Sorry for firing the issue here, but I don't know where to push it.
Currently, when creating a subscription through Payjp (monthly/yearly payment) we only can specific the customerId and the planId without cardId.
The scenario is if a customer has two cards name card A and card B, then he wants to buy a subscription with card B but the default card is card A, the problem will occur since he will be charge by card A instead of card B.
I know that we can create different customerId for each card but our system is running on production 4 years ago, so we can't remove all old cards of our customers.
So please help me to figure out if there is a valid solution in the solutions list below:
- Create a
new customerId list from old list cards. We already stored card id (car_325xxxxxxxxxxxxx) in our DB, so can we use the old card id list to create a new customers list?
- Create a
new customerId list and a new cards list from old list cards?
- Or create subscription by
cardId instead customerId?
I read all the Payjp API docs but can't find anyway...
Thanks and best regards.
Sorry for firing the issue here, but I don't know where to push it.
Currently, when creating a subscription through Payjp (monthly/yearly payment) we only can specific the
customerIdand theplanIdwithoutcardId.The scenario is if a customer has two cards name
card Aandcard B, then he wants to buy a subscription withcard Bbut the default card iscard A, the problem will occur since he will be charge bycard Ainstead ofcard B.I know that we can create different customerId for each card but our system is running on production 4 years ago, so we can't remove all old cards of our customers.
So please help me to figure out if there is a valid solution in the solutions list below:
new customerId listfromold list cards. We already stored card id (car_325xxxxxxxxxxxxx) in our DB, so can we use the old card id list to create a new customers list?new customerId listand anew cards listfromold list cards?cardIdinsteadcustomerId?I read all the Payjp API docs but can't find anyway...
Thanks and best regards.