PHP dynamical duplicate Mysqli results set - php

Question

I am attempting to populate 0 to many drop downs with the same results from a database table depending on a previous selection. it is working fine when 0 and 1 are selected but not when i am attempting to insert the result set into subsequent select elements. i am assuming it is a problem with the $row array position.
$homePlayers = "SELECT first_name, last_name, player_id FROM players WHERE team_name LIKE '$homeTeam%'";
$homePlayersQuery = mysqli_query($dbc, $homePlayers);
if (!$homePlayersQuery) {
echo 'err';
} else {
for ($i = 1; $i <= $homeTeamScore; $i++) {
echo "<select name='select-home-scorer-$i'>";
while ($row = mysqli_fetch_array($homePlayersQuery)) {
echo current($row);
echo "<option value='" . $row['player_id'] . "'>" . $row['first_name'] . " " . $row['last_name'] . "</option>";
}
echo "<option value='og'>Own Goal</option></select><br/>";
}
}
}

Answer 1

The problem you experience is that mysqli_fetch_array() reads data of the result set once. When you have read the last record, subsequent calls to mysqli_fetch_array() will return false.
You have 2 choices.
Firstly, you can read the records into an array and repeatedly parse the array.
while ($row = mysqli_fetch_array($homePlayersQuery)) {
$resultsArray[] = $row;
}
for ($i = 1; $i <= $homeTeamScore; $i++) {
foreach($resultsArray as $resultItem) {
do_something_here();
}
}
}
As a second option, you could rewind the pointer to the mysql result set so that you can start reading the results again.
for ($i = 1; $i <= $homeTeamScore; $i++) {
// Rewind the pointer to the start of the results
mysqli_data_seek($result, 0);
while ($row = mysqli_fetch_array($homePlayersQuery)) {
$resultsArray[] = $row;
}
}