Access maven properties defined in the pom

Question

How do I access maven properties defined in the pom in a normal maven project, and in a maven plugin project?

What kind of information are you interested in? If it's just versions or similar things, then it's probably a better idea to [put it in the manifest and read it from there](http://stackoverflow.com/questions/2712970/how-to-get-maven-artifact-version-at-runtime/2713013#2713013). — Joachim Sauer, Jul 16 '12 at 08:31
There is a lot of post in Stack Overflow about this : http://stackoverflow.com/search?q=read+pom+in+java, at least, take a look before post. SO edit helpers should show them. Google too. — Jean-Rémy Revy, Jul 17 '12 at 09:27
Thanks guys. I now realize this is really bad practice. Gonna try other approach. — SparedWhisle, Jul 18 '12 at 03:01
possible duplicate of [Get MavenProject from just the POM.xml - pom parser?](http://stackoverflow.com/questions/4381460/get-mavenproject-from-just-the-pom-xml-pom-parser) — oers, Jul 18 '12 at 10:03

score 88 · Answer 1 · edited Aug 13 '19 at 00:21

Use the properties-maven-plugin to write specific pom properties to a file at compile time, and then read that file at run time.

In your pom.xml:

<properties>
     <name>${project.name}</name>
     <version>${project.version}</version>
     <foo>bar</foo>
</properties>

<build>
    <plugins>
        <plugin>
            <groupId>org.codehaus.mojo</groupId>
            <artifactId>properties-maven-plugin</artifactId>
            <version>1.0.0</version>
            <executions>
                <execution>
                    <phase>generate-resources</phase>
                    <goals>
                        <goal>write-project-properties</goal>
                    </goals>
                    <configuration>
                        <outputFile>${project.build.outputDirectory}/my.properties</outputFile>
                    </configuration>
                </execution>
            </executions>
        </plugin>
    </plugins>
</build>

And then in .java:

java.io.InputStream is = this.getClass().getResourceAsStream("my.properties");
java.util.Properties p = new Properties();
p.load(is);
String name = p.getProperty("name");
String version = p.getProperty("version");
String foo = p.getProperty("foo");

That worked, but made Eclipse fall into **infinite build / deploy-to-Tomcat loop** (build -> generate my.properties -> resource changed -> build), so I had to change phase to `compile`. Not sure if it's proper solution though. — Nikita Bosik, Mar 29 '16 at 15:05
@atoregozh suggested using Guava Resource's instead for loading the properties file. OK. But reverted change to keep this example minimal with standard API's. — Leif Gruenwoldt, Oct 05 '16 at 02:02

score 27 · Answer 2 · edited May 23 '17 at 11:54

27

Set up a System variable from Maven and in java use following call

System.getProperty("Key");

edited May 23 '17 at 11:54

Community

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answered Jul 16 '12 at 08:32

Santosh

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This will only work if you are running the resulting build from Maven, perhaps using the maven-exec-plugin. If you are just using Maven to compile the code, this will not work. – David Pashley Jul 13 '15 at 06:27
@DavidPashley, thats true. In the compile only case, I guess the variable needs to be set from the compiler plugin. – Santosh Jul 14 '15 at 06:32

score 7 · Answer 3 · edited Dec 13 '18 at 23:13

7

This can be done with standard java properties in combination with the maven-resource-plugin with enabled filtering on properties.

For more info see http://maven.apache.org/plugins/maven-resources-plugin/examples/filter.html

This will work for standard maven project as for plugin projects

edited Dec 13 '18 at 23:13

OneCricketeer

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answered Jul 16 '12 at 08:32

Mark Bakker

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score 6 · Answer 4 · edited May 23 '17 at 12:09

6

Maven already has a solution to do what you want:

Get MavenProject from just the POM.xml - pom parser?

btw: first hit at google search ;)

Model model = null;
FileReader reader = null;
MavenXpp3Reader mavenreader = new MavenXpp3Reader();

try {
     reader = new FileReader(pomfile); // <-- pomfile is your pom.xml
     model = mavenreader.read(reader);
     model.setPomFile(pomfile);
}catch(Exception ex){
     // do something better here
     ex.printStackTrace()
}

MavenProject project = new MavenProject(model);
project.getProperties() // <-- thats what you need

edited May 23 '17 at 12:09

Community

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answered Jul 16 '12 at 08:29

SWoeste

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Single-link answers are discouraged here (mostly because they can become useless when the linked-to resource goes away). You should at least summarize the information from the linked to post here. – Joachim Sauer Jul 16 '12 at 08:31
I accidently posted the wrong link, its a stackoverflow post, that should not go away ;) – SWoeste Jul 16 '12 at 08:33
But i agree whit you, just a link is a bit unpleasant. So based on the source from the link i added some comments and the method call you are searching for (last line). – SWoeste Jul 16 '12 at 08:41
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This is really a very bad practice, just use the maven resource plugin! – Mark Bakker Jul 16 '12 at 08:43
1

Why should this be bad practice if he wants to access the properties from java code? – SWoeste Jul 16 '12 at 08:45
Because the maven resource plugin does do it out of the box – Mark Bakker Jul 16 '12 at 08:58
2

But only if you assume that he wants to/could change his pom(s). I agree whith you that it might be bad to have the properties directly in the pom file instead of an external file. That file could be read with the java properties class - sure. But what in the other case? Then the maven-model is the best solution to do the trick ... but before David Dai dont tell us what exactly he wants to do we all are only guessing. – SWoeste Jul 16 '12 at 09:04
1

It was my understanding too that we have one mvn pom file (containing properties) and one java application (from where we want to read details of that pom file). Without actually running maven. – Andreas Dolk Jul 16 '12 at 09:04

score 6 · Answer 5 · answered Sep 28 '21 at 10:15

Update for Leif Gruenwoldt answer:

It's important to use

this.getClass().getClassLoader().getResourceAsStream("maven.properties");

instead of just

 this.getClass().getResourceAsStream("maven.properties");

Especially, if you write your maven.properties file right to project.build.outputDirectory:

<plugin>
   <groupId>org.codehaus.mojo</groupId>
   <artifactId>properties-maven-plugin</artifactId>
   <version>1.0.0</version>
   <executions>
      <execution>
         <phase>generate-resources</phase>
         <goals>
            <goal>write-project-properties</goal>
         </goals>
         <configuration>
            <outputFile>${project.build.outputDirectory}/maven.properties</outputFile>
         </configuration>
      </execution>
   </executions>
</plugin>

Explanation is right there.

I tried your code and it generate the properties under the path classes, but when I try to retrieve it gives me null — Robs, Jan 20 '23 at 08:13

Naor Bar · Answer 6 · 2022-01-11T10:56:49.200

I use the build-info goal in spring-boot-maven-plugin:

In my pom.xml:

<plugin>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-maven-plugin</artifactId>
    <executions>
    ...
      <execution>
        <id>build-info</id>
        <goals>
          <goal>build-info</goal>
        </goals>
      </execution>
    </executions>
    <configuration>
    ...
    </configuration>
</plugin>

And in my code:

@Autowired
BuildProperties buildProperties;

...

@GetMapping
public Map<String, String> getVersion() {
    return Map.of(
            "build.name", buildProperties.getName(), 
            "build.version", buildProperties.getVersion(), 
            "build.date", buildProperties.getTime().toString());
}

More information about this plugin goal can be found here: https://docs.spring.io/spring-boot/docs/current/maven-plugin/reference/htmlsingle/#goals-build-info

score -7 · Answer 7 · answered Jul 16 '12 at 08:27

-7

You can parse the pom file with JDOM (http://www.jdom.org/).

answered Jul 16 '12 at 08:27

alien01

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Access maven properties defined in the pom

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