Best practices to work with file paths in Java

Question

I m trying to load properties file to create a configuration. namely foo.properties.

What the best place to store this config file. Within the package where i will use initialize the using class? on the root level of the src?

Also, while accessing the files, i am using the location of the file in the file system: ie: c:\configs\foo.properties.

Is there a better practice for this? such as using the path relative to the project rather than the file system location? if yes, how can i reference the location of a property file within a project tree. such as:

src/
   /com.foo.bar
   /com.foo.car
   /com.foo.zar
   foo.properties

lets say i want to read the properties file to memory, how do i access this foo.properties without using the OS location/ or the location in the file system.

Answers to your questions are heavily dependent on what kind of app you're talking about, how it will be started, and what kind of configuration you're talking about. — Ryan Stewart, Oct 13 '11 at 23:27

score 2 · Answer 1 · edited May 23 '17 at 12:24

2

The way that I've always seen it done is to place the properties file in src/main/resources and then access it with getResourceAsStream().

This would be a good question to look at: Load a resource contained in a jar

edited May 23 '17 at 12:24

Community

1
1

answered Oct 13 '11 at 23:28

Jon7

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score 1 · Accepted Answer · answered Oct 13 '11 at 23:31

If you want keep something close to your current structure then you can just add a resources folder, like so:

src/
   /com.foo.bar
   /com.foo.car
   /com.foo.zar
resources
   /foo.properties

But I highly recommend following Maven directory structures, they really do give you a lot of flexibility:

src/
    /main
        /java
            /com.foo.bar
            /com.foo.car
            /com.foo.zar
        /resources
            /foo.properties

Regardless of what layout you go with I would recommend putting your resources in a folder structure matching one of your Java packages, that way you can easily load them as resources, avoiding your absolute vs relative file path conundrum. Something like this:

src/
    /resources/
       /com/foo/bar/foo.properties

And given a class com.foo.bar.MyClass you can do something like this:

InputStream is = MyClass. getResourceAsStream("com/foo/bar/foo.properties");

Good answer. I would like to stress that it is more important that the config files are _on the classpath_ when you run your application than where they locate in your source tree. The Maven default structure supports that because files in `src/main/resources` and `src/test/resources` get copied to the classpath (and to the JAR artifact if any) alongside with the java classes. — swiedenfeld, Dec 27 '16 at 16:52
What if you have more dynamic resources? An example: A "/resources/image" directory, where you store and retrieve images based on an externally provided image name. Wouldn't including the "/resources/image" directory in the .jar force you to rebuild your application every time there is a change to the set of contained images? — bnunamak, May 13 '17 at 07:15

score 0 · Answer 3 · edited Dec 27 '16 at 15:32

I think the location really depends on your project. I usually like having a specific folder to store configuration files.

As to the way you access the file, using absolute paths is a really bad ideia: Imagine you want to distribute your application as a Jar, deploy it on a web server, or simply move it to another place completly!

Take a look at:

http://docs.oracle.com/javase/8/docs/api/java/lang/ClassLoader.html#getResourceAsStream-java.lang.String-

If you have the properties file in your classpath (as part of your source, for instance), then you can load it in the same way independently of where you are (even inside a jar).

Best practices to work with file paths in Java

3 Answers3