Uploading file using POST request in Node.js

Question

I have problem uploading file using POST request in Node.js. I have to use request module to accomplish that (no external npms). Server needs it to be multipart request with the file field containing file's data. What seems to be easy it's pretty hard to do in Node.js without using any external module.

I've tried using this example but without success:

request.post({
  uri: url,
  method: 'POST',
  multipart: [{
    body: '<FILE_DATA>'
  }]
}, function (err, resp, body) {
  if (err) {
    console.log('Error!');
  } else {
    console.log('URL: ' + body);
  }
});

do you have your form with option enctype="multipart/form-data"? — monkeyinsight
– monkeyinsight, Commented Aug 16, 2014 at 23:17
I'm not using any form. It's server request. I'm sending file from browser to server using sockets and later I have to send that file to another server using POST request. — Łukasz Jagodziński
– Łukasz Jagodziński, Commented Aug 17, 2014 at 10:44

Leonid Beschastny · Accepted Answer · 2014-08-17 00:00:33Z

135

Looks like you're already using request module.

in this case all you need to post multipart/form-data is to use its form feature:

var req = request.post(url, function (err, resp, body) {
  if (err) {
    console.log('Error!');
  } else {
    console.log('URL: ' + body);
  }
});
var form = req.form();
form.append('file', '<FILE_DATA>', {
  filename: 'myfile.txt',
  contentType: 'text/plain'
});

but if you want to post some existing file from your file system, then you may simply pass it as a readable stream:

form.append('file', fs.createReadStream(filepath));

request will extract all related metadata by itself.

For more information on posting multipart/form-data see node-form-data module, which is internally used by request.

answered Aug 17, 2014 at 0:00

Leonid Beschastny

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17 Comments

Doug Donohoe Over a year ago

When I was learning node and the request module, I was confused as to why the form could be modified after the post method was called. Buried in the request docs is the explanation - the form "can be modified until the request is fired on the next cycle of the event-loop".

Vitor Freitas Over a year ago

I keep getting '[Error: write after end]' when using form and form.append, anyone knows why?

Leonid Beschastny Over a year ago

@VitorFreitas you should call req.form() and fill it with all appropriate data synchronously right after calling request.post. It's important to do it during the same event loop tick, otherwise your request may be already send and underlying stream closed.

David Over a year ago

The request was deprecated, do you have a alternative?

Leonid Beschastny Over a year ago

@David got is a good alternative

|

Clavin · Accepted Answer · 2017-03-25 18:39:33Z

An undocumented feature of the formData field that request implements is the ability to pass options to the form-data module it uses:

request({
  url: 'http://example.com',
  method: 'POST',
  formData: {
    'regularField': 'someValue',
    'regularFile': someFileStream,
    'customBufferFile': {
      value: fileBufferData,
      options: {
        filename: 'myfile.bin'
      }
    }
  }
}, handleResponse);

This is useful if you need to avoid calling requestObj.form() but need to upload a buffer as a file. The form-data module also accepts contentType (the MIME type) and knownLength options.

This change was added in October 2014 (so 2 months after this question was asked), so it should be safe to use now (in 2017+). This equates to version v2.46.0 or above of request.

Philip Kirkbride · Accepted Answer · 2016-12-19 13:26:17Z

4

Leonid Beschastny's answer works but I also had to convert ArrayBuffer to Buffer that is used in the Node's request module. After uploading file to the server I had it in the same format that comes from the HTML5 FileAPI (I'm using Meteor). Full code below - maybe it will be helpful for others.

function toBuffer(ab) {
  var buffer = new Buffer(ab.byteLength);
  var view = new Uint8Array(ab);
  for (var i = 0; i < buffer.length; ++i) {
    buffer[i] = view[i];
  }
  return buffer;
}

var req = request.post(url, function (err, resp, body) {
  if (err) {
    console.log('Error!');
  } else {
    console.log('URL: ' + body);
  }
});
var form = req.form();
form.append('file', toBuffer(file.data), {
  filename: file.name,
  contentType: file.type
});

edited Dec 19, 2016 at 13:26

Philip Kirkbride

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answered Aug 17, 2014 at 10:41

Łukasz Jagodziński

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3 Comments

Leonid Beschastny Over a year ago

There is a simpler way to convert ArrayBuffer to Buffer, using build-in Buffer constructor from an array of octets: var buffer = new Buffer(new Uint8Array(ab));

michaelAdam Over a year ago

Where did the "file" in file.data, file.name, and file.type come from in your last function? I don't see that variable mentioned anywhere else.

Łukasz Jagodziński Over a year ago

I'm using Meteor and community package for file management. However if you are using pure node then you can use file system functions to get all the info about the file and its data nodejs.org/api/fs.html

chrismarx · Accepted Answer · 2018-05-09 22:45:12Z

4

You can also use the "custom options" support from the request library. This format allows you to create a multi-part form upload, but with a combined entry for both the file and extra form information, like filename or content-type. I have found that some libraries expect to receive file uploads using this format, specifically libraries like multer.

This approach is officially documented in the forms section of the request docs - https://github.com/request/request#forms

//toUpload is the name of the input file: <input type="file" name="toUpload">

let fileToUpload = req.file;

let formData = {
    toUpload: {
      value: fs.createReadStream(path.join(__dirname, '..', '..','upload', fileToUpload.filename)),
      options: {
        filename: fileToUpload.originalname,
        contentType: fileToUpload.mimeType
      }
    }
  };
let options = {
    url: url,
    method: 'POST',
    formData: formData
  }
request(options, function (err, resp, body) {
    if (err)
      cb(err);

    if (!err && resp.statusCode == 200) {
      cb(null, body);
    }
  });

edited May 9, 2018 at 22:45

chrismarx

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answered Mar 25, 2017 at 17:17

Marwen Landoulsi

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2 Comments

Fabio says Reinstate Monica Over a year ago

Please edit your answer and add some explanation or comment as to how your code works. This would help other users decide whether your answer is interesting enough to be considered. Otherwise people have to analyse your code (which takes time) even to have a vague idea whether this might be what they need. Thank you!

user985366 Over a year ago

5 years later someone is going to want an explanation and you're not going to be around or not going to bother. That's why Fabio asked you to put the explanation in the response, and not on request.

Atul · Accepted Answer · 2022-02-08 19:06:11Z

0

I did it like this:

// Open file as a readable stream
const fileStream = fs.createReadStream('./my-file.ext');

const form = new FormData();
// Pass file stream directly to form
form.append('my file', fileStream, 'my-file.ext');

answered Feb 8, 2022 at 19:06

Atul

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Comments

Xu Xiang · Accepted Answer · 2020-09-23 09:22:15Z

-1

 const remoteReq = request({
    method: 'POST',
    uri: 'http://host.com/api/upload',
    headers: {
      'Authorization': 'Bearer ' + req.query.token,
      'Content-Type': req.headers['content-type'] || 'multipart/form-data;'
    }
  })
  req.pipe(remoteReq);
  remoteReq.pipe(res);

answered Sep 23, 2020 at 9:22

Xu Xiang

1

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