I've searched the archives and have not found an answer to this question. I have a profile whose activation I'd like to trigger based on the "non-emptiness" of an environment variable.
When I run mvn help:active-profiles with that environment variable set, it reports that no profiles are active. My settings.xml file, where the profile is defined, looks like this (in part): <?xml version="1.0" encoding="UTF-8"?> <settings 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/settings-1.0.0.xsd "> [stuff snipped here] <profiles> <profile> <id>Development Server</id> <activation> <property> <name>env.GLASSFISH_HOME</name> </property> </activation> <properties> <appserver.autodeploy.dir>${env.GLASSFISH_HOME }/domains/domain1/autodeploy</appserver.autodeploy.dir> </properties> </profile> </profiles> </settings> I followed the example of the maven.apache.org page on build profiles<http://maven.apache.org/guides/introduction/introduction-to-profiles.html>, which says, in part: * Note* : Environment variable FOO would be set like env.FOO . ...but this only works if I do: mvn help:active-profiles -Denv.GLASSFISH_HOME=foobar ...i.e. it is treated (by the help:active-profiles plugin, at any rate) as a regular property, not an environment variable. How, if not that way as documented, do I activate a profile based on the presence of a non-empty environment variable value? Thanks, Laird P. S.: The output of mvn --version is below: Maven version: 2.0.8 Java version: 1.6.0_04 OS name: "windows xp" version: "5.1" arch: "x86" Family: "windows"
