Python Rest client api to upload a file

Question

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': 'abc@gmail.com', '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.

Possible duplicate of [Python Requests - Post a zip file with multipart/form-data](https://stackoverflow.com/questions/18208109/python-requests-post-a-zip-file-with-multipart-form-data) — stovfl, Oct 24 '18 at 19:38
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. 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. — Malar Kandasamy, Oct 25 '18 at 06:57
***to 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 and `result.json()` you get. — stovfl, Oct 25 '18 at 07:43
'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"} — Malar Kandasamy, Oct 25 '18 at 10:10

score 5 · Answer 1 · answered Oct 24 '18 at 19:18

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']

It's just an example pulled from the site. You would need to modify it for your purposes. — eatmeimadanish, Oct 25 '18 at 14:46

score 3 · Answer 2 · answered Oct 25 '18 at 06:51

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': 'abc@gmail.com', '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")

Here, 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)} — Malar Kandasamy, Oct 25 '18 at 06:59

Python Rest client api to upload a file

2 Answers2