I have a web project with following structure:
app
pom.xml
–modules:
app-api (contains only OpenAPI documentation used to generate REST endpoints)
pom.xml
app-service (contains the actual business logic with services/controllers/etc)
pom.xml
I am trying to manually run SonarCloud.
I put the following settings in the parent pom.xml
<sonar.organization>company-account</sonar.organization>
<sonar.host.url>https://sonarcloud.io</sonar.host.url>
<sonar.projectKey>project-name</sonar.projectKey>
All of the above are taken according to what the Sonar web portal is recommending.
I added the SONAR_TOKEN variable as well.
At the moment I am getting:
"Project ‘…’ can’t have 2 modules with the following key: …
I tried a bunch of other combinations, like putting the maven properties in the service submodule.
It either does not recognize the project or gives me a stackoverflow error.
I tried passing the project key as a parameter:
mvn verify org.sonarsource.scanner.maven:sonar-maven-plugin:sonar -Dsonar.projectKey=the-project-key
It just gives another error:
Project not found. Please check the ‘sonar.projectKey’ and ‘sonar.organization’ properties, the ‘SONAR_TOKEN’ environment variable, or contact the project administrator
Do you have any idea what am I doing wrong?
Related
So I have basic multi-module Spring Boot project. The goal, that I had was to build executable jar and pass additional properties with the help of -Dloader.path=....
For some reason (if I understand purpose of this argument) loader.path is being ignored completely.
My project structure is following:
\-
|--conf
|---default
|--pets-api
|--pets-app (this module contains the Main-Class)
|--pets-domain
|--pets-infrastructure
Since no custom active profile is being passed it uses "default". Jar contains application-default.propeties file, that has single configuration server.servlet.context-path=/v1.
/conf/default location has 2 properties files:
application.properties
random.properties - this is bind to #ConfigurationProperties(prefix = "...") inside application
When I run it normally everything is fine java -jar pets-app-0.0.1-SNAPSHOT.jar. It just uses application-default.properties file and that is it.
Now when I am trying to utilize -Dloader.path argument as in java -Dloader.path=PATH/TO/conf/default -jar pets-app-0.0.1-SNAPSHOT.jar it starts application same as before, as if I am not adding 2 more file to classpath.
What is used:
Java 17
Spring Boot 2.6.12
Gradle
Did anyone come across this as well? Any suggestion on how to resolve it?
PS. If there is need to see the code, I can upload it to GitHub.
I'm working on a multi module spring-boot project to build a REST API. Here is my project structure:
Parent project (packaging is pom)
core module (#SpringBootApplication + handle path like / or /status)
restControllerA module (Handle path like /routeA/*)
restControllerB module (Handle path like /routeB/*)
Everything is working in this project :)
In another non Spring project I would like to reuse a service of restControllerB. This service return the result of the request body validation.
First I try to add the restControllerB.jar as a dependency to this new project... But this jar does not contain its depedencies (who are in the fatJAR "core.jar"). When I run the project, I get a lot of ClassNotFoundException.
How can I manage to reuse this service as a dependency ? I thought to create a validator module which implements the validatorService interface, but I'm not sure if it is the best solution.
After few hours googling, It seems that creating an external librairy is the right choice. I create an external module and add it as a dependecy to restControllerB.
I am running Jenkins ver. 2.60.2 and it doesn't seem possible, within a Maven Job, to define a local repository not in /usr/share/tomcat7/.m2.
Here are my attempts:
I created a Global Maven settings.xml and a Settings file with the Config File Management Plugin, that contains:
<settings>
<localRepository>/srv/maven/.m2/repository</localRepository>
...
</settings>
I Created a new Maven Project. Tried to make the Job see that file by attempting all of the following:
a) Defining either Settings file or Global settings file (I created two identical files) within the build step:
b) Adding a Pre-step Provide Configuration files, and then using the variable MY_SETTINGS either in the Goals and options or MAVEN_OPTS:
c) Use the Provide Configuration files within the build environment (and using the MY_SETINGS in the same way as in the previous step.:
However, none of these seems to work. The job always fails, trying to use the default maven repository location (/usr/share(tomcat7/.m2) - which I have no idea how to re-define:
provisioning config files...
copy managed file [MYFILE settings] to file:/srv/webapps/jenkins/jobs/testJob/workspace#tmp/config3408982272576109420tmp
provisioning config files...
copy managed file [MYFILE settings] to file:/srv/webapps/jenkins/jobs/testJob/workspace#tmp/config2203063037747373567tmp
Parsing POMs
using global settings config with name MYFILE settings
Replacing all maven server entries not found in credentials list is true
Deleting 1 temporary files
ERROR: Failed to create /usr/share/tomcat7/.m2
Finished: FAILURE
Do you know how to make this work within a Maven Job type in Jenkins?
I'm currently writing a custom maven plugin for generating a XML file in a multi-module maven project.
My maven structure is pretty standard: one parent project and a module by project components in the parent project folder:
-- Parent
-- module A
-- module B
-- module C
I need to list, by module, a set of classes flagged by a custom annotation.
I already wrote a set of custom annotations and an annocation processor to create a XML file at compile time in the corresponding module output directory (${project.build.outputDirectory}) .
Now i need to merge each module XML into one file, but i don't know how to access each modules from within my maven plugin except having each path set as parameters (i don't like this method).
Any idea on how to do this ?
Does maven plugins can traverse project modules ?
Thank you in advance.
To get the list list of all projects you can use:
List<MavenProject> projectList = MavenSession.getProjectDependencyGraph().getSortedProjects()
If one of your goals is correctly executed you will get everything you need. Every MavenProject contains a getBaseDir() etc.
After some researches, it seems that MavenProject.getCollectedProjects() will return the list of projects beeing manipulated by a goal execution in a multi-module project.
i created the basic Spring +Jdbc seeing the following Spring + JDBC example
For this i used the Spring Template Project
i added all the classes inside src/main/java/
But now as i run the project by rightclicking the App.java Class ->run as->run on server.
it gives me the following error.
Requested Resource is not found.
url above is /app/WEB-INF/classes/com/shr/common/App.java
Now do i need to add the classes to WEB-INF/classes folder?
how should i run my project. please help me.
That is not a web application. You have to start it as Java Application (run as -> Java Application)