from php curl to python urllib (POST to Upload API Problem)

Question

I tried to convert a php api code to python:

This is the php code:

// Variables to Post
$local_file = "/path/to/file"; 
$file_to_upload = array(
 'file'=>'@'.$local_file, 
'convert'=>'1', 
'user'=>'YOUR_USERNAME', 
'password'=>'YOUR_PASSWORD'
); 

// Do Curl Request
$ch = curl_init(); 
curl_setopt($ch, CURLOPT_URL,'http://example.org/dapi.php'); 
curl_setopt($ch, CURLOPT_POST,1); 
curl_setopt($ch, CURLOPT_POSTFIELDS, $file_to_upload); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$result=curl_exec ($ch); 
curl_close ($ch); 

// Do Stuff with Results
echo $result;

And this is my Python Code:

url = 'http://example.org/dapi.php'
file ='/path/to/file'
datei= open(file, 'rb').read()

values = {'file' : datei ,
     'user' : 'username',
     'password' : '12345' ,
     'convert': '1'}

data = urllib.urlencode(values)
req = urllib2.Request(url, data)
response = urllib2.urlopen(req)
the_page = response.read()

print the_page

It uploads my files but the response is an Error so that something has to be wrong with my python code. But I can´t see my mistake.

What error? Also, I assume '@'. formats a string in PHP, if not, what does it do? What does CURLOPT_RETURNTRANSFER do? — agf
– agf, Commented Jul 27, 2011 at 1:13
yep, that @. is concatenating a string on local file and in python your are reading as it is the file and sending it. It's the only difference I can really see. — Hassek
– Hassek, Commented Jul 27, 2011 at 1:32

W0bble · Accepted Answer · 2011-07-27 11:14:56Z

1

After trying a lot of opportunities. I found my solution using pycurl:

import pycurl
import cStringIO

url = 'http://example.org/dapi.php'
file ='/path/to/file'


print "Start"
response = cStringIO.StringIO()
c = pycurl.Curl()
values = [('file' , (c.FORM_FILE,  file)),
      ('user' , 'username'),
      ('password' , 'password'),
      ('convert', '1')]


c.setopt(c.POST, 1)
c.setopt(c.URL,url)
c.setopt(c.HTTPPOST,  values)
#c.setopt(c.VERBOSE, 1)
c.setopt(c.WRITEFUNCTION, response.write)
c.perform()
c.close()
print response.getvalue()
print "All done"

answered Jul 27, 2011 at 11:14

W0bble

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Comments

twasbrillig · Accepted Answer · 2014-11-02 13:19:08Z

1

There is no simple way to upload a file using multipart/form-data encoding. There are some snippets you can use, though:

[http://pymotw.com/2/urllib2/index.html#module-urllib2] [http://code.activestate.com/recipes/146306-http-client-to-post-using-multipartform-data/]

A simpler way would be to use a library. Some good libraries that I use are:

edited Nov 2, 2014 at 13:19

twasbrillig

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answered Jul 27, 2011 at 6:02

a9.

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1 Comment

W0bble Over a year ago

Thank you for the ideas. I´ll check them.

cwallenpoole · Accepted Answer · 2011-07-27 04:19:21Z

0

Your issue is with this line: datei= open(file, 'rb').read(). For urllib2.Request to upload a file, it needs an actual file object, so the line should be: datei= open(file, 'rb'). open(...).read() returns a str instead of the file object.

edited Jul 27, 2011 at 4:19

answered Jul 27, 2011 at 4:14

cwallenpoole

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1 Comment

cwallenpoole Over a year ago

@isbadawi Sorry about that. Trying to do too much at once.

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