从PHP Curl获得POST Zapier webhook响应

Question

I´m using Zapier Webhooks for integration between a Web Portal and an API. Both, the web portal and the API was developed with php. I´m using curl to send a request to a catch hook in Zapier, sending the Post Data. Then in my zap, the second step toke the catched fields and send it as form data to and PUT webhook to my API URL.

The problem is that in my portal, when I print the response of this call it always show something like this:

{"status": "success", "attempt": "5a81c6d1-bb9b-4afe-9ece-0cba4a0a52b0", "id": "cec1978a-c98f-4521-89f3-83a4041c15a4", "request_id": "5a81c6d1-bb9b-4afe-9ece-0cba4a0a52b0"}

But I need the real response of the webhook in the second step, showed in my zapier task as Data Output.

Someone knows why this is happening and how can I get the real response?

Thanks

Answer 1

David here, from the Zapier Platform team.

What you're describing isn't currently possible. When Zapier receives a hook, it lets the sender know the hook was successful and they need not retry; this isn't customizable.

In this situation, it sounds like you need a simple webserver, so you can customize the in and out behavior of your data? Or skip Zapier entirely and send the webhook directly to the destination.