How do i stream a variable to a file Node JS?

Question

I need to save a response to a file. The response is a zip file returned from a server which is recieved as a blob. I need to save to blob as a zip file on my local machine. The application itself is Electron, and the file needs to be stored in the background (without bothering the user). Filetype is zip ( and needs to be unzipped after this.

const writer = fs.createWriteStream(path.join(dir, 'files.zip'));
                writer.on('pipe', (src) => {
                  console.log('Something is piping into the writer.');
                });
                writer.end('This is the end\n');
                writer.on('finish', () => {
                  console.log('All writes are now complete.');
                });

writer.pipe(new Blob([response.data]));

Best i can do gives a 1kb corrupt file. I have read the node docs, but i can't get it to work.

ANY response is greatly appreciated, and please elaborate if you can. I feel like i need to use some type of buffer.

Answer 1

So i figured finally it out.

Instead of a return type of Blob i had to return an arraybuffer. Next step is using the JSzip library and convert my function to async.

Final result:

//Create JSZip object to read incoming blob
const zip = new JSZip;

try {
  //Await unpacked arrayBuffer
  const zippedFiles = (await zip.loadAsync(response.data, { createFolders: true })).files;

  for (const file of Object.values(zippedFiles)) {
    if (file.dir) {
      await mkdirAsync(path.join(dir, file.name), { recursive: true });
    } else {
        file
          .nodeStream()
          .pipe(fs.createWriteStream(path.join(dir, file.name)))
          .on("finish", console.log);
      }
    }
  } catch (e) {
   throw MeaningfulError;
 }

In short: This function takes an arraybuffer (of type .zip) and unpacks it to your system.

• response.data is the arraybuffer (which needs to be unpacked).
• dir is the directory in which the content needs to be unpacked.

And that's all you need!