I have a rest api that offers an upload file post functionality. To test this, I simply did this on my python test code:
fo = open('file.zip', 'rb')
rb_file = {'confile': ('file.zip', fo, 'multipart/form-data')}
resp = requests.post(url,files=rb_file)
fo.close()
This request returns a uuid response {ce9f2d23-8ecd-4c60-9d31-aef0be103d44} that is needed for initiating another post run for a scan.
From Swagger, manually passing this uuid to the scan post request generates the following curl:
curl -X POST "http://....../scan" -H "Content-Type: multipart/form-data" -F "FilID=ce9f2d23-8ecd-4c60-9d31-aef0be103d44"
My question is how to convert this curl to a python request code noting that I no longer have the file on my local machine. The server seems to be expecting a -F rather than a param. How do I pass a file that doesn't exist on my local machine to http request in this case? All I have is the filID. I tried running as param but that doesn't find the resource.
try this !
import requests
headers = {
'Content-Type': 'multipart/form-data',
}
files = {
'FilID': (None, 'ce9f2d23-8ecd-4c60-9d31-aef0be103d44'),
}
response = requests.post('http://....../scan', headers=headers, files=files)
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