Converting cURL request to Python

Question

I'm trying to turn this cURL request into a Python code. I want to eventually be able to save this to a CSV file but I need to get connected first.

curl --compressed -H 'Accept: application/json' -H 'X-Api-Key: 123abc' 'https://us.market-api.kaiko.io/v2/data/trades.v1/exchanges/cbse/spot/btc-usd/aggregations/count_ohlcv_vwap?interval=1h'

I started with this:

import requests
import json

key='api-key'

url = 'https://us.market-api.kaiko.io/v2/data/trades.v1/exchanges/'
s = requests.Session()
s.auth = (key)

headers = {
   *not sure how to do this*
}
r = requests.get(url, headers=headers)

The docs say this needs to be in the header:

Accept: application/json
Accept-Encoding: gzip:

How do I include the API key? how do I save the data once its returned?

You could just put the curl request in a subprocess.check_output command and go from there. — I break things
– I break things, Commented Jan 19, 2021 at 19:23

Axiumin_ · Accepted Answer · 2021-01-19 19:25:37Z

X-Api-Key would be a request header, so you can include it in your headers variable, like this:

headers = {
   "X-Api-Key": key,
   "Accept": "application/json",
   "Accept-Encoding": "gzip"
}

(took the others ones from your current curl request)

You can get the data by using r.text, like this:

print(r.text)

Your code should look like this:

import requests
import json

key='api-key'

url = 'https://us.market-api.kaiko.io/v2/data/trades.v1/exchanges/'

headers = {
   "X-Api-Key": key,
   "Accept": "application/json",
   "Accept-Encoding": "gzip"
}

r = requests.get(url, headers=headers)
print(r.text)

If you want to get a json object instead, you can use r.json()

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