I have a Maven project that downloads some test files into its build directory ./target/files
. These files should then be available to a servlet, which I can easily achieve by hardcoding the full path as an <init-param>
of the servlet:
<servlet>
<servlet-name>TestServlet</servlet-name>
<servlet-class>my.package.TestServlet</servlet-class>
<init-param>
<param-name>filepath</param-name>
<param-value>/home/user/testproject/target/files</param-value>
</init-param>
</servlet>
How can I avoid hardcoding the full path and use a dynamic parameter replacement instead? I tried the following, but it did not work:
<param-value>${project.build.directory}/files</param-value>
Add to your pom section:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-war-plugin</artifactId>
<configuration>
<webResources>
<resource>
<filtering>true</filtering>
<directory>src/main/webapp</directory>
<includes>
<include>**/web.xml</include>
</includes>
</resource>
</webResources>
<warSourceDirectory>src/main/webapp</warSourceDirectory>
<webXml>src/main/webapp/WEB-INF/web.xml</webXml>
</configuration>
</plugin>
See Maven: Customize web.xml of web-app project for more details