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I'm using Maven for desktop build application. I read about Maven standard directory layout and I have this project structure for now:

App
|-- pom.xml
`-- src
    |-- main
        |-- java
        |   `-- java classes
        |-- resources
        |   `-- images
        |       `-- app images
        `--config
           `--config.xml

I want to find a way to load my resources and config files. I read some articles and topics and found this (simplified example from my code):

//class for loading config
public class Preferences {
    public Preferences() {
        InputStream image = Preferences.class.getResourceAsStream("images/image.png");
        InputStream config = Preferences.class.getResourceAsStream("config.xml");
    }        
}

But image and config variables contains null. I was trying different variants of loading (from root folder, with this.getClass() instead of Preferences.class, and others), but it's always null. I really don't understand this resource loading system and I didn't find any good documentation about it. It would be nice, if somebody gives a good explanation about this mechanism (or just give a link on docs). So, the main question is: How can I load my resources and config files?

cb4
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LMnet
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6 Answers6

13

You can use getResourceAsStream() method of java.lang.Class as you have done, but you have to add / before the path.


This question is tricky.

1. Two methods with same name

First of all, exist two methods of same name and same signature in these two classes:

java.lang.Class
java.lang.ClassLoader

They have the same name: getResource(String) (and getResourceAsStream(String) is alike).

2. They accept params of different format

Then, the param of them has different format:

  • The method java.lang.Class.getResouce<asStream>() accepts path with and without the leading /, resulting in different resources searching strategies. If a path has no /, Java will search the resource in the package/folder where the .class file resides. If it has /, Java will begin the searching from classpath root.
  • The method java.lang.ClassLoader.getResource<asStream>() accepts only path without /, because it always search from classpath. In a classpath based path, / is not a valid character. *

    *: As this answer states: this.getClass().getClassLoader().getResource("...") and NullPointerException

How to add a folder to classpath? In Eclipse we resolve to the context menu of a project: "Build path" - "Configure build path..." and add some folder to build path.

3. When it comes to Maven

At last, if a project is a Maven project, by default src/main/resources is in the classpath, so we can use

Class.getResource("/path-to-your-res");

or,

ClassLoader.getResource("path-to-your-res");

, to load anything under src/main/resources.

If we want to add another resources folder, as you have mentioned, it is done in pom.xml. And they are added into classpath as well, done by Maven. No extra config is needed.

4. Example

For example, if your config.ini is under src/main/resources/settings, myAvatar.gif under src/main/images, you can do:

In pom.xml:

<build>
    <resources>
        <resource>
            <directory>src/main/images</directory>
        </resource>
    </resources>
</build>

In code:

URL urlConfig = MyClass.class.getResource("/settings/config.ini"); //by default "src/main/resources/" is in classpath and no config needs to be changed.
InputStream inputAvatar = MyClass.class.getResourceAsStream("/myAvatar.gif"); //with changes in pom.xml now "src/main/images" is counted as resource folder, and added to classpath. So we use it directly.

We must use / above.

Or, with ClassLoader:

URL urlConfig = MyClass.class.getClassLoader().getResource("settings/config.ini"); //no leading "/"!!!
InputStream inputAvatar = MyClass.class.getClassLoader().getResourceAsStream("myAvatar.gif"); //no leading "/"!!!
WesternGun
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  • Without the `/` you get an error `Cannot invoke "java.net.URL.toExternalForm()" because "location" is null` ... this caught me out for a while. – BugHunterUK Nov 23 '21 at 20:25
7

I think I found the solution. As Juned Ahsan and mR_fr0g write, I need to use ClassLoader class, instead of this.getClass().getResource(). But, it works only for resource folder. But maven allows to add other folders as resource folders. I was just needed to add this section to pom.xml:

<build>
    <resources>
        <resource>
            <directory>src/main/config</directory>
        </resource>
    </resources>
</build>

And working java code is:

InputStream image = this.getClass().getClassLoader().getResourceAsStream("images/image.png");
InputStream config = ClassLoader.getSystemResourceAsStream("config.xml");
LMnet
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    Just a small sidenote, by adding config to your resources, maven will treat it as a resource. So it will also be included in compiled jars, which might be something you don't want to do with the config files. – Erwin Jun 16 '14 at 09:47
  • @Erwin, then how to read from the config directory without setting it as a resource one? – Turkhan Badalov Oct 08 '17 at 21:25
3
public Preferences() {
    InputStream image = this.getClass().getClassLoader().getResourceAsStream("image.png");
    InputStream config = this.getClass().getClassLoader().getResourceAsStream("config.xml")
} 
mR_fr0g
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2

How about using this appraoch:

InputStream file = ClassLoader.getSystemResourceAsStream("res.txt");

or

InputStream file = Thread.currentThread().getContextClassLoader().getResourceAsStream("MyProperty.properties");

As you currently have it there, that will look for the MyProperty.properties file at the top of your classpath. The could be in your src/main/resources directory or other src folder -- it would depend on how your application (jar/war) is built.

If you are building a jar then you should be able to unpack it and see your properties file at the top level of the jar. If you are building a war, maybe it should be in the WEB-INF/classes directory. Again, it depends on how it was built.

Juned Ahsan
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1

Try to build the project before. If you just put the files in the resources folder then you need to build the project in order to copy the new resource files into target folder

0

It's an old answered question, but I found that you need to use the include tag set or it won't find the resource when you call.

ClassLoader.getSystemResource("config.xml").

I'm using the latest maven (Oct 2020) on JDK14

Here's the working xml. Hope it helps as I tried it as stated in the solution here and it would not work until my xml looked as follows.

<build>                
    <resources>
        <resource>
            <directory>src/main/resources</directory>
            <includes>
                <include>**/*.xml</include> 
<!-- or whatever other file extensions you want to include -->
            </includes>
        </resource>
    </resources>        
</build>