How do I format a Curl command to work with Zapier Webhooks?

Question

I'm trying to use a curl request in Zapier. I'm unable to get my formatting right for this particular situation.

curl -X POST https://proxy.twilio.com/v1/Services/KSXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX/Sessions \
--data-urlencode "UniqueName=MyFirstSession" \
-u ACXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX:your_auth_token

Do you mean the webhook request action? What exactly have you tried, and what's the problem with it? — jonrsharpe, Jun 21 '19 at 22:31
@jonrsharpe essentially I want to do this: https://stackoverflow.com/questions/50789542/converting-a-curl-command-to-run-in-a-zapier-webhook/56710842#56710842 But with the twilio code — Evan Lodge, Jun 21 '19 at 22:44
everything abode should map right into the zapier form. the curl command has a url, a method, a form parameter, and a basic auth pair. If you're not getting it, can you update your question to show what you've tried and what error message it shows? — xavdid, Jun 28 '19 at 18:20

score 0 · Answer 1 · answered Jun 24 '19 at 05:39

I think that something like this is what you want you obviously just need to fill in the correct data. Are you using the webhook as the trigger or an action? Let me know if you have any questions. Top of request The bottom of the request Aslo be sure to repace KSXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX and put in your your_auth_token.

How do I format a Curl command to work with Zapier Webhooks?

1 Answers1