2 ms·

Thanks for this explanation. So if I understand correctly, `someOp` has thrown an error but this doesn't affect `someOtherOp`? So `someOtherOp` will end up bein

Thanks for this explanation. So if I understand correctly, `someOp` has thrown an error but this doesn't affect `someOtherOp`? So `someOtherOp` will end up being called twice?

inglor 6y ago
Correct, the easy workaround is not to use that transaction API and write your own disposer instead of using withTransaction.