#vivsriva01
1 messages · Page 1 of 1 (latest)
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- vivsriva01, 5 days ago, 13 messages
You'd use the Customer API to create the customer (https://stripe.com/docs/api/customers) in combination with the Stripe-Account header (https://stripe.com/docs/connect/authentication#stripe-account-header) set to the connected account
Thank you, but how do I create a customer for a connected account? I don't see this information on the page.
that's how you do it - you use the API with the Stripe-Account header and the Customer API
Or are you trying to do this in a no-code way?
I am looking at doing it via API.
how would I specify the connect account?
I don't see any parameter that defines this.
Deleted that message since it had your test secret key in it (this is a public channel)
oh, ok, didn't realize it. Thank you.
That looked like python, so you'd do somethign like this:
stripe.Customer.create(
description="My First Test Customer",
stripe_account='{{CONNECTED_ACCOUNT_ID}}',
)
There was an example of this in the second link I sent you https://stripe.com/docs/connect/authentication#stripe-account-header
ok, I see this. I am not probably explaining myself correctly. Ok, let me try with some details. Let's say I am using connect platform, and I have connected-ac A and connected-ac B as two connected account that have onboarded with me. Now, connected-ac A has a a customer with which I need to setup payment. When I create an invoive with direct charge, then the customer should see that it comes from connected-ac A.
to create this invoice, I need to first create a customer of connected-ac A. Is this not right?
Yes, thta is correct
ok, thanks. Then what parameters would have to be passed to create customer x who is a customer of connected-ac A.
somewhere in that payload, we may need to identify the connected-ac A, which I am not sure how to specify.
Have you tried the snippet I already shared with you? That's exactly how you'd create a Customer on a connected acount - you'd pass in stripe_account with the account ID you want
ok, got it. I will try.