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I have written below for zipping a directory :

const archiver = require('archiver');
let createArchive = function(sourceDir, destDir) {
    const archive = archiver('zip');
    const stream = fs.createWriteStream(destDir);
    return new Promise((resolve, reject) => {
    archive
      .directory(sourceDir, false)
      .on('error', err => reject(err))
      .pipe(stream)
    ;

    stream.on('close', () => resolve());
    archive.finalize();
  });
}

Before zipping my directory looked like this :

  • excelArchive.zip
  • test.txt
  • test2.gz

When I unzipped the archive (called yayy.zip) it had these files in it:

  • excelArchive.zip
  • test.txt
  • test2.gz
  • yayy.zip

There is an invalid file called yayy.zip inside it. How can I avoid this?

miken32
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Pawan Saxena
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1 Answers1

2

Instead of directory, you can use the glob method to specify the files to zip. It takes a second argument where you can specify files to ignore.

const archiver = require('archiver');
let createArchive = function(sourceDir, destDir) {
    const archive = archiver('zip');
    const stream = fs.createWriteStream(destDir);
    return new Promise((resolve, reject) => {
    archive
      .on('error', err => reject(err))
      .pipe(stream)
      .glob(
        '**/*',
        {cwd: sourceDir, ignore: filename}
      )
    ;

    stream.on('close', () => resolve());
    archive.finalize();
  });
}
miken32
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  • What is filename here, i think i want to avoid yayy.zip which was not existing earlier. It got created by default inside the zip, so filename param won't work ( as it is non-existent at the time of generation ). Also can you help me undertsand why it happened in first place. – Pawan Saxena Sep 16 '21 at 02:33
  • It's obviously the filename you want not to include in the zip file. i.e. the name of the zip file itself. – miken32 Sep 16 '21 at 16:29