unzipper ignores empty directory

Question

I am a Japanese Web Developer.

I am not good at English, sorry.

https://www.npmjs.com/package/unzipper

I am using this library.

Here is my code.

// unzip module
import fs from 'fs-extra'
import unzipper from 'unzipper'
import Promise from 'bluebird'

export default ({ inputFilePath, outputFilePath }) => {
  return new Promise(async (resolve, reject, onCancel) => {
    try {
      await streamPromise(fs.createReadStream(inputFilePath).pipe(unzipper.Extract({ path: outputFilePath })))
        .then(resolve)
    } catch (error) {
      reject(error)
    }
  })
}

const streamPromise = (stream) => {
  return new Promise((resolve, reject) => {
    stream.on('end', () => {
      resolve('end')
    })
    stream.on('finish', () => {
      resolve('finish')
    })
    stream.on('error', (error) => {
      reject(error)
    })
  })
}

But directory which has no file inside, won't be copied.

Any way to fix this?

score 1 · Accepted Answer · answered May 29 '19 at 03:24

You want to unzip a zip file using unzipper.
You want to also export the directories without files.

If my understanding is correct, how about this modification? I think that there are several answers for your situation. So please think of this as just one of them.

Modification point:

In this modification, when the file type is "Directory", a new Directory is created. By this, the directory without files can be exported.

Modified script:

Please modify as follows.

// unzip module
import fs from 'fs-extra'
import unzipper from 'unzipper'
import Promise from 'bluebird'
import path from 'path' // Added

// Modified
export default ({ inputFilePath, outputFilePath }) => {
  return new Promise(async (resolve, reject, onCancel) => {
    try {
      await streamPromise(fs.createReadStream(inputFilePath).pipe(unzipper.Parse()), outputFilePath)
        .then(resolve)
    } catch (error) {
      reject(error)
    }
  })
}

// Modified
const streamPromise = (stream, outputFilePath) => {
  return new Promise((resolve, reject) => {
    stream.on("entry", entry => {
      const fileName = entry.path;
      const type = entry.type;
      if (type == "Directory") {
        fs.mkdirSync(path.join(outputFilePath, fileName));
      } else if (type == "File") {
        entry.pipe(fs.createWriteStream(path.join(outputFilePath, fileName)));
      }
    });
    stream.on("end", () => {
      resolve("end");
    });
    stream.on("finish", () => {
      resolve("finish");
    });
    stream.on("error", error => {
      reject(error);
    });
  });
};

Reference:

unzipper

In my environment, I could confirm that the modified script worked. But in your environment, the script didn't work, I apologize.

Thank you so much. You are a genius. It worked in my environment too. — Kuru, May 29 '19 at 04:33
@akao Thank you for replying. I'm glad your issue was resolved. — Tanaike, May 29 '19 at 07:41

unzipper ignores empty directory

1 Answers1

Modification point:

Modified script:

Reference: