Uploading and Processing files with Flask Rest API

Question

I want to create a REST API using Flask. I have two files at any given time, I will call my API and send these two files, a python script will run on the API side and send me back a JSON Response. Now this is the overall idea of my application , but I don't know how to do all this. As I do not want to save those two files, just take them as an Input and send the result back, I don't think i will need to save the files. But I do not know how to send files to the API, logic part is ready, I just need to know how to send two files, process them and send back the response in just one API call. Thanks in advance

I have just setup the basic API till now.

from flask import Flask

app = Flask(__name__)

@app.route('/')
def index():
    return 'Server works'


if __name__== '__main__':
    app.run(debug=True)

funkyfresh12 · Accepted Answer · 2020-05-15 17:41:45Z

Found this on the Flask documentation:

import os
from flask import Flask, flash, request, redirect, url_for
from werkzeug.utils import secure_filename

UPLOAD_FOLDER = '/path/to/the/uploads'
ALLOWED_EXTENSIONS = {'txt', 'pdf', 'png', 'jpg', 'jpeg', 'gif'}

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER


def allowed_file(filename):
    return '.' in filename and \
           filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS

@app.route('/', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        # check if the post request has the file part
        if 'file' not in request.files:
            flash('No file part')
            return redirect(request.url)
        file = request.files['file']
        # if user does not select file, browser also
        # submit an empty part without filename
        if file.filename == '':
            flash('No selected file')
            return redirect(request.url)
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file',
                                    filename=filename))
    return '''
    <!doctype html>
    <title>Upload new File</title>
    <h1>Upload new File</h1>
    <form method=post enctype=multipart/form-data>
      <input type=file name=file>
      <input type=submit value=Upload>
    </form>
    '''

In your API backend, you will receive the file by using file = request.files['file']. The name "file" comes from the name of the input tag in the HTML Form you are using to send the file to your backend.

    '''
    <!doctype html>
    <title>Upload new File</title>
    <h1>Upload new File</h1>
    <form method=post enctype=multipart/form-data>
      <input type=file name=file>
      <input type=submit value=Upload>
    </form>
    '''

In this example, the backend is saving the uploaded files to UPLOAD_FOLDER. This example is also using some exception handling to make sure that the user is uploading the correct type of file types.

[EDIT] I misread your question. If you just want to send back a JSON response instead of JSON containing the files content you would do this:

return jsonify({"response": "success"})

Finally, here's a link to the full Flask documentation

Nathaniel Cobbinah · Accepted Answer · 2025-04-04 05:53:12Z

from flask_restx import Resource, fields
from flask import request
import sys
import os
from werkzeug.datastructures import FileStorage
from werkzeug.utils import secure_filename   

BASEDIR = os.path.abspath(os.path.dirname(__file__))

upload_parser = api.parser()
upload_parser.add_argument('file', location='files',
                               type=FileStorage, required=True) 

@api.route("/your_route_name")
@api.expect(upload_parser)
class MyResource(Resource):
    @api.marshal_with(<output_dataformat_schema_defined_here>, skip_none=True)
    def post(self):
       uploaded_file = request.files['file']

       if uploaded_file:
          secured_uploadedfilename = 
                    secure_filename(uploaded_file.filename)
          uploaded_file.save(os.path.join(os.path.join(BASEDIR, 
             "your_foldername"), secured_uploadedfilename))
          return {"message": "File successfully uploaded"}
       else:
          return {"message": "File upload unsuccessful"}

Source: Flask restx, Flask

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