How to post 'multipart/form-data' content without file via python requests?

Question

I want to post password to server. Here's the server's source code(mainly):

<FORM ENCTYPE="multipart/form-data" METHOD=POST> 
Password:<INPUT NAME="pw1" TYPE="password">
Password:<INPUT NAME="pw2" TYPE="password">   
<INPUT TYPE="submit" VALUE="Confirm">

Below is my code:

import requests
url = 'http://192.168.0.1/pw'
file = {"pw1":"abc","pw2":"abc"}
r = requests.post(url, files = file)

From wireshark, I get:

--4d4bb99654064447b31a7afb787e5dbc
Content-Disposition: form-data; name="pw1"; filename="pw1"

abc
--4d4bb99654064447b31a7afb787e5dbc
Content-Disposition: form-data; name="pw2"; filename="pw2"

abc
--4d4bb99654064447b31a7afb787e5dbc--

What I expect is that there's no filename="pw1"/"pw2" i.e.,

--4d4bb99654064447b31a7afb787e5dbc
Content-Disposition: form-data; name="pw1"

abc
--4d4bb99654064447b31a7afb787e5dbc
Content-Disposition: form-data; name="pw2"

abc
--4d4bb99654064447b31a7afb787e5dbc--

The telegram can be recognized by server only in this way. How to solve handle the post request? Do I have to use files if I want to use requests library in this case? Thank you for any help.

Vikas Ojha · Accepted Answer · 2017-08-28 15:51:23Z

3

For the form html you have posted, you dont have to send the input elements in files dict. The following code snippet should work -

import requests
url = 'http://192.168.0.1/pw'
data = {"pw1":"abc","pw2":"abc"}
files = {'file': ''}
r = requests.post(url, data=data, files=files)

answered Aug 28, 2017 at 15:51

Vikas Ojha

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2 Comments

xin Over a year ago

Yes it works. Thank you. From the guideline here, there's no such format as you posted. So is there a more detailed manual for a novice?

Vikas Ojha Over a year ago

I am glad it worked for you. You can accept the answer then.

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