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Each pom.xml have:
<groupId>com.vendor</groupId>
<artifactId>product</artifactId>
<version>1.0.13.0-dev</version> <!-- THIS FIELD -->
<packaging>war</packaging>
version piece is useful to bunch with application so user can see and report them.
But I dislike code duplication and look for a way to avoid putting property file with version info to VCS (Git/SVN) because each time I bump version I must commit into two places (pom.xml and version.properties).
Which way can I make version properties from pom.xml programmatically available in Java application?
326
Open your pom.xml file and add below under <build>. Note: I’ve added 3 plugins below below. maven-resources-plugin : The Resources Plugin handles the copying of project resources to the output directory. The main resources are the resources associated to the main source code.
Alternatively, maven stores the Implementation-Version in the MANIFEST.MF file when building a jar. If you dont mind not having a version until you run it from a jar, you can use this to get the application version. Java allowes for easy access to this using:
The upside of this approach is that you always have a version number even when you start your application from your IDE, also by putting your variable in info.app.version, if you use Spring boot actuator, your version will be automatically available under the /info endpoint. However, this approach does require quite a bit of configuration.
Spring. Spring Boot. There are a few ways to get your application version with Spring Boot. One approach is using Maven resource filtering to add the version number as an enviroment variable in the application.yml during the build. Another approach is to use the Implementation-Version stored in the manifest file.
Found a solution to use the maven-war-plugin to be more elegant.
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-war-plugin</artifactId>
<version>2.3</version>
<configuration>
<webResources>
<resource>
<targetPath>WEB-INF/views/jsp</targetPath>
<directory>${basedir}/src/main/webapp/WEB-INF/views/jsp</directory>
<includes>
<include>footer.jsp</include>
</includes>
<filtering>true</filtering>
</resource>
</webResources>
</configuration>
</plugin>
footer.jsp then had:
Application Version: ${project.version}
My personal production solution below.
pom.xml:
<?xml version="1.0" encoding="utf-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<version>1.0.13.0-dev</version>
<build>
<resources>
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>
</resources>
...
base.jsp:
<%@taglib prefix="spring" uri="http://www.springframework.org/tags"%>
<!DOCTYPE html>
<html>
<head></head>
<body>
<a href='<c:url value="/"/>'>
<spring:eval expression="@props.getProperty('version')"/>
</a>
<jsp:include page="${contentPage}" />
</body>
</html>
proj.properties:
version=${project.version}
applicationContext.xml:
<bean id="props" class="org.springframework.beans.factory.config.PropertiesFactoryBean" autowire="byName" >
<property name="location" value="classpath:proj.properties"/>
</bean>
<bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="location" value="classpath:proj.properties"/>
</bean>
or just
<util:properties id="props" location="classpath:proj.properties"/>
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