Get pom.xml property from CommandLine
command-line, java, maven, pom.xml, properties
Solution
I know the question is old but I spent some time looking for this.
To filter output you may use flags "-q -DforceStdout" where "-q" prevents output and "-DforceStdout" forces outputting result of plugin. E.g.:
BUILD_VERSION=$(mvn help:evaluate -Dexpression=project.version -q -DforceStdout)
echo $BUILD_VERSION
will result in printing version of project from POM.
Second important problem I had was accessing "properties" which is explained in Nick Holt comment. To access properties you just access them directly
<project ...>
<version>123</version>
(...)
<properties>
(...)
<docker.registry>docker.registry.com</docker.registry>
(...)
</properties>
(...)
</project>
WRONG
mvn help:evaluate -Dexpression=project.properties.docker.registry -q -DforceStdout
OK
mvn help:evaluate -Dexpression=docker.registry -q -DforceStdout
Problem
I am currently writing a script for our Gitlab CI that automatically uploads files to an NFSShare folder in the network. Since I want to organize the builds and we're using maven, I thought I could "easily" get the project name from the pom.xml. Is there a way to get the properties available from within a pom.xml through a command-line tool or something? My only other way I could think of was "regex-grepping the value by hand" - not a very clean solution in my opinion. I already found the the properties plugin, but it only seem to ADD new properties through actual .properties files... Any help would be much appreciated!