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build-helper-maven-plugin adding additional source

I am trying to add an additional source folder to my current maven project by using build-helper-maven plugin.
This source folder contains some common classes, like utility classes.
For that, here is my relevant pom.xml
<plugin>
<groupId>org.codehaus.mojo</groupId>
<artifactId>build-helper-maven-plugin</artifactId>
<version>1.9.1</version>
<executions>
<execution>
<id>add-source</id>
<phase>generate-sources</phase>
<goals>
<goal>add-source</goal>
</goals>
<configuration>
<sources>
<source>C:/Users/CommonIncludes/src</source>
</sources>
</configuration>
</execution>
</executions>
</plugin>
Eclipse is showing the following error :
Build path entry is missing.
Project->Right Click->Java build path->Source->
Project/Users/CommonIncludes/src(missing)
Here the additional source location : "C:/Users/CommonIncludes/src" is outside of the workspace of the current project. But eclipse always treating this as a location from current project.
I am using Eclipse 4.3 and m2e.
How can I overcome this error through MAVEN, so that Eclipse can identify the linked source from correct location.? Or is there any alternate way to do this using MAVEN?
Any help will be greatly appreciated. Thanks in advance.

Found it in an alternate way..works great.!
Steps include
1. Removed build-helper-maven-plugin from pom .
2. Created another maven project and added the common classes in it. Added maven-source-plugin in this pom to generate sources.
3. To the same pom, added maven-dependency-plugin to copy this generated sources to the desired location (My project's src/main/java).
4. Run a maven build for common classes project.
Now the common code in my project. Thanks.

I had a lot of inconsistent trouble with this plugin.
by far the simplest workaround in my case was simply to apply the addition via the standard 'resources' options.
hope this helps someone.
<build>
<resources>
...
<resource>
<directory>${project.build.directory}/generated-sources</directory>
<targetPath>${project.build.outputDirectory}</targetPath>
</resource>
</build>

maven copy a particular dependency jar to a specific directory in the .war file

I am trying out a Simple Java Web Start project based on the Oracle Tutorial. I am using maven to package it as a webapp and deploy it to application server. The full source code is available here
https://github.com/KiranMohan/dynamic-tree-javaws-sample-project
The maven project structure is like
parent
|--lib
|--webapp
The webapp module is a maven war module. It is required to package lib.jar at the root of webapp.war. NOT under WEB-INF/lib.
How to achieve this in maven?

I found that the right way to do this is to use the maven-dependency-plugin.
Since "lib.jar" is not used in the compile phase of "webapp" module, it is only a package time dependency. Using maven-dependency-plugin, I can copy lib.jar to any required directory during the prepare-package phase. The maven-war package would then include the lib.jar in the .war package.
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-dependency-plugin</artifactId>
<executions>
<execution>
<id>copy</id>
<phase>prepare-package</phase>
<goals>
<goal>copy</goal>
</goals>
<configuration>
<artifactItems>
<artifactItem>
<groupId>[ group id ]</groupId>
<artifactId>[artifact id]</artifactId>
<version>[ version ]</version>
<outputDirectory>${project.build.directory}/${project.artifactId}</outputDirectory>
</artifactItem>
</artifactItems>
<!-- other configurations here -->
</configuration>
</execution>
</executions>
</plugin>
Update:
There is a webstart-maven-plugin that does a better job of packaging javaws applications. See my sample project
https://github.com/KiranMohan/dynamic-tree-javaws-sample-project
for details

Well how i said in the comments for sake of readability here is a part of the answer:
Since Maven will always store the dependencies of a web project under its WEB-INF/lib folder by default i (i am no Maven expert ...) would try to place my lib.jar inside the /target folder of the project before the phase package is executed.
NOTE: I havent tried it out so you will have to adjust the paths - expecially the output path so your lib.jar is placed properly to be packed into the root of the war (e.g. if you open your war there will be a lib.jar next to folders such as WEB-INF).
<!--
lets assume the root of my project would be under C:/devjba/projectX this equals the maven
variable ${project.basedir}.
from there the output-directory would be located under C:/devjba/projectX/target which equals the
maven variable ${project.build.directory}. This is the location a .war would be placed in after
the build
lets assume the required jar lib.jar is located under C:/devjba/projectX/misc which would equal to
the expression: ${project.basedir}/misc
-->
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>2.6</version>
<executions>
<execution>
<id>foo</id>
<!-- may adjust to another phase before package but make sure your plugin is bound to a phase
because otherwise it wont be invoked during build! Now its bound to the first phase of the
default lifecycle for packaging type war -->
<phase>process-resources</phase>
<goals>
<!-- use the copy-resources goal of this plugin - it will copy resources :) -->
<goal>copy-resources</goal>
</goals>
<configuration>
<!-- this points to /target of the current project, you may adjust it to wherever it must be placed to be packed into the root of the war (just try&error) -->
<outputDirectory>${project.build.directory}</outputDirectory>
<resources>
<resource>
<!-- this points to a folder /misc under the project root where we expect the lib.jar -->
<directory>${project.basedir}/misc</directory>
<!-- unless you specify what to include anything of the above directory will be included -->
<includes>
<include>lib.jar</include>
</includes>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>
As i said i have no experience in signing JARs at all but there is a plugin called maven-jarsigner-plugin which i guess will do the job (i would sign it, then move it, then package the war) with a manual - i recomend you try to configure it according to my "example configuration of the maven-resource-plugin and post a new question directly containing your two plugin configurations. Dont forget to link to this question in that case. And also leave this question open so someone with a better approach may correct my way).

Maven: creating 2 war files each with different web.xml?

I need to create two WAR files with two different web.xml files using Maven. The WAR files are the same otherwise.
The WAR files are created as expected, however, the second file has the same web.xml file as the first.
How can I force the Maven WAR plugin to use the second file for the second WAR? I cannot use profiles since I need both files created in one pass.
My current code:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-war-plugin</artifactId>
<version>2.2</version>
<configuration>
<warSourceDirectory>${basedir}/WEB</warSourceDirectory>
<warName>main</warName>
<webXml>${basedir}/CONF/web.xml</webXml>
</configuration>
<executions>
<execution>
<id>hist-war</id>
<phase>package</phase>
<goals>
<goal>war</goal>
</goals>
<configuration>
<warName>hist</warName>
<warSourceDirectory>${basedir}/WEB</warSourceDirectory>
<webXml>${basedir}/CONF/hist.web.xml</webXml>
</configuration>
</execution>
</executions>
</plugin>

Use the overlay feature of the Maven WAR plugin.
You need two WAR projects. The first is your original one, with its web.xml file in the correct place. This doesn't have any specific needs in terms of configuration. Just ensure that it builds as you expect.
The second only contains its POM and the alternate web.xml in the WEB-INF directory. This project's POM must contain a dependency on the first project with type specified as war and scope specified as runtime. You should not need to explicitly configure the Maven WAR plugin.

seems there are a variety of gotchas using the war plugin. in this case, i would just use 2 separate sub-modules to build the appropriate wars.

Maven web resources question

I have a Maven web project and I have some CSS and Javascript files under the src\main\webapp\ folder. I constantly make changes to those files and would like to see my changes quickly. If I run maven install, it takes ages due to project dependencies. Sometimes all I want to change is one line of code in my CSS file and do not want to recompile everything else. I have a maven plugin that publishes my output war file to my JBoss instance. Ideally, I would like to run a maven execution script that will quickly copy my web resources to the output folder and reploy the changed war file without recompiling everything else.
I tried invoking the generate-resources goal but that doesn't seem to look in the src\main\webapp directory as it is expecting my resources to be under the src\main\resources folder. What am I missing here?
Thanks

If you want to add more resources to be copied during the generate-resources plugin, you can change the resources folders used by your build. The project.build.resources property controls which folders are searched for resources. You could add:
<project>
...
<build>
...
<resources>
<resource>
<directory>src/main/resources</directory>
</resource>
<resource>
<directory>src/main/webApp</directory>
<includes>
<include>*.css</include>
<include>*.js</include>
You would then run mvn resources to copy the files.
This approach is that these files will always be copied during the resources phase of any build. You can get around this by using the copy-resources goal instead of resources. In this case you would use the following configuration:
<build>
<plugins>
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>2.5</version>
<executions>
<execution>
<id>copy-web-resources</id>
<!-- here the phase you need -->
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>${project.build.directory}/webApp</outputDirectory>
<resources>
<resource>
<directory>src/main/webApp</directory>
<includes>
<include>*.css</include>
<include>*.js</include>
You could then run mvn resources:copy-resources to copy the files.

I think you could accomplish this by using the war:war goal. This should generate a war file in the output folder for you without re-compiling the source.

How to change the default output from a Maven 2 / Cobertura instrument goal?

when i instrument my classes using Maven 2 using the command
mvn cobertura:instrument
The output (the instrumented classes) are put in \target\generated-classes. Is there a way to change the output location to \target\classes?
I checked the instrumentation tasks of the cobertura-maven plugin but this does not give me a solution sofar.

You have not said why you want to overwrite the default location, but I assume it is so that you can use the instrumented classes from another project, or perhaps include them in a web archive or something similar.
I added the following to my pom.xml
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<executions>
<execution>
<id>instrumented-classes</id>
<goals>
<goal>jar</goal>
</goals>
<phase>package</phase>
<configuration>
<classifier>instrumented</classifier>
<classesDirectory>${project.build.directory}/generated-classes/cobertura</classesDirectory>
</configuration>
</execution>
</executions>
</plugin>
This makes maven generate an additional jar file called projectname-instrumented.jar
It is then possible to depend on this jar file from any other pom (including for example a web module) using
<depends>
<group>mygroup</group>
<project>projectname</project>
<version>1</version>
<classifier>instrumented</classifier>
</depends>
I did not test this 100% but have used similar mechanisms in the past

As far as I understand, the instrumented classes are only needed by cobertura for report generation. If you create them in target/classes, they will overwrite the original class files.
If you need the instrumented files in a jar as a result, you can configure the maven-jar-plugin to pick up the files from the target/generated-classes directory instead of or in addition to the files from the standard ${build.project.outputDirectory}.
Edit
Have a look at the maven-jar-plugin description. To only use target/generated-classes, the following addition to your POM should work - try it and modify it to your needs:
<project>
...
<build>
<plugins>
...
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.3</version> <!-- replace with correct version nbr! -->
<configuration>
<includes>
<include>${project.build.directory}/generated-classes/**/*.class</include>
</includes>
<excludes>
<exclude>${project.build.directory}/classes/**/*.class</include>
</excludes>
</configuration>
</plugin>
...
</plugins>
</build>
...
</project>
${project.build.directory} points to your target folder, ${project.build.ouputDirectory} to target/classes. I do not know if you can simply set ${project.build.ouputDirectory} to a new value - have a look at the this chapter of the maven book, maybe you find some hints
Edit 2
Alternativly or additionally you can use maven to copy the files from target/generated-classes to target/classes after coberture:instrument has finished. This question has one answer with an example POM (fragment), you just have to identify the correct phase (process-resources is definitely too early for your case)

Did you try "mvn cobertura:instrument install"? It will generate a jar file including all the cobertura version classes.
If you want to change back original version, just run the command without "cobertura:instrument".

I just implemented the solution proposed by Andreas_D, modified my pom and uses the maven-resources-plugin. So on some stage of my build the Cobertura generated files are copied to the /target/classes directory.

You can configure it using <classesDirectory>[Your DIR]</classesDirectory>

In cobertura-maven-plugin version 2.4 this is still not supported. I've just created an improvement ticket, patch is attached to the ticket.