Get XERO API Access code error: unsupported_grant_type

Question

I’m trying to get accessToken from Xero

This is their instruction:
https://developer.xero.com/documentation/guides/oauth2/auth-flow/#3-exchange-the-code

POST https://identity.xero.com/connect/token
authorization: "Basic " + base64encode(client_id + ":" + client_secret)
Content-Type: application/x-www-form-urlencoded

grant_type=authorization_code
&code=xxxxxx
&redirect_uri=https://myapp.com/redirect

This is how I do it:

tokenb4 = "{0}:{1}".format(CLIENT_ID, CLIENT_SECRET)
basic_token = base64.urlsafe_b64encode(tokenb4.encode()).decode()
response = requests.post(
    "https://identity.xero.com/connect/token",
    headers={
        "Authorization": "Basic {0}".format(basic_token),
        "Content-Type": "application/x-www-form-urlencoded"
    },
    params="grant_type=authorization_code 
            &code=MYCODE 
            &redirect_uri=https://developer.xero.com/"
)

Now I keep getting error 400
{"error":"unsupported_grant_type"}

redirect_uri is the same with the one I leave in the App in Xero demo company.
I’ve googled some, seems like it’s a format issue but I can’t see what’s wrong with my code.

Much appreciation if anyone could help

Asked By: maggie3003

||

Source

Answer 1

ref:Python requests module sends JSON string instead of x-www-form-urlencoded param string

I changed the params to

data={
        "grant_type": "authorization_code",
        "code": "MYCODE",
        "redirect_uri": "https://developer.xero.com/"
    }

And it works! Hooray!

Answered By: maggie3003

Answer 2

I was getting the same error. It turns out i was converting my body to a json object. Removing that code got it to work.

Answered By: Sasha Muller

Get XERO API Access code error: unsupported_grant_type

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