File Loading Using getClass().getResource()
When working with resource files, like property files, a common approach is to use getClass().getResource(path) to load them. However, differences arise when executing code from the command line compared to within an IDE like Eclipse.
Behavior within Eclipse
In Eclipse, the getClass().getResource(path) method effectively loads files from the source folder. If the resource file (e.g., Test.properties) is placed in the same package as the Java file, it can be accessed within Eclipse without any issues.
Issue with Command Line Deployment
When deploying the application outside of Eclipse using the command line, the resource file may not be present in the classpath. Consequently, getClass().getResource(path) will fail to load the file, resulting in a null pointer exception.
Solution
To resolve this issue, ensure that the resource file is included in the classpath. This can be achieved by placing the file in the same directory as the compiled class file or by packaging it into a JAR file.
Alternative Approach
Instead of using getClass().getResource(path), consider using getClass().getResourceAsStream(path) to create an InputStream for the resource file. This method avoids the need to convert the URI to a File and handles file loading from various sources, such as the file system, a JAR file, or over a network.
Finally, be mindful of the argument passed to the getClass().getResourceAsStream method. Using Foo.class.getResourceAsStream("Test.properties") will load the file from the same package as Foo, while Foo.class.getResourceAsStream("/com/foo/bar/Test.properties") will load the file from the specified package.
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