Python Rest client api to upload a file
Solution 1
requests library worked with below changes in my code :
import requests
from requests.auth import HTTPBasicAuth
import json
from pathlib import Path
file_ids = ''
headers={'Username': '[email protected]', 'apikey':'123-456'}
# Upload file
f = open('C:/Users/ADMIN/Downloads/abc.zip', 'rb')
files = {"file": ("C:/Users/ADMIN/Downloads/abc.zip", f)}
resp = requests.post("https:// ../analytics/upload_file", files=files, headers=headers )
print resp.text
print "status code " + str(resp.status_code)
if resp.status_code == 201:
print ("Success")
data = json.loads(resp.text)
file_ids = data['file_ids']
print file_ids
else:
print ("Failure")
Solution 2
Use urllib3 module.
https://urllib3.readthedocs.io/en/latest/user-guide.html
Files & binary data
For uploading files using multipart/form-data encoding you can use the same approach as Form data and specify the file field as a tuple of (file_name, file_data):
with open('example.txt') as fp:
file_data = fp.read()
r = http.request(
'POST',
'http://httpbin.org/post',
fields={
'filefield': ('example.txt', file_data),
})
json.loads(r.data.decode('utf-8'))['files']
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Malar Kandasamy
Updated on October 25, 2020Comments
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Malar Kandasamy over 3 years
I am using Python 2.7. My Rest server side API works fine and I am able to upload a zip file using Postman. I am trying to upload a zip file using Rest client api. I tried requests package, but it is unable to send the files. I am getting an error : missing file argument.
This is my python server side code :
@ns.route('/upload_file', strict_slashes=False) class Upload(Resource): @api.expect(upload_parser) def post(self): args = upload_parser.parse_args() file_nameup = args.file.filename
Here is the rest api client code :
import requests from requests.auth import HTTPBasicAuth import json headers={'Username': '[email protected]', 'apikey':'123-e01b', 'Content-Type':'application/zip'} f = open('C:/Users/ADMIN/Downloads/abc.zip', 'rb') files = {"file": f} resp = requests.post("https://.../analytics/upload_file", files=files, headers=headers ) print resp.text print "status code " + str(resp.status_code) if resp.status_code == 200: print ("Success") print resp.json() else: print ("Failure")
This is my error : {"message":"Input payload validation failed","errors":{"file":"Missing required parameter in an uploaded file"} status code 400 Failure
In postman, I passed a zip file with in body with "file" as key and value as abc.zip file. It worked fine. I tried to use httplib library, but it fails as my post url does not contain port number. This the error with httplib :
python HttpClientEx.py Traceback (most recent call last): File "HttpClientEx.py", line 4, in h = http.client.HTTPConnection(url) File "c:\python27\Lib\httplib.py", line 736, in init (self.host, self.port) = self._get_hostport(host, port) File "c:\python27\Lib\httplib.py", line 777, in _get_hostport raise InvalidURL("nonnumeric port: '%s'" % host[i+1:]) httplib.InvalidURL: nonnumeric port: '// ....net/analytics/upload_file'
How to invoke rest url post and upload a file using urllib library. Please suggest any other ways to upload file in rest client. Thanks.
I found another duplicate post :
Python Requests - Post a zip file with multipart/form-data
The solution mentioned there did not work. I found that you need to provide the full path of the file, otherwise it will not work.
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stovfl over 5 yearsPossible duplicate of Python Requests - Post a zip file with multipart/form-data
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Malar Kandasamy over 5 yearsThe solution mentioned there did not work. I found that you need to provide the full path of the file, otherwise it will not work. example: files = {"file": ("C:/Users/ADMIN/Downloads/abc.zip", f)} if I specify files = {"file": ("abc.zip", f)}, its not working. Even though my program and zip are in the same folder.
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stovfl over 5 yearsto provide the full path: This makes no sense. The first parameter in the
tuple
is used from the server, how to name the uploaded file. I could not imagin how a local path are usefull for a server at all. its not working: This is not helpfull, edit your Question and show requests header andresult.json()
you get. -
Malar Kandasamy over 5 years'file' is the parameter name referred in my server rest api code. Refer python server side code above: file_nameup = args.file.filename What you said is correct. Somehow, the api requires a full path of zip file path, thats why I took long time to find the issue. This is my result : {"message": "File uploaded successfully", "file_ids": "123.zip"}
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Malar Kandasamy over 5 yearsHere, note I removed Content-Type from my header, also I am providing full file name with path in files = {"file": ("C:/Users/ADMIN/Downloads/abc.zip", f)}
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eatmeimadanish over 5 yearsIt's just an example pulled from the site. You would need to modify it for your purposes.