Cherrypy Upload File

I'm want to POST a file from a python3 client to cherrypy. I'm using the requests library. My client code: import requests url = 'http://127.0.0.1:8080/upload' files = {'file.zip'

Solution 1:

It's usefull to get information from response. When you send a request, you receive a response. From this response you can get information about HTTP code where 200 means OK and 400 means bad request. That's the text you can see in your cherrypy log: POST /upload HTTP/1.1" 400. To get more information, print the response text using print(r.text)

#!/usr/bin/env python# -*- coding: UTF-8 -*-import requests

url = 'http://127.0.0.1:9090/upload'
files = {'ufile': open('file.txt', 'rb')}

r = requests.post(url, files=files)

print(r)
print(r.text)

If you use the code above with the code bellow, it's working example of uploading file to cherrypy server.

#!/usr/bin/env python# -*- coding: UTF-8 -*-import os
import cherrypy

config = {
    'global' : {
        'server.socket_host' : '127.0.0.1',
        'server.socket_port' : 9090,
        'server.thread_pool' : 8,
        'server.max_request_body_size' : 0,
        'server.socket_timeout' : 60
    }
}


classApp:

    @cherrypy.exposedefupload(self, ufile):
        upload_path = os.path.normpath('/path/to/project/data/')
        upload_file = os.path.join(upload_path, ufile.filename)
        size = 0withopen(upload_file, 'wb') as out:
            whileTrue:
                data = ufile.file.read(8192)
                ifnot data:
                    break
                out.write(data)
                size += len(data)
        out = '''
length: {}
filename: {}
mime-type: {}
''' .format(size, ufile.filename, ufile.content_type, data)
        return out


if __name__ == '__main__':
    cherrypy.quickstart(App(), '/', config)

Replace the path /path/to/project/data/ to path that fits your project.