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Editorial Team
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Asked: 2026-05-27T19:07:27+00:00

I have a Java project and used the standard maven archetype to create the

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I have a Java project and used the standard maven archetype to create the dir structure.
It looks like this:

|-src/main/java 
|-src/main/resources
|-target/classes
|- ...

Now one of my classes uses a .properties file to read in some settings. I placed it in src/main/resources and read it through File propertiesFile = new File("./src/main/resources/starter.properties");.
When I use the eclipse run-configuration, everything works fine. But recently I tried to start the same Java-class from my console using java some.package.Class, and since the .class-file is located in target/classes I got the message, that ./src/main/resources/starter.properties couldn’t be found.

What am I doing wrong? Is the .properties file not supposed to be located in the resources-folder or do I have to use an other way to load it?

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1 Answer

  1. Editorial Team

    The two previous answers are correct, but I wanted to give a bit more context.

    This file is in two places. It starts off in /src/main/resources and when you build the project, Maven copies it to /target/classes.

    At runtime, you shouldn’t access the copy that is in your source code. Otherwise, your software would need access to the source code in order to run. Rather, you should access the copy that is in your deliverable. At execution time, you can safely assume that you will find it on the classpath. It’s in the same place as your compiled classes, so if it weren’t on the classpath, you wouldn’t have been able to run the program in the first place. This is why you should use getResourceAsStream() as mentioned by the other answerers.

    (Though for production software, I do recommend Spring’s Resource abstraction for accessing these kinds of things.)

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