In Maven, how output the classpath being used?

Question

For my current purposes I have a Maven project which creates a war file, and I want to see what actual classpath it is using when creating the war.

Answer 1

The command mvn dependency:list will list the classpath with all the jars used for compilation, runtime and test in the following format:

INFO] --- maven-dependency-plugin:2.8:list (default-cli) @ MyProject ---
[INFO]
[INFO] The following files have been resolved:
[INFO]    org.netbeans.api:org-openide-filesystems:jar:RELEASE80:compile
[INFO]    org.netbeans.api:org-netbeans-modules-queries:jar:RELEASE80:compile
[INFO]    org.netbeans.api:org-netbeans-api-progress:jar:RELEASE80:compile
[INFO]    org.netbeans.api:org-openide-dialogs:jar:RELEASE80:compile
[INFO]    org.apache.derby:derby:jar:10.11.1.1:compile
[INFO]    org.netbeans.api:org-openide-windows:jar:RELEASE80:compile

The only requirement is that the compilation is finished. It doesn't work if the compilation isn't ran.

An other command is The command mvn dependency:tree.

Answer 2

To get the classpath all by itself in a file, you can:

mvn dependency:build-classpath -Dmdep.outputFile=cp.txt

Or add this to the POM.XML:

<project>
  [...]
  <build>
    <plugins>
      <plugin>
        <groupId>org.apache.maven.plugins</groupId>
        <artifactId>maven-dependency-plugin</artifactId>
        <version>2.9</version>
        <executions>
          <execution>
            <id>build-classpath</id>
            <phase>generate-sources</phase>
            <goals>
              <goal>build-classpath</goal>
            </goals>
            <configuration>
              <!-- configure the plugin here -->
            </configuration>
          </execution>
        </executions>
      </plugin>
    </plugins>
  </build>
  [...]
</project>

From: http://maven.apache.org/plugins/maven-dependency-plugin/usage.html

Answer 3

Added the missing arg for scope to Janiks answer. Now it is complete. You are welcome.

mvn -q exec:exec -Dexec.classpathScope="compile" -Dexec.executable="echo" -Dexec.args="%classpath" 

Answer 4

or call "mvn -e -X ...." and check the output...

Answer 5

This is a single command solution but does compile the code.

mvn -e -X -Dmaven.test.skip=true clean compile | grep -o -P '\-classpath .*? ' | awk '{print $2}'

• It's based in Philip Helger's previous answer (Thanks by the way)

Shell script example usage

MAVEN_CLASSPATH=$(mvn -e -X -Dmaven.test.skip=true clean compile | grep -o -P '\-classpath .*? ' | awk '{print $2}')

I used a variation of it in a shell script to run a standalone main() (for Hibernate schema generation) this way

#/bin/bash

MAVEN_TEST_CLASSPATH=$(mvn -e -X clean package | grep -o -P '\-classpath .*?test.*? ')

java $MAVEN_TEST_CLASSPATH foo.bar.DBSchemaCreate

File output example

mvn -e -X -Dmaven.test.skip=true clean compile | grep -o -P '\-classpath .*? ' | awk '{print $2}' > maven.classpath

Answer 6

This command outputs the classpath on Mac and Linux:

mvn -q exec:exec -Dexec.executable=echo -Dexec.args="%classpath"

Having the result printed and not saved into a file might be useful, for instance, when assigning the result to a variable in a Bash script. This solution runs on Mac and Linux only, but so do Bash shell scripts.

In Windows (e.g. in BAT files), where there is no echo executable, you will need something like this (untested):

mvn -q exec:exec -Dexec.executable=cmd -Dexec.args="/c echo %classpath"

---

Alternatively, you can just execute java program with the classpath:

mvn -q exec:exec -Dexec.executable=java -Dexec.args="-cp %classpath Main"

Or even like that (it will use the correct classpath automatically):

mvn -q exec:java -Dexec.mainClass="Main" 

However, both these alternative approaches suffer from Maven adding its error messages when your program fails.