POST request fails on flask

Question

I'm trying to reproduce the code from this page, the complete github code is here:

The application works just fine on the browser, but I'm not able of reproduce the POST request from python.

I have tried with the same data showed on the payload when I use the browser

PEOPLE =  {"fname": "DDoug",
        "lname": "FarDrell"}

url = "http://localhost:5000/api/people"
data = requests.post(url,data=json.dumps(PEOPLE) )

but I get the following errror:

data.text

'{\n  "detail": "Invalid Content-type (), expected JSON data",\n  "status": 415,\n  "title": "Unsupported Media Type",\n  "type": "about:blank"\n}\n'

I tried like this as well:

url = "http://localhost:5000/api/people"
data = requests.post(url,data=json.dumps(PEOPLE) )

BUT got this error:

'{\n  "detail": "Invalid Content-type (application/x-www-form-urlencoded), expected JSON data",\n  "status": 415,\n  "title": "Unsupported Media Type",\n  "type": "about:blank"\n}\n'

djnz · Accepted Answer · 2019-06-11T00:16:05.280

4

Add the Content-Type into your post headers to specify you're sending JSON data:

requests.post(url,data=json.dumps(PEOPLE), headers={'Content-Type': 'application/json'})

You can also use the json parameter to achieve the same result:

requests.post(url, json=json.dumps(PEOPLE))

edited Jun 11 '19 at 00:16

answered Jun 10 '19 at 23:48

djnz

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Some examples don"t have that, how do you know is necessary? – Luis Ramon Ramirez Rodriguez Jun 10 '19 at 23:55
1

By default the request headers is `None`, so will not be read as JSON. You can also (as a shorthand) use the `json=` parameter to avoid `headers={...`. I will update my answer to include this. – djnz Jun 11 '19 at 00:14

POST request fails on flask

1 Answers1