zip a file with maven to a NO java application directory - node.js

Question

I don´t know if this is possible . This is what I want to do.
I have some jenkins pipelines that build a VUE.js application using node js.
using the "nmp run build" command. The result of this build, is the directory name "Static" and a index.html file. After that I zip those 2 into a file.zip which I upload to Artifactory, so later it can be downloaded unziped and put it into a docker file to build an image and later a container in azure (ACI).
I want to implement some versioning now of those zips which I already done with other apps but in Java with maven + POM + Maven+Metadata+Plugin + jenkins+ artifactory where I have 2 jobs. 1 job to build with maven and push to artifactory the file.war, and other job to choose the file.war from artifactory with "build with parameters" option.
I read something about using also maven for creating a zip file even though is not a java app here and do the same for node js.
So, Is it possible with maven to zip a directory and a file even thought they are not a java app and include job number to version this zip file and pushing into artifactory?, if not, which is the best approach to do the same as I did with Java and maven for versioning, but for VUE.js applications in a jenkins pipeline that push a zip along with build number into artifactory and then using the "build with parameters" option to choose the zip I want?
thank you!

Answer 1

You mentioned, that you are already using npm to build your Vue app. There are libraries for npm available for zipping. You could e.g. do something like this:
const fs = require('fs');
const archiver = require('archiver');
const archive = archiver('zip', {
zlib: { level: 9 }
});
const filename = "./output.zip";
const fileOutput = fs.createWriteStream(filename);
fileOutput.on('close', function() {
console.info('ZIP file created. ' + archive.pointer() + ' total Bytes.');
});
archive.pipe(fileOutput);
archive.directory('./input-directory', '/');
archive.on('error', function(error) {
throw error;
});
archive.finalize();
This assumes, that the output of your build process is stored in a dist folder on your file system and that you have access to the filesystem.
There are more examples of archiver in the official docs: https://github.com/archiverjs/node-archiver

Answer 2

You may build a zip file using the maven-assembly-plugin with an assembly descriptor.
The assembly descriptor goes in the project's src/assembly directory, named something like my-zip-format.xml. You'll need to customize the content to include the files needed but this is the idea.
<assembly xmlns="http://maven.apache.org/plugins/maven-assembly-plugin/assembly/1.1.2"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/plugins/maven-assembly-plugin/assembly/1.1.2 http://maven.apache.org/xsd/assembly-1.1.2.xsd">
<id>my-zip</id>
<formats>
<format>zip</format>
</formats>
<fileSets>
<fileSet>
<directory>path/to/input/dir</directory>
<outputDirectory>name-of-output-dir</outputDirectory>
<directoryMode>0750</directoryMode>
<fileMode>0640</fileMode>
<lineEnding>unix</lineEnding>
</fileSet>
</fileSets>
</assembly>
Then, tell the POM to use the assembly:
....
<plugins>
<plugin>
<artifactId>maven-assembly-plugin</artifactId>
<version><pluginVersionHere></version>
<executions>
<execution>
<id>make-zip</id>
<phase>package</phase>
<goals>
<goal>single</goal>
</goals>
<configuration>
<descriptors>
<descriptor>src/assembly/my-zip-format.xml</descriptor>
</descriptors>
</configuration>
...
Documentation goes into a lot more detail, and there are plenty of SO questions/answers about assembly plugin nuances as well.