Same file upload will upload again with a different name, instead of beeing cache like where only the name is returned

[banli17]

hi, i found same file upload will return different name，and this will create many files, i want upload same file. return the same file，please help me， i found this code

IncomingForm.prototype._uploadPath = function(filename) {
  var buf = crypto.randomBytes(16);
  var name = 'upload_' + buf.toString('hex');

  if (this.keepExtensions) {
    var ext = path.extname(filename);
    ext     = ext.replace(/(\.[a-z0-9]+).*/i, '$1');

    name += ext;
  }

  return path.join(this.uploadDir, name);
};

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/cc @tunnckoCore @GrosSacASac

[banli17]

but if i upload same file , it will recreate a new file with the same name。how can i avoid it。

i want to if the file is exist，not create

[tunnckoCore]

You don't need to override internal functions, there's not documented options.filename in the v2 (canary).

formidable/src/Formidable.js

Lines 60 to 70 in 5110ef8

	const hasRename = typeof this.options.filename === 'function';
	if (this.options.keepExtensions === true && hasRename) {
	this._rename = (part) => {
	const resultFilepath = this.options.filename.call(this, part, this);
	return this._uploadPath(part, resultFilepath);
	};
	} else {
	this._rename = (part) => this._uploadPath(part);
	}

const { formidable } = require('formidable');

const form = formidable({
  keepExtensions: true, // that must be `true`
  filename: (part) => {
    // it has `part.name` and etc
    console.log(part);
  }
});

And yes, if you implement to have same name, it will probably recreate/rewrite the existing file.

One workaround could be to cache the filename in some variable outside, and check if exists (fs.existsSync()), then not call the parse (or formidable() at all) if it exists.

Sorry about that. Not possible otherwise, for the moment.

[banli17]

thx @tunnckoCore

[banli17]

but it sames can not rename according file's md5，part has no unique flag to rename file。

part Stream {
  _events: [Object: null prototype] {},
  _eventsCount: 0,
  _maxListeners: undefined,
  readable: true,
  headers: {
    'content-disposition': 'form-data; name="file"; filename="tpl.jpg"',
    'content-type': 'image/jpeg'
  },
  name: 'file',
  filename: 'tpl.jpg',
  mime: 'image/jpeg',
  transferEncoding: 'binary',
  transferBuffer: ''
}

const form = formidable({
            multiples: true,
            keepExtensions: true,
            uploadDir: getUploadPath('../upload_cache'),
            filename: (part) => {
                console.log(Object.keys(part))
                console.log('part', part)
            }
        });

[tunnckoCore]

Sorry, I can't understand your last comment. Don't feel bad to use something like Google Translate if your English isn't good! 😉 There's nothing wrong :)

After a few hours of trying how I can implement what I was thinking and proposing... I gave up, I can't come to a solution, without changes in the codebase. Even if the filename or the contents is always the same it will always overwrite its contents... I still have a feeling I can do it, but I'm not 100% sure and does not worth the time.

Fundamentals & codebase are messy, and currently we can't do some things, sorry.

[banli17]

Sorry, I will use Google Translate. I want to upload the same image only once, and if it exists, it will return the name of the previously generated image (similar to cache) instead of regenerating. I'm not sure if this kind of cache is better？

[tunnckoCore]

Yes, I thought about that but can't think of a solution for now, sorry... :(

[banli17]

thank you @tunnckoCore