Home Forums The XBRL API Getting error – 'Bad or expired token'

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    • #157736
      Sriram Srinivasan

      Hello everyone,

      I’m trying to access XBRL data from Python using the API. I’m new to Python and working with APIs so I’m going through some teething pains and would be glad to have some assistance.

      As a first step, I had to get the access token, which I successfully managed to do in Python using the below code (all sensitive data is masked).

      import requests
      API_ENDPOINT = “https://api.xbrl.us/oauth2/token”

      data = {‘grant_type’:’password’,
      ‘password’: ‘xxxxxxxxxx’}

      # sending post request and saving response as response object
      r = requests.post(url = API_ENDPOINT, data = data)

      # extracting response text
      pastebin_url = r.text

      This gave me output including an access token and a refresh token. I’m now trying to use the access token to pull data.

      This is the code I have been working with:

      “import requests

      # api-endpoint
      URL = “https://api.xbrl.us/api/v1/fact/search?concept.local-name=Assets&period.fiscalyear=2017&period.fiscal.period=1Q&Entity.cik=0000001&fields=fact.value”

      PARAMS = {‘Host’:’api.xbrl.us’,
      ‘Authorization’: ‘Bearer<xxxxxxx-xxxxx-xxx-xxx-xxxxx>’}

      # sending get request and saving the response as response object
      r = requests.get(url = URL, params = PARAMS)

      # extracting data in json format
      data = r.json()

      This returns the error: ‘Bad or expired token’. I have tried getting a fresh access token and running the code again but I’m still getting the same error.

      Is there anything wrong with my code? Will be grateful for any guidance. Thanks in advance for your help.

    • #157795
      Marc Ward

      Hi Sriram,

      You’re almost there! The token needs to submitted in the header of the request. You don’t need to submit any parameters since they’re in the url.

      Add this to your code:

      headerarray = {'Authorization' : 'Bearer xxxxxxx-xxxxx-xxx-xxx-xxxxx'}

      Then change your request line to:

      r = requests.get(url, headers=headerarray)

      You should receive the results of the query now.

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