Create Dropdown Menu On Each Loop Assign Database Value To Option Element
Im working on a page for a sports team where the coach can select his team. What im trying to do is: 1) Print different positions 2) Assign next to position name, players who are O
Solution 1:
This could be your problem
foreach ($position as $pposition) {
$result = mysql_query("SELECT `player_id`,`name`,`surname` FROM `player_info` WHERE `position` = '$pposition'") or die(mysql_error());
while ($row = mysql_fetch_array($result)) { //create arrays
$id[] = $row['player_id'];
$playerName[] = $row['name'];
$playerLastName[] = $row['surname'];
// print position and open select element
print $pposition;
echo '<select>';
foreach ($playerName as $name) { //assign playername to position element
echo '<option>' . $name;
'</option>';
echo '</select>';
echo '<br />';
} //close assign player nae to position loop
} //end while loop
} //end opening for each lo
I have removed foreach ($position as $playerPosition) {
Solution 2:
You can do it like this. the logic is something different
function get_players_by_position($position)
{
$query = "SELECT
`player_id`,
`name`,
`surname`
FROM `player_info`
WHERE `position` = $position
ORDER BY name ";
$result = mysql_query($query);
$data = array();
while ($row = mysql_fetch_array($result)) {
$data[] = $row;
}
return $data;
}
foreach($position as $pposition){
$data = get_players_by_position($pposition);
echo $pposition;
echo '<select>';
foreach($data as $row){
echo '<option '.$row['id'].'>' . $row['name'].'</option>';
}
echo '</select>';
echo '<br />';
unset($data);
}
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