Problem with python curl while submitting file

Question

#!/usr/bin/python
import pycurl
import re
import StringIO


#CONSTANTS
URL = "http://www.imagehost.org"
FILE = "/datos/poop1.jpg"
POST_DATA = [("a", "upload"), ("file[]", (pycurl.FORM_FILE, FILE))]



buffer = StringIO.StringIO()


c = pycurl.Curl()
c.setopt( c.URL, URL )
c.setopt( c.POST, 1 )
c.setopt( c.POSTFIELDS, POST_DATA )
##c.setopt( c.HTTPPOST, POST_DATA )
c.setopt( c.USERAGENT,'Curl')
c.setopt( c.WRITEFUNCTION, buffer.write)
c.setopt(pycurl.VERBOSE, 1)

c.perform()
c.close()

#c.setopt(c.PROXY, proxyHostAndPort)
#c.setopt(c.PROXYUSERPWD, proxyAuthentication)

parse = buffer.getvalue()


pattern = re.compile('/<td nowrap="nowrap">(.+)<\/td>\s*<td class="link"><input.+value="([^"]+)" \/><\/td>/i')

result = re.search(pattern, parse)
print result

The problem is in the way on how to do the post.

c.setopt( c.POSTFIELDS, POST_DATA ) does not accept lists, so what should I do instead of adding a list?

And c.setopt( c.HTTPPOST, POST_DATA ) drops :

Traceback (most recent call last): 
  File "pymage", line 26, in <module>
c.perform() pycurl.error: (26, 'failed creating formpost data')

Update:

-----------------------------15758382912173403811539561297\r\nContent-Disposition: form-data; name="a"\r\n\r\nupload\r\n-----------------------------15758382912173403811539561297\r\nContent-Disposition: form-data; name="file[]"; filename="Datos_Pegados_0a17.jpg"\r\nContent-Type: image/jpeg\r\n\r\nÿØÿà

That's what I get using tamper data.

Interesting part of the postfield:

form-data; name="a"\r\n\r\nupload\r\n

form-data; name="file[]"

So... you say the POST_DATA should be 'a=upload&file[]=FILE'?

Update2:

<form method="post" action="/" enctype="multipart/form-data" onsubmit="javascript:Upload(); return true;">

<input type="hidden" name="a" value="upload" />

<td class="left">File:</td>
td class="right"><input name="file[]" type="file" size="20" /></td>

That's the code...

Now it's working the form-data configuration, but it's not uploading the file I believe

c.setopt( c.POSTFIELDS, 'a=upload&file[]=/datos/poop1.jpg' )

I'm getting this:

* About to connect() to www.imagehost.org port 80 (#0)
*   Trying 74.63.87.74... * connected
* Connected to www.imagehost.org (74.63.87.74) port 80 (#0)
> POST / HTTP/1.1
User-Agent: Curl
Host: www.imagehost.org
Accept: */*
Content-Length: 32
Content-Type: application/x-www-form-urlencoded

< HTTP/1.1 200 OK
< Transfer-Encoding: chunked
< Date: Wed, 25 Mar 2009 06:53:49 GMT
< Content-Type: text/html
< Server: nginx/0.7.11
< Set-Cookie: userhash=7c09b97cc70c8c133c850a3e744b416e; expires=Thu, 25-Mar-2010 06:53:49 GMT; path=/; domain=.imagehost.org; httponly
< 
* Connection #0 to host www.imagehost.org left intact
* Closing connection #0

Care to be a little more specific? What error are you getting? — Paolo Bergantino
– Paolo Bergantino, Commented Mar 25, 2009 at 2:07
You may get better responses if you describe what you are trying to do, and the results/errors that you get when you attempt to do it. — vezult
– vezult, Commented Mar 25, 2009 at 2:22
c.setopt( c.POSTFIELDS, POST_DATA ) does not accept lists, so what should I do instead of adding a list? And c.setopt( c.HTTPPOST, POST_DATA ) drops : Traceback (most recent call last): File "pymage", line 26, in <module> c.perform() pycurl.error: (26, 'failed creating formpost data') — matta
– matta, Commented Mar 25, 2009 at 2:41

GrahamW · Accepted Answer · 2011-02-09 09:55:40Z

The pycurl documentation is not so clear on this, but the HTTPPOST option can take a list of tuples, each exactly two elements long. Each tuple's first argument is the form field name, the second is the value.

However, the value may be a tuple as well. This tuple must contains pairs of data relating to that field: ( form_option, option_value, form_option, option_value, etc., etc.)

For example, a multi-part form with three fields, the last being a file upload - we can set the filename and the mime type:

c = pycurl.Curl()
c.setopt(c.URL, base_url + 'upload.cgi')
c.setopt(c.HTTPPOST,[ ("fieldname1", "value1"), 
                      ("fieldname2", "value2"), 
                      ("uploadfieldname", 
                                 (c.FORM_FILE, local_filename, 
                                  c.FORM_CONTENTTYPE, "application/x-gzip"))
                    ])

You can find the options in the documentation for the C API for curl_formadd(): http://curl.haxx.se/libcurl/c/curl_formadd.html however, from the pycurl source, it looks like only FORM_FILE, FORM_FILENAME, FORM_CONTENTTYPE and FORM_COPYCONTENTS are supported.

Sean Reifschneider · Accepted Answer · 2011-08-12 06:50:18Z

2

Error 26 (in "pycurl.error: (26, 'failed creating formpost data')") means that the filename you specified to upload does not exist. I was running into this same error, and that was definitely the problem. See the curl source code for the location that it generates error 26.

answered Aug 12, 2011 at 6:50

Sean Reifschneider

1,26110 silver badges16 bronze badges

Comments

Van Gale · Accepted Answer · 2009-03-25 05:17:28Z

1

I believe the argument for POSTFIELDS needs to be a simple URL encoded string, for example:

POST_DATA = 'a=foo&b=bar'

Next, I'm not sure about your FILE stuff. Check out this mail message for an example.

answered Mar 25, 2009 at 5:17

Van Gale

44k9 gold badges75 silver badges81 bronze badges

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