querying database with array variable [duplicate] - php

Question

This question already has answers here:
create dropdown menu on each loop assign database value to option element
(2 answers)
Closed 9 years ago.
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)
If 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

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);
}

Answer 2

In your original code, you could remove the two foreach loops in the while loop. Use $pposition in place of $playerPosition and your code should work as expected. Here's your code, with my suggested changes, in your procedural style:
$position = array(
"THP",
"HKR",
"LH",
"LK4",
"LK5",
"FLH"
);
echo '<form name="selectPlayers" method="post">';
foreach ($position as $pposition) {
$result = mysql_query("SELECT `player_id`,`name`,`surname` FROM `player_info`
WHERE `position` = '$pposition'") or die(mysql_error());
echo '<select name="'. $pposition . '">';
while ($row = mysql_fetch_array($result)) {
print $pposition;
echo '<option value="' . $row['player_id'] . '">'.
$row['name'] .' '. $row['surname'].
'</option>';
} //end while loop
echo '</select>';
echo '<br />';
} //end opening for each loop
echo '</form>';