I have the following Maven project structure.
project
-- src
-- main
-- java
-- models
-- resources
I want to create and deploy a jar project-models.jar containing everything inside the folder models and nothing else. Since I'm not very familiar with Maven, I'd really appreciate if you could provide me some example.
models belongs to resources (They should neither be compiled nor tested, should they?)
See How to create an additional attached jar artifact from the project:
Specify a list of fileset patterns to be included or excluded by adding <includes>/<include> or <excludes>/<exclude> and add a classifier in your pom.xml.
Note: the jar-plugin must be defined in a new execution, otherwise it will replace the default use of the jar-plugin instead of adding a second artifact. The classifier is also required to create more than one artifact.
Related
I have a Spring Integration requirement , where I need to externalize the libraries and properties file from my war file. I am able to achieve this through Maven assembly plugin, where i create a zip file which may contain
lib/*.jar
properties/{artifactId}/*.properties
The reason I am adding the artifact Id to the path is, I will be creating 100s of wars in future and would need to distinguish between them.
This wars will not contain Web.xml and the Initializer is part of one of my libraries file.
THe Initializer should know the artifactId in order to load the correct properties.
With maven, the maven artifact details gets published to
META-INF/maven/${groupId}/${artifactId}/pom.properties
META-INF/maven/${groupId}/${artifactId}/pom.xml
If I could move these files to
META-INF/maven/pom.properties
META-INF/maven/pom.xml
My application would be able to read the artifact id from pom.properties.
I need help in achieve this.
Or if there are any other approach please help in solving the issue.
The Maven archiver component does that: see here at addMavenDescriptor element. It doesn't seem to be possible to customize the paths of these files.
But I guess every property you need can just be placed in a specific file and so you just have to create a resource file (properties like) containing all the information you want and let Maven filter that file for you.
My maven top level project refers to a common-db project. In this project I have a spring file which defines the DB parameters.
However, I want the top-level project to define the DB parameters through the profile and inject these into the spring config file in /src/main/resources.
The top-level project only does the filtering on its own /src/main/resources files and ignores those located in the JAR dependencies.
How can I do this?
So you want to depend on common-db but then modify its contents to change the parameters in the config file? Ok, if you really want to do that, you could do something convoluted where you use dependency:unpack to expand the common-db jar, then overwrite / filter its contents, and then use a custom jar:jar execution to re-jar up the dependency and ship it with your application.
But, wow - why would you jump through all these hoops? Like #hoaz suggested, just place your application-specific config in the same classpath location so that it is loaded before common-db's default configuration. This is the convention followed by many, many Java libraries.
I am working on a java maven project with several modules.
I am facing issues with sharing configuration files from one module to its dependency.
For instance i have a module named utils which holds a log.properties file and i would like to use it in another module named gui. What is the best practice to do this ?
Currently we put the log.properties in a config directory as Maven standard layout suggest it, and it is not included in the jar file. Is it correct ? Should I put it in resources instead ?
I use assembly plugin to copy it to a common config directory, this works well, but when I try to build each module individually the config file cannot be reached. How can i solve this ?
Thanks for your help,
Pierre.
You should put your configuration in src/main/resources/config/. This way it will be included in the jar by default. The maven convention is that only src/main/java and src/main/resources are contained in the final jar by default.
Making property files directly accessible to other modules is not a good practice. You should provide a service in the module owning the configuration that is the only place where those files are accessed. This service will be able to give configurations to other modules. Otherwise you violate the single responsibility principle.
You can configure maven to include the src/main/config directory in the built artifact by specifying it in the resources section of the pom as described here.
I have a couple of java modules set up in IDEA and I am wanting to mavenize them. These java modules use classes from one another.
I was not quite sure how I should take up this and I decide to add modules on a maven project using IDEA. Hence first I created a maven project, let's name it pm1 which has a class let's name it TempClass1. Now this class can be used in other maven project. Hence I added another maven module - pm11 and tried to use TempClass1 with in pm11. It worked and I notices that IDEA had added module dependency of pm1 in pm11. So whole structure looks as -
But now when I do mvn test from pm11 then it fails with error message package package1 does not exist and it looks to me that it is because package1 is in a different maven project. And I am not sure how I could use classes which reside in a different maven project. I hope I am clear in my question.
You can use classes of other maven projects, as long as there's a proper maven dependency defined in pom.xml. Ensure that the dependency is defined and its' scope is either undefined or relevant (You may have problems if the scope is provided for example).
We are using Maven(3.0.3) as build tool and we need to have different version for different environments (DEV , TEST, QA ) . If we pass version property value during build time based on environment , the installed POM doesn't have the passed property values instead it still has the ${app-version} string.
I saw already there is a bug for this http://jira.codehaus.org/browse/MNG-2971
Is there any other alternative ,because we cannot different POM file for different environments ,which will be hard to maintain..
Thanks
Vijay
Create different artifacts for the environments and use the parameter as a classifier. The pom is the same for all three artifacts but the classifier separates them.
Apparently Maven does not make any variable/property substitution when installing the POM. It is installed as is, that is the principle. You'd better not read any properties from POM (unless this is e.g. version number), bout you should configure your properties in external file (one per stage, e.g. dev.properties, test.properties, ...) and then configure Maven profiles (again, one per stage) and invoke Maven like mvn -Pdev depending on what you want to build. In profile you can package your final application with whatever properties you like (e.g. with the help of build-helper-maven-plugin:add-resource or maven-antrun-plugin + copy rule).
Alternatively you can filter your resources. For example, you can filter your Spring context XML file, which refers the properties file (so you package all property files, but Spring will refer only some specific). Or you can filter another properties file from which you will learn what is the "main" properties file to use (double indirection).
You should create the archives for your different targets within a single build and use as already mentioned the classifier to separate those artifacts from each others.