Help Looping an Option Form

[topflight]

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.

[flyhoney]

<?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>

[topflight]

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.

[flyhoney]

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));

[topflight]

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.

[topflight]

Also three is right becuase I currently have three members in the database.

[flyhoney]

I dont know your database schema.  What is the primary key for your pilots table?  You need to use that in the line:

$pilots[$pilot['PRIMARY_KEY']] = $pilot['lname'] . ',  ' . $pilot['fname'];

[topflight]

id I have my primay key set up as ID so what do I do now?

[topflight]

any other help.

[topflight]

?

[topflight]

Any Help would be appreicated

[timmah1]

Wouldn't this do it

 

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