CODEWITHSUNDEEP
javascriptnode.jstypescriptexpress
FlameDra
FlameDra

Reputation: 2087

Zip image stream using archiver and send as express response

This is on Node/Express/Typescript. I'm trying get an image on my file system, stream it to a zip file, and then stream this zip file to the client. I have to make sure every step is streamed since this will expand to zipping up multiple files, which need to be streamed to the client as a zip.

I have the following code:

import express, { Application, Request, Response } from "express";
import fs from "fs";
import stream from "stream";
import archiver from "archiver";

app.get("/images", async (req: Request, res: Response) => {
    const r = fs.createReadStream("appicon.png");
    const ps = new stream.PassThrough();

    // stream the image
    stream.pipeline(
        r,
        ps,
        (err) => {
            if (err) {
                console.log(err);
                return res.sendStatus(400);
            }
        }
    );


    // zip the image and send it
    let archive = archiver("zip");

    archive.on("end", () => {
        console.log(archive.pointer() + " total bytes");
        console.log("archiver finalized");
    })

    archive.on('error', (err) => {
        return res.status(500).send({
            message: err
        });
    })

    res.attachment('output.zip');
    ps.pipe(archive); 
    archive.pipe(res);



    archive.finalize();

});

However, when I access my /images route, I get an output.zip file which is empty.

I feel like I'm messing up the order of my pipes somehow.

What am I missing?

Upvotes: 4

Views: 4223

Answers (1)

FlameDra
FlameDra

Reputation: 2087

I figured out the issue. Here is the code that works:

app.get("/images", async (req: Request, res: Response) => {
    const r = fs.createReadStream("appicon.png");
    const ps = new stream.PassThrough();

    stream.pipeline(
        r,
        ps,
        (err) => {
            if (err) {
                console.log(err);
                return res.sendStatus(400);
            }
        }
    );

    //r.pipe(ps); // alternative way to do it without pipeline


    // zip the image and send it
    let archive = archiver("zip");

    archive.on("end", () => {
        console.log(archive.pointer() + " total bytes");
        console.log("archiver finalized");
    })

    archive.on('error', (err) => {
        return res.status(500).send({
            message: err
        });
    })

    // name the output file
    res.attachment("output.zip");

    // pipe the zip to response
    archive.pipe(res);

    // add the image from stream to archive
    archive.append(ps, {name: "image.png"});

    archive.finalize();

});

I had to use archive.append(ps, {name: "image.png"}); to add my image stream to the zip archive.

Upvotes: 4

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