PHP mysqli wrapper: passing by reference with __call() and call_user_func_array()

Question

I\'m a long running fan of stackoverflow, first time poster. I\'d love to see if someone can help me with this. Let me dig in with a little code, then I\'ll explain my probl

Answer 1

I'm guessing this is because the original function signature specifies that it expects references, whereas your __call cannot do so. Therefore, try not using __call but explicitly adding the bind_result with the same function signature as the original.