Problem with "select" and displaying info...

[dniwebdesign]

I have this code that gives me a selection box of a list of players from a certain team. The team is selected from a pull down box on an earlier page.

The problem is that it cannot find the number of a player. Everyway I try to do it it cannot find it. However if I move "number AS number" to the beginning just after the "select" call it find the number and stores it as the firstname.

What is wrong here?

  <form name="findplayer" method="post" action="teams.php?ax=editteam">
   
  <div align="center"> 
    <?php
$query = mysql_query("select firstname AS firstname, lastname AS lastname, number AS number from hockeystats_players WHERE team='$team' ORDER BY lastname");
echo "<select name=playersname>";
if (!isset($firstname)) { echo "<option value=null selected>Choose a Player</option>n";}
while(list($firstname,$lastname) = mysql_fetch_row($query))
{
echo "<option value=$number";
if ($firstname == $firstname) { echo ""; }
echo ">$lastname, $firstname</option>";
}
echo "</select>";?>
    <br>
    <input type="submit" name="Submit" value="Submit">
  </div>


</form>

My tables are set up as:

number team firstname lastname gp goals
--------- --------- ------------- ------------ --- -------
16 thunder dawson irvine 50 32

there are more fields but I don't need those selected yet. Just the name and number so I can select the players without confusion, incase there are two with the same name.

[Nightfire]

First off

$query = mysql_query("select firstname AS firstname, lastname AS lastname, number AS number from hockeystats_players WHERE team='$team' ORDER BY lastname"); 

That's wrong, it should be

$query = mysql_query("SELECT firstname, lastname, number FROM hockeystats_players WHERE team='$team' ORDER BY lastname ASC"); 

Just added the caps also, to make things clearer

if (!isset($firstname)) { echo "<option value=null selected>Choose a Player</option>n";} 

That bit will always echo the option, so why take it off the if statement. So just have

echo "<option value=null selected>Choose a Player</option>\n"; 

Now this bit

while(list($firstname,$lastname) = mysql_fetch_row($query))
{
echo "<option value=$number";
if ($firstname == $firstname) { echo ""; }
echo ">$lastname, $firstname</option>";
} 

Just simply change it to

while($row = mysql_fetch_array($query))
{
echo "<option value='" .$row['number']. "'>". $row['lastname']. ", " .$row['firstname']. "</option>";
} 

I really don't see what this

if ($firstname == $firstname) { echo ""; } 

is meant to do?

[missing-score]

$query = mysql_query("SELECT number, firstname, lastname FROM hockeystats_players WHERE team='".$team."' ORDER BY lastname ASC"); 

This may not fix the problem, but it does cut out some unrequired code. You do not have to use AS if you want the data selected as its original name.

Also, with ORDER BY, you should use ASC or DESC (ascending or descening)

[dniwebdesign]

Thanks nightfire