CODEWITHSUNDEEP
pythoncurlfile-uploadkeyerror
mrmoment
mrmoment

Reputation: 757

KeyError in Python HTTP server when handling curl file upload requests

I have a HTTP server in Python which handles file upload requests from curl

class HTTPRequestHandler(BaseHTTPRequestHandler):
    def do_POST(self):
            if re.search('/api/v1/addphoto', self.path):
                    form_data=cgi.FieldStorage()
                    file_data=form_data['photofile'].value 
                    # Next I will save this file 
                    # fp =open('some/file','wb')
                    # fp.write(file_data)
                    # fp.close()

Now I have to use curl to send the request, and the command is 

curl -i -F name=photofile -F [email protected] http://server_ip:port/api/v1/addphoto

But the server report error 

File "./apiserver.py", line 21, in do_POST
    file_data=form_data['photofile'].value
File "/usr/lib/python2.7/cgi.py", line 541, in __getitem__
    raise KeyError, key
KeyError: 'photofile'

What is the problem here?

Upvotes: 0

Views: 2403

Answers (2)

mrmoment
mrmoment

Reputation: 757

OK, I find the reason and how to do it from here.

The reason is "You can't embed the CGI code in the server code like that. The whole point of CGIHTTPServer (and CGI in general) is that it executes separate scripts in designated directories and returns the output. Because you had overridden the do_POST method, the code to parse the form data was never called." (Actually I print the form_data and it is empty - "None, None, []").

And to fix it, use passed in parameters (details)

 form = cgi.FieldStorage(
        fp=self.rfile, 
        headers=self.headers,
        environ={'REQUEST_METHOD':'POST',
                 'CONTENT_TYPE':self.headers['Content-Type'],
                 })

Upvotes: 0

NoamG
NoamG

Reputation: 1161

KeyError is raised when no key found in the dictionary.

I'm not familiar with cgi.FieldStorage but I guess you're expected to get a dictionary and the key 'photofile' is missing. Just print the entire form_data and see it's content.

You can use form_data.get('photofile', '') to get default value and avoid the exception.

BTW, for writing to a file you can use with statement as a context manager.

with open('some/file','wb') as fp:
    fp.write(file_data)

Upvotes: 1

Related Questions