image.png");
InputStream config = ClassLoader.getSystemResourceAsStream("config.xml");

public Preferences() {
InputStream image = this.getClass().getClassLoader().getResourceAsStream("image.png");
InputStream config = this.getClass().getClassLoader().getResourceAsStream("config.xml")
}

How about using this appraoch:
InputStream file = ClassLoader.getSystemResourceAsStream("res.txt");
or
InputStream file = Thread.currentThread().getContextClassLoader().getResourceAsStream("MyProperty.properties");
As you currently have it there, that will look for the MyProperty.properties file at the top of your classpath. The could be in your src/main/resources directory or other src folder -- it would depend on how your application (jar/war) is built.
If you are building a jar then you should be able to unpack it and see your properties file at the top level of the jar. If you are building a war, maybe it should be in the WEB-INF/classes directory. Again, it depends on how it was built.

Try to build the project before.
If you just put the files in the resources folder then you need to build the project in order to copy the new resource files into target folder

It's an old answered question, but I found that you need to use the include tag set or it won't find the resource when you call.
ClassLoader.getSystemResource("config.xml").
I'm using the latest maven (Oct 2020) on JDK14
Here's the working xml. Hope it helps as I tried it as stated in the solution here and it would not work until my xml looked as follows.
<build>
<resources>
<resource>
<directory>src/main/resources</directory>
<includes>
<include>**/*.xml</include>
<!-- or whatever other file extensions you want to include -->
</includes>
</resource>
</resources>
</build>