I am working on a Spring Boot application. I need to parse an XML file (countries.xml) on start. The problem is that I do not understand where to put it so that I could access it. My folders structure is
ProjectDirectory/src/main/java
ProjectDirectory/src/main/resources/countries.xml
My first idea was to put it in src/main/resources, but when I try to create File (countries.xml) I get a NPE and the stacktrace shows that my file is looked in the ProjectDirectory (so src/main/resources/ is not added). I tried to create File (resources/countries.xml) and the path would look like ProjectDirectory/resources/countries.xml (so again src/main is not added).
I tried adding this with no result
@Override
public void addResourceHandlers(final ResourceHandlerRegistry registry) {
registry.addResourceHandler("/resources/**").addResourceLocations("/resources/");
registry.setOrder(Ordered.HIGHEST_PRECEDENCE);
super.addResourceHandlers(registry);
}
I know that I can add src/main/ manually, but I want to understand why is it not working as it has to. I also tried examples with ResourceLoader - with the same no result.
Could anyone suggest what the problem is?
UPDATE: Just for future references - after building the project, I encountered problem with accessing file, so I changed File to InputStream
InputStream is = new ClassPathResource("countries.xml").getInputStream();
Just use Spring type ClassPathResource.
File file = new ClassPathResource("countries.xml").getFile();
As long as this file is somewhere on classpath Spring will find it. This can be src/main/resources
during development and testing. In production, it can be current running directory.
EDIT: This approach doesn't work if file is in fat JAR. In such case you need to use:
InputStream is = new ClassPathResource("countries.xml").getInputStream();