Need help, getting errors

tekhneek

I'm trying to write a program that lets a gamer pre-register (addplayer.php) for a LAN party, then display the people who have registered (viewplayer.php) in a drop down menu list, the addplayer.php works fine, but I can't seem to get the query to select and display the info to work, maybe you guys can help, here's the code.
html>
head>
title>/title>
/head>body>
select name="player_list">
?php
$db = mysql_connect("localhost", "******", "wayne224");
$Query = mysql_query("SELECT player_name FROM bawm_players", $db);
$NumRows = mysql_num_rows($Query);
while ($Count < $NumRows)
{
$Display = mysql_fetch_array($Query, MYSQL_ASSOC);

echo "option value='" . $Display['id'] . "'>" . $Display['player_name'] . "/option>";
}
?>
/select>
/body>
/html>

www.bawmlan.com/phpplayground/playerscript/addplayer.php
www.bawmlan.com/phpplayground/playerscript/viewplayer.php
please help, thanks.

oh yeah and if you view source on viewplayer.php it has 2 errors, i'm sorry i've only been programming in php for a week so i'm still kind of new

TheDefender

1st thing 1st... Always enclose your code using the PHP button above, so things like your tags don't get messed up when posting:

Secondly, try changing your code to the below code:

<html> 
<head> 
<title></title> 
</head><body> 
<select name="player_list"> 
<?php 
$db = mysql_connect("localhost", "******", "wayne224"); 
$Query = mysql_query("SELECT player_name FROM bawm_players", $db); 

while ($Display = mysql_fetch_array($Query, MYSQL_ASSOC)){ 
echo "<option value=\"" . $Display['id'] . "\">" . $Display['player_name'] . "</option>"; 
} 
?> 
</select> 
</body> 
</html>

chads2k2

Hey things look ok ... though I re-wrote the code with some opening and closing tags. What I added was a selecting a database. I think your not selecting your database.

mysql_select_db("databasename",$db);

So here is your code with everything edited.

//IGNORE THIS CODE.  Read next post from me.  My bad.

Hope that will help! Thanks so much for posting on the board!

Chad R. Smith

chads2k2

Sorry I found a bug. My bad.

<html>
<head>
<title></title>
</head>

<body>
<form action="where_am_i_going.php" method="POST">
<select name="player_list">
<?php
	$db = mysql_connect("localhost", "******", "wayne224");
	mysql_select_db("database_name",$db);

$sql = "SELECT player_name FROM bawm_players";
$result = mysql_query($sql);
$num_rows = mysql_num_rows($sql);
if($num_rows < 0)
{
	echo "Sorry but nobody has signed up yet.";
	exit();
}
   while ($row = mysql_fetch_array($result)) {
		$id = $row['id'];
		$player_name = $row['player_name'];
		echo "<option value='$id'>$player_name</option>";
	}
?>
</select>
</body>
</html>

tekhneek

thanks alot guys! I really want to get very good at this. Thanks so much for replying

chads2k2

It is all good. Take care and keep coding!

Chad