在 Node.js 中使用 POST 请求上传文件

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);
  }
});

Answer 1

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 .

Answer 2

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 .

Answer 3

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
});

Answer 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);
    }
  });

Answer 5

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');

Answer 6

 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);