Issue
I have a script that calls a POST endpoint but getting a 400 error. Meanwhile, the corresponding cURL request is successful.
First, here is the cURL:
curl -X 'POST' \
'http://localhost:8080/api/predict?Key=123testkey' \
-H 'accept: application/json' \
-H 'Content-Type: multipart/form-data' \
-F '[email protected];type=image/jpeg'
And translated to requests:
import requests
url = 'http://localhost:8080/api/predict?Key=123testkey'
headers = {
'accept': 'application/json',
'Content-Type': 'multipart/form-data',
}
params = {'Key' : '123testkey'}
files = {'image': open('156ac81cde4b3f22faa4055b53867f38.jpg', 'rb')}
response = requests.post(url, files=files, params=params, headers=headers)
Have also tried using a URL that does not include the key, since the key is already specified in params:
import requests
url = 'http://localhost:8080/api/predict'
headers = {
'accept': 'application/json',
'Content-Type': 'multipart/form-data',
}
params = {'Key' : '123testkey'}
files = {'image': open('156ac81cde4b3f22faa4055b53867f38.jpg', 'rb')}
response = requests.post(url, files=files, params=params, headers=headers)
I thought this should be simple but I consistently get the 400 error with requests no matter what I try. Any suggestions?
Edit: have also tried 'image/jpeg' instead of 'image' to no avail.
Edit: replacing the "image" key with "file" unfortunately didn't work either
Edit: It works in postman desktop just fine, and generates the following code. However, this code also throws an error.
The generated code from postman:
import requests
url = "http://localhost:8080/api/predict?Key=123test"
payload={}
files=[
('file',('images19.jpg',open('156ac81cde4b3f22faa4055b53867f38.jpg','rb'),'image/jpeg'))
]
headers = {
'Accept': 'application/json',
'Content-Type': 'multipart/form-data'
}
response = requests.request("POST", url, headers=headers, data=payload, files=files)
print(response.text)
And the error from the previously generated code from postman:
{"detail":"There was an error parsing the body"}
Any help figuring out what is going on would be much appreciated!
Solution
Your issue is in the variable files you need to add with the key 'file' instead of 'image' that's the difference between your curl and your python code, also remove the header because when you pass the file parameter the request set the proper header for send files. for example:
import requests
url = 'http://localhost:8080/api/predict?Key=123testkey'
params = {'Key' : '123testkey'}
files = {'file': open('156ac81cde4b3f22faa4055b53867f38.jpg', 'rb')}
response = requests.post(url, files=files, params=params)
Answered By - Jose Lora