[BUG] Package extension.yml file

[joshpalis]

What is the bug?

Building and running the opensearch-sdk artifact would look for the extension.yml file within the opensearch-sdk/build/distributions/opensearch-sdk-1.0.0-SNAPSHOT/src/test/resources/ directory instead of the opensearch-sdk/src/test/resources directory

How can one reproduce the bug?

Run ./gradlew build in the opensearch-sdk directory to produce the artifact. Navigate to opensearch-sdk/build/distributions/ and untar the generated artifact using tar -xvf opensearch-sdk-1.0.0-SNAPSHOT.tar. Navigate to opensearch-sdk-1.0.0-SNAPSHOT/ and run ./bin/opensearch-sdk

This will produce the following error :
Exception in thread "main" java.io.FileNotFoundException: src/test/resources/extension.yml (No such file or directory)

In order to fix this issue, we must create the directory within the package and generate the extension.yml file here so that the artifact has access.

EDIT: Following discussion, the extensions.yml file will be moved to the config directory where opensearch.yml is currently stored.

[dbwiddis]

This is not as simple as just changing the directory paths.

Both "file paths" are accessible with the current setup but neither file path will work in a compiled jar file.

The file should be moved to src/main/resources which will put it at the root of the classpath. You'll then have to use some variant of getResource() or getResourceAsStream() from the classloader, such as:

// reads from the root of the classpath so will find something in src/main/resources
InputStream in = this.getClass().getClassLoader().getResourceAsStream("extension.yml");
// or get the filepath
String ymlPath = this.getClass().getClassLoader().getResource("extension.yml").getPath();

You can also get it from the class itself using NIO, but this is relative to the package, so if you're in the org.opensearch.sdk package you need to navigate up 3 directories:

// reads from the root of the package
Path ymlPath = Paths.get(this.getClass().getResource("../../../extension.yml").toURI());
// or
File ymlFile = new File(this.getClass().getResource("../../../extension.yml").toURI());
// I think if you lead with a slash you can work around the relative package directory things, so try
Path ymlPath = Paths.get(this.getClass().getResource("/extension.yml").toURI());
// or
String ymlPath = this.getClass().getResource("../../config/dev/example.xml").getPath();

or variations of the above.

One other complication for using the classpath: anyone else can have a same-named file on the class path and they will conflict. So a generic name like extension.yml is going to be problematic. To work around this:

• put it in a subdirectory in src/main/resources that a user is unlikely to also have in their own classpath, e.g., src/main/resources/org/opensearch/sdk (which is essentially like putting it in the package anyway!)
• give it a more verbose name like opensearch_sdk_extension.yml

[dbwiddis]

See this commit for implementation for a different YAML file: 821d27f

[dbwiddis]

The changes in #109 may make this obsolete. Review discussion on that PR before spending any time on this.