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Help Looping an Option Form


topflight

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I am trying to create a simple option form and have it loop at the name in the database, so that all the names will show in the option menu. This is for a form I am working on here is my code

 

<?php
session_start();
if(isset($_SESSION['SESS_LOGGEDIN']) == FALSE){
      echo "You must be logged in";
}
else {


include 'db.php';
$login = $_SESSION['login'];
$lname = $_SESSION['SESS_LNAME'];
$fname = $_SESSION['SESS_FNAME'];

$connect = mysql_connect($db_host,$db_username,$db_password)
  or die("MySQL Said:".mysql_error());
  
$database = mysql_select_db($db_database,$connect)
  or die("MySQl Said:".mysql_error());

$checkl = mysql_query("SELECT * FROM `pilots` WHERE lname='$lname' AND fname='$fname'")
  or die("MySQL Said:".mysql_error());

$edata=mysql_fetch_assoc($checkl);

$link = mysql_query("SELECT * FROM `pilots` WHERE status >'0' ORDER BY login ASC")
  or die("MySQL Said:".mysql_error());
  

  

  

if($bm=='1'){

?>

<form method="post" action="?page=addstaff_phrase" name="adds"/>
<select name="pilots"> 
<?php while($data = mysql_fetch_assoc($link))?>{<option><?php echo"$fname";?> <? echo"$lname";?> - <?php echo"$pid"; ?> </option>
</select>
</form>
<?php
}


}
  $pid = $data['login'];
  $fname = $data['fname'];
  $lname = $data['lname']; 



?>

I know this sounds simple but I am confuse on how to do it please help thanks.

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<?php
$query = "SELECT * FROM `pilots` WHERE lname='$lname' AND fname='$fname'";
$result = mysql_query($query) or die("MySQL Said:".mysql_error());

$pilots = array();
while ($pilot = mysql_fetch_assoc($result)) {
    $pilots[$pilot['id']] = $pilot['lname'] . ',  ' . $pilot['fname'];
}
?>

<select name="pilot">
<?php foreach ($pilots as $id => $name): ?>
    <option value="<?php echo $id ?>"><?php echo $name ?></option>
<?php endforeach ?>
</select>

I have changed my code to this:

<?php
session_start();
if(isset($_SESSION['SESS_LOGGEDIN']) == FALSE){
      echo "You must be logged in";
}
else {



include 'db.php';
$login = $_SESSION['login'];
$lname = $_SESSION['SESS_LNAME'];
$fname = $_SESSION['SESS_FNAME'];

$connect = mysql_connect($db_host,$db_username,$db_password)
  or die("MySQL Said:".mysql_error());
  
$database = mysql_select_db($db_database,$connect)
  or die("MySQl Said:".mysql_error());

$checkl = mysql_query("SELECT * FROM `pilots` WHERE lname='$lname' AND fname='$fname'")
  or die("MySQL Said:".mysql_error());

$edata=mysql_fetch_assoc($checkl);

$result = mysql_query("SELECT * FROM `pilots` ORDER BY login ASC")
  or die("MySQL Said:".mysql_error());
  

$bm = $edata["bm"];

  

if($bm=='1'){

$pilots = array();
while ($pilot = mysql_fetch_assoc($result)) {
    $pilots[$pilot['id']] = $pilot['lname'] . ',  ' . $pilot['fname'];
}
?>

<select name="pilot">
<?php foreach ($pilots as $id => $name): ?>
    <option value="<?php echo $id ?>"><?php echo $name ?></option>
<?php endforeach ?>
</select>
<?
} else {echo'<font color="#FF0000"> Insufficient Rights </font>';}?>
<?
} 
?>

And it is still on displaying one name in the drop down menu.

Are you sure you are getting more than one row from the query?

 

Add this and make sure

<?php
$result = mysql_query("SELECT * FROM `pilots` ORDER BY login ASC")
  or die("MySQL Said:".mysql_error());

die("num results: " . mysql_num_rows($result));

Now my code look like this:

<?php
session_start();
if(isset($_SESSION['SESS_LOGGEDIN']) == FALSE){
      echo "You must be logged in";
}
else {



include 'db.php';
$login = $_SESSION['login'];
$lname = $_SESSION['SESS_LNAME'];
$fname = $_SESSION['SESS_FNAME'];

$connect = mysql_connect($db_host,$db_username,$db_password)
  or die("MySQL Said:".mysql_error());
  
$database = mysql_select_db($db_database,$connect)
  or die("MySQl Said:".mysql_error());

$checkl = mysql_query("SELECT * FROM `pilots` WHERE lname='$lname' AND fname='$fname'")
  or die("MySQL Said:".mysql_error());

$edata=mysql_fetch_assoc($checkl);


$result = mysql_query("SELECT * FROM `pilots` ORDER BY login ASC")
  or die("MySQL Said:".mysql_error());

die("num results: " . mysql_num_rows($result));

$bm = $edata["bm"];

  

if($bm=='1'){

$pilots = array();
while ($pilot = mysql_fetch_assoc($result)) {
    $pilots[$pilot['id']] = $pilot['lname'] . ',  ' . $pilot['fname'];
}
?>

<select name="pilot">
<?php foreach ($pilots as $id => $name): ?>
    <option value="<?php echo $id ?>"><?php echo $name ?></option>
<?php endforeach ?>
</select>
<?
} else {echo'<font color="#FF0000"> Insufficient Pilot Privileges </font>';}?>
<?
} 
?>

And on the page it says num rows three. I am trying to display all the member in the MySQL Database in an option menu.

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