Reputation: 801
I am trying to send an excel file to server in POST Request in multipart/form-data content-type. I am getting an error:
too many values to unpack
What could be the reason? Below is the request what I am trying:
#Data = get_data('C:\foo.xls')
#print Data
Data = open('C:\foo.xls', 'rb')
print Data
headers = {
'access-control-allow-origin': '*',
'accept': 'application/json',
'content-type': 'multipart/form-data',
'authorization': 'Basic xxxxxxxxx'
}
R = requests.post('http://testserver:8080/v1/readyapi/executions/'+executionId+'/files', headers=headers, params=params, files=Data)
print R.content
here is the error:
Traceback (most recent call last):
(body, content_type) = self._encode_files(files, data)
File "C:\Python27\lib\site-packages\requests\models.py", line 132, in _encode_files
for (k, v) in files:
ValueError: too many values to unpack
I could not figure out by myself. tried few things, didn't work. Can someone please advise?
Upvotes: 1
Views: 19210
Reputation: 1
Try this, right MIME type is important (https://wiki.selfhtml.org/wiki/MIME-Type/%C3%9Cbersicht)
headers = {
'access-control-allow-origin': '*',
'accept': 'application/json',
'content-type': 'multipart/msexcel',
'authorization': 'Basic xxxxxxxxx'
}
Upvotes: 0
Reputation: 11665
Try the following code
Data = open('C:\foo.xls', 'rb')
headers = {
'access-control-allow-origin': '*',
'accept': 'application/json',
'content-type': 'multipart/form-data',
'authorization': 'Basic xxxxxxxxx'
}
files = {"file_name": Data}
url = 'http://testserver:8080/v1/readyapi/executions/'+executionId+'/files'
R = requests.post(url, headers=headers, params=params, files=files)
print R.content
You must have to pass the files argument as a dictionary or you can also try like below
files = {'file': ('foo.xls', open('foo.xls', 'rb'), 'application/vnd.ms-excel', {'Expires': '0'})}
Reference: http://docs.python-requests.org/en/master/user/quickstart/#post-a-multipart-encoded-file
Upvotes: 2