I have a Java EE project packaged as an EAR using Maven, it contains two WAR modules (one web one mobile) in following structure:
EAR
-Entities.jar
-EJB.jar
-Test.jar
-Web.war
-Mobile.war
I have a directory within the Web.war/WEB-INF which contains some .xhtml files that are common to both .wars
Is there any way to instruct Maven to copy this directory into the other .war automatically during build?
eg. I would like..
WebApp-Web/src/main/webapp/WEB-INF/emails/*
to build into
WebApp/WebApp-web/WEB-INF/emails
AND
WebApp/WebApp-mobile/WEB-INF/emails
(I am using the maven-war-plugin btw)
If you add your shared resources into the "parent project" (or elsewhere), you could define the relative path in maven-war-plugin's webResources tag (for both web projects):
<plugin>
<artifactId>maven-war-plugin</artifactId>
<version>2.6</version>
<configuration>
<failOnMissingWebXml>false</failOnMissingWebXml>
<webResources>
<resource>
<directory>../src/main/resources</directory>
<targetPath>WEB-INF</targetPath>
</resource>
</webResources>
</configuration>
</plugin>
Another way would be to use the maven-resources-plugin to copy the files from a relative path at build-time, but I believe the maven-war-plugin approach is better.
Using maven-resources-plugin: (source how-to-get-a-war-package-with-resources-copied-in-web-inf)
I only changed the directory, to make it relative to its parent (ie: ../src/main/resource)
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-resources-plugin</artifactId>
<version>2.4.2</version>
<executions>
<execution>
<id>default-copy-resources</id>
<phase>process-resources</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<overwrite>true</overwrite>
<outputDirectory>${project.build.directory}/${project.artifactId}-${project.version}/WEB-INF/</outputDirectory>
<resources>
<resource>
<directory>../src/main/resources</directory>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>
If you're using a web container that implements Servlet 3.0 or newer then you can make use of the web-fragment feature to build these common files into a jar file that is added to the WEB-INF/lib directory of each web application.
Make web resources such as your xhtml files available by placing them in the META-INF/resources directory of the jar file. Any web.xml configuration that might be associated with these xhtml files can be placed in a META-INF/web-fragment.xml file.
Related
I've got a project A which consists of several modules. This is the "product" itself (server/client application). Furthermore there is another project B which is a specific customer project.
When the client+server *.jar of project A are assembled, resources of project B need to be integrated. I.e. copied into a resource folder.
I'm using the maven-resources-plugin to copy some of the property files:
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>2.7</version>
<executions>
<execution>
<phase>generate-resources</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>${basedir}/target/classes/</outputDirectory>
<overwrite>true</overwrite>
<resources>
<resource>
<directory>../../commonDashboard/src/main/resources/</directory>
<includes>
<include>commonDashboard.*.properties</include>
<include>db.*.properties</include>
</includes>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>
This approach requires a relative dependency between both projects. I want to get rid of the relative path and use a path definition which is provided as VM Options in IntelliJ. So when running the assembly modules I need to access the VM options. I'm not quite sure if this is possible at all. I guess it's not, right? Do you know any other approach to get rid of this dependency between both projects? I don't want to add the options into the maven assembly file itself.
I'm working with a simple project, with the webapp directory in the standard location (src/main/webapp). For some reason, it looks like the maven-shade-plugin doesn't include that in the final generated jar - it will only include artifacts src/main/resources by default.
I have tried to use the following configuration, with no luck:
<configuration>
<artifactSet>
<includes>
<include>*</include>
<include>*:*</include>
</includes>
</artifactSet>
</configuration>
Any suggestions on how to do that without having to move src/main/webapp into src/main/resources/webapp?
Thank you.
Eugen.
Have you tried to update your build section with your resource path ?
<build>
<resources>
<resource>
<directory>src/main/webapp</directory>
</resource>
</resources>
</build>
EDIT
As already said, shade purpose is not to war packages.
You need to put it in src/main/resources/webapp or specify the appropriate location explicitly for it to be caught up properly. Annoying, but it works.
For those asking why this kind of thing makes sense to do, it's a typical setup for a jar with an embedded web server like Jetty.
I use the resource plugin to move the webapp to the classes folder.
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>3.0.2</version>
<executions>
<execution>
<id>copy-resources</id>
<phase>compile</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>${basedir}/target/classes/webapp</outputDirectory>
<resources>
<resource>
<directory>src/webapp/dist</directory>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>
This way I can place the folder with my webapp anywhere, define the destination directory and it is included in my shaded jar.
It sounds you would like to package sources into a package. There exists a maven-source-plugin which will create a jar which contain the source files. Furthermore this is usually done during a release cycle. As already mentioned the maven-share-plugin intention is to package jar files together (their contents; keyword: ueberjar).
I would use the war plugin http://maven.apache.org/plugins/maven-war-plugin/faq.html#attached to create a jar of a webapp project. You have already set <packaging>war</packaging>?
What's the point in adding your webapp resources into a JAR? That's WAR-like.
Change your packaking type to war and it will work.
We are building a jar file from external(to the project) classes.
That works fine but we have not been able to figure out how to also include the external source files. I have tried using the "< includes >" tag but only end up with a manifest file in the final jar when used. I have looked at using the maven-resources-plugin but either I used it wrong or it doesn't work in my case. Here is a copy of our of code:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.4</version>
<executions>
<execution>
<!-- <phase>generate-resources</phase> -->
<phase>clean</phase>
<goals>
<goal>jar</goal>
</goals>
<configuration>
<classesDirectory>${itendant.path}/web/rocket/WEB-INF/classes</classesDirectory>
<finalName>${itendant.jar.name}</finalName>
<outputDirectory>${itendant.jar.path}</outputDirectory>
</configuration>
</execution>
</executions>
</plugin>
Sources? Do you mean external dependencies? These should also be managed with maven, using mvn deploy:deploy-file as described on http://maven.apache.org/plugins/maven-deploy-plugin/usage.html , and imported in your pom.xml.
If you really mean external resources, then a proper resources declaration would be:
<project>
...
<build>
...
<resources>
<resource>
<directory> [your folder here] </directory>
</resource>
</resources>
...
</build>
...
</project>
You can have multiple tags if you have multiple resource directories, of course. Also note that building anything during clean is questionable, as clean is not run every build - package would be a better option.
I am experimenting with Maven and I am trying to mavenize a project originally build with shell scripts.
With the Maven rule-of-thumb: one project, one artifact, I created the following structure:
<PROJECT>
<MODULE-1>
<MODULE-2>
<MODULE-3>
..
<MODULE-N>
<RESOURCES>
<DISTRIB>
The RESOURCES module is structured as follows:
<RESOURCES>/src/main/resources/<MODULE-1>/bin
<RESOURCES>/src/main/resources/<MODULE-1>/lib
<RESOURCES>/src/main/resources/<MODULE-1>/doc
<RESOURCES>/src/main/resources/<MODULE-2>/bin
<RESOURCES>/src/main/resources/<MODULE-2>/lib
<RESOURCES>/src/main/resources/<MODULE-2>/doc
...
<RESOURCES>/src/main/resources/<MODULE-N>/bin
<RESOURCES>/src/main/resources/<MODULE-N>/lib
<RESOURCES>/src/main/resources/<MODULE-N>/doc
The reason for doing it this way was that the resources above are needed at runtime, not compile-time and they are mostly property files, config files and shell scripts to invoke the various jar-files. For the final resources step, I wanted to combine the subdirectories into one global bin/lib/doc directory. However, I do not see an option in the assembly descriptor to strip of the prefix of the modules to get to what I want:
<RESOURCES>/target/resources/bin
<RESOURCES>/target/resources/lib
<RESOURCES>/target/resources/doc
where bin would contain all the files found in the /src/main/resources//bin directory, /src/main/resources//bin directory etc. Similarly for lib, doc.
My question: should I have a:
<MODULE-1>/src/external/resources/bin
<MODULE-1>/src/external/resources/lib
<MODULE-1>/src/external/resources/doc
...
<MODULE-N>/src/external/resources/bin
<MODULE-N>/src/external/resources/lib
<MODULE-N>/src/external/resources/doc
structure instead, that I would then access through a dependency-set? These files should not be part of the jar-file of the various modules and therefore cannot be stored in src/main/resources of their respective projects. Or is what I want to achieve doable by using some other maven plugin instead?
In the DISTRIB module I would combine the output of the RESOURCES module with the JAR-files and dependencies to create a directory structure that would then be used with a packaging tool (Solaris package and WiX installer).
Any help would be appreciated!
In the pom fort he resources module set the source directory to something other than /src/main/resources/ so that it does not copy them to the target folder. Then use the Maven Resources Plugin to copy the resource files to target/bin and target lib etc.
e.g.
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>2.4.3</version>
<executions>
<execution>
<id>copy-resources</id>
<phase>generate-sources</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>target/bin/</outputDirectory>
<resources>
<resource>
<directory>src/external/resources/bin</directory>
<filtering>false</filtering>
</resource>
</resources>
</configuration>
</execution>
<execution>
<id>copy-resources</id>
<phase>generate-sources</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>target/lib/</outputDirectory>
<resources>
<resource>
<directory>src/external/resources/lib</directory>
<filtering>false</filtering>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>`
When I create an executable jar with dependencies (using this guide), all properties files are packaged into that jar too. How to stop it from happening? Thanks.
UPDATE: I tried to exclude them using the Maven resources plugin, but then my application won't find the properties files when I run it in Eclipse (right click on the module -> Run As -> Java Application)
UPDATE: Thanks for your useful answers. I think I'd better spend time to learn Maven, for now I just choose the simplest solution.
To exclude any file from a jar / target directory you can use the <excludes> tag in your pom.xml file.
In the next example, all files with .properties extension will not be included:
<build>
<resources>
<resource>
<directory>src/main/resources</directory>
<excludes>
<exclude>*.properties</exclude>
</excludes>
<filtering>false</filtering>
</resource>
</resources>
</build>
By convention, the directory src/main/resources contains the resources that will be used by the application. So Maven will include them in the final JAR.
Thus in your application, you will access them using the getResourceAsStream() method, as the resources are loaded in the classpath.
If you need to have them outside your application, do not store them in src/main/resources as they will be bundled by Maven. Of course, you can exclude them (using the link given by chkal) but it is better to create another directory (for example src/main/external-resources) in order to keep the conventions regarding the src/main/resources directory.
In the latter case, you will have to deliver the resources independently as your JAR file (this can be achieved by using the Assembly plugin). If you need to access them in your Eclipse environment, go to the Properties of your project, then in Java Build Path in Sources tab, add the folder (for example src/main/external-resources). Eclipse will then add this directory in the classpath.
This calls exactly for the using the Maven JAR Plugin
For example, if you want to exclude everything under src/test/resources/ from the final jar, put this:
<build>
<plugins>
<!-- configure JAR build -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.3.1</version>
<configuration>
<excludes>
<exclude>src/test/resources/**</exclude>
</excludes>
</configuration>
</plugin>
...
Files under src/test/resources/ will still be available on class-path, they just won't be in resulting JAR.
Put those properties files in src/test/resources. Files in src/test/resources are available within Eclipse automatically via eclipse:eclipse but will not be included in the packaged JAR by Maven.
Exclude specific pattern of file during creation of maven jar using maven-jar-plugin.
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.3</version>
<configuration>
<excludes>
<exclude>**/*.properties</exclude>
<exclude>**/*.xml</exclude>
<exclude>**/*.exe</exclude>
<exclude>**/*.java</exclude>
<exclude>**/*.xls</exclude>
</excludes>
</configuration>
</plugin>
Do you mean to property files located in src/main/resources? Then you should exclude them using the maven-resource-plugin. See the following page for details:
http://maven.apache.org/plugins/maven-resources-plugin/examples/include-exclude.html
Another possibility is to use the Maven Shade Plugin, e.g. to exclude a logging properties file used only locally in your IDE:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>${maven-shade-plugin-version}</version>
<executions>
<execution>
<phase>package</phase>
<goals>
<goal>shade</goal>
</goals>
<configuration>
<filters>
<filter>
<artifact>*:*</artifact>
<excludes>
<exclude>log4j2.xml</exclude>
</excludes>
</filter>
</filters>
</configuration>
</execution>
</executions>
</plugin>
This will however exclude the files from every artifact, so it might not be feasible in every situation.
When I create an executable jar with dependencies (using this guide), all properties files are packaged into that jar too. How to stop it from happening? Thanks.
Properties files from where? Your main jar? Dependencies?
In the former case, putting resources under src/test/resources as suggested is probably the most straight forward and simplest option.
In the later case, you'll have to create a custom assembly descriptor with special excludes/exclude in the unpackOptions.
here is another solution to exclude all files in resources folder, the final configuration looks like:
<build>
<!-- exclude all files in resources-->
<resources>
<resource>
<directory>src/main/resources</directory>
<excludes>
<exclude>**/**</exclude>
</excludes>
<filtering>false</filtering>
</resource>
</resources>
<!-- other configurations/plugins in the pom.xml-->
</build>
or we can use includes to only package some file or folder.
But this method has a side effect. IDE will also exclude the resource files in target/classes folder. maven-jar-plugin only affect jar file.
I found a better solution to execludes resourses folder using maven-jar-plugin, here we use includes:
<build>
<plugins>
<!-- configure JAR build -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.3.1</version>
<configuration>
<includes>
<include>**/*.class</include>
</includes>
</configuration>
</plugin>
...