Out-of-tree build with maven. Is it possible?

Question

I start learning packaging for several distros (currently Cygwin and Debian).

They have requirement to build system to allow out-of-tree build (syno

Answer 1

You could do like this to get it in your current working directory:

<project xmlns="http://maven.apache.org/POM/4.0.0"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>

    <groupId>com.stackoverflow</groupId>
    <artifactId>Q13173063</artifactId>
    <version>1.0-SNAPSHOT</version>

    <name>${project.artifactId}-${project.version}</name>

    <properties>
        <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
        <buildDir>${user.dir}</buildDir>
    </properties>

    <build>
        <directory>${buildDir}</directory>
    </build>
</project>

Then you can issue

mvn -f my-app/pom.xml compile

And it will give you your classes in the current working directory.

And easily change to another output directory:

mvn -f my-app/pom.xml -DbuildDir=/tmp/build compile

Answer 2

It might be as simple as having a

<build>
    <directory>/your/build/directory</directory>
</build>

in your pom.xml. /your/build/directory need not be in the source tree and can be parameterized using the usual ${...} syntax.

Cheers,