notice: array to string conversion in php

Question

prepare(\"SELECT p_id FROM players_to_team WHERE t_id = ?\");

    $t         


        

Answer 1

Try replacing $player[] = array(); by $player = array(); at the beginning (line 2).

This is because that you declare an array at the index 0 of this variable which is told to be an array because of the []. You therefore try to place an array in your array, making it multidimensional.