Python Flask：发送文件和变量

Question

I have two servers where one is trying to get a file from the other. I am using Flask get requests to send simple data back and forth (strings, lists, JSON objects, etc.).

I also know how to send just a file, but I need to send an error code with my data.

I'm using something along the following lines:

Server 1:

req = requests.post('https://www.otherserver.com/_download_file', data = {'filename':filename})

Server 2:

@app.route('/_download_file', methods = ['POST'])
def download_file():
    filename = requests.form.get('filename')

    file_data = codecs.open(filename, 'rb').read()
    return file_data

Server 1:

with codecs.open('new_file.xyz', 'w') as f:
    f.write(req.content)

...all of which works fine. However, I want to send an error code variable along with file_data so that Server 1 knows the status (and not the HTTP status, but an internal status code).

Any help is appreciated.

Answer 1

One solution that comes to my mind is to use a custom HTTP header.

Here is an example server and client implementation.

Of course, you are free to change the name and the value of the custom header as you need.

server

from flask import Flask, send_from_directory

app = Flask(__name__)

@app.route('/', methods=['POST'])
def index():
    response = send_from_directory(directory='your-directory', filename='your-file-name')
    response.headers['my-custom-header'] = 'my-custom-status-0'
    return response

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

client

import requests

r = requests.post(url)

status = r.headers['my-custom-header']

# do what you want with status

UPDATE

Here is another version of the server based on your implementation

import codecs

from flask import Flask, request, make_response

app = Flask(__name__)

@app.route('/', methods=['POST'])
def index():
    filename = request.form.get('filename')

    file_data = codecs.open(filename, 'rb').read()
    response = make_response()
    response.headers['my-custom-header'] = 'my-custom-status-0'
    response.data = file_data
    return response

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