create dropdown menu on each loop assign database value to option element - php

Question

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 ONLY relevant to the position. (i.e if the positions name is flanker only flankers should be displayed in the drop-down menu)
My logic for the above problem is:
Assign position names to array called $position Loop over array
querying database, with position having a different value after each
loop.
Selected players that match the position gets assigned to an array
called $playerName
Print the $position variable
Create dropdown menu for each position
assign value from $playername array to the option element in the drop
down menu.
Now there should be different positions with dropdown menus, next to
them, containing player names relevant to position.
//create position array
$position = array(
"THP",
"HKR",
"LH",
"LK4",
"LK5",
"FLH"
);
echo '<form name="slectPlayers" method="post">';
foreach ($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[] = $row['surname'];
// print position and open select element
foreach ($position as $playerPosition) {
print $playerPosition;
echo '<select>';
foreach ($playerName as $name) { //assign playername to position element
echo '<option>' . $name;
'</option>';
echo '</select>';
echo '<br />';
} //close assign player nae to position loop
} //end print position and open select loop
} //end while loop
} //end opening for each loop
echo '</form>';
unfortunately for me either my logic is wrong or my code is incorrect. This is the ouput I get: (Note only the name Tendai is displayed in all dropdown meus no other names appear)
Ive been struggling with this one all morning if someone could point me in the right direction it would be greatly appreciated
(note to modirators the picure above does not contain real names and is only a fictional database)

Answer 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) {

Answer 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);
}