This topic was automatically generated from Slack. You can find the original thread here.
I previously reported an issue where, if multiple accounts are connected for the same app under one external user, there’s no way to choose which account to use when running a tool. The SDK only lets you specify the external user and the app, not the specific account.
Because of this, I’m forced to create a separate “external user” in Pipedream for every account a single user wants to connect, even when they’re the same person. This is frustrating, since you bill by the number of external users, which means I’m getting charged more than I should just to work around this limitation.
To make matters worse, your product does allow connecting multiple accounts of the same app to a single user. However, it just doesn’t let me select the specific account when running an action, just the action id and user. This feels inconsistent and creates unnecessary overhead and cost.
Got it, the idea for additional users is billing for each user + their apps combination, not sure if having multiple accounts for the same app was thought of, I’ll check with the team
I understand the point, but it’s an inconsistency that seems overlooked. I can create two different accounts for the same app and user, yet I can’t use it.
The issue is that, since my app does support multiple accounts for the same app and user, I am now forced to register every new account as a completely different “external user” for Pipedream (because using the same one does work properly). This workaround is costing me money because of how pricing works on Pipedream.
Now, if the same user adds 4 different accounts, even for different apps, I am forced to register it as 4 different external users, just to be able to support it
Personally in terms of api design I would expect the connected account id to be specified outside the configurable props alongside the external user id (or even in place of the external user id, why do you even need it, if you already have the connected account?)