Simple form help needed!

[Seaholme]

Hi,

 

I'm struggling to solve what seems to be potentially quite a simple problem with my form! It's for a really basic web-based racing game, and basically the issue is that in the form, the option I've named "teamid" isn't sending any information forwards so $newteamid in the next section is always =0, which is the default in the MySQL database. I just don't know how to set it! I want the team id from the first part to send through into the second part. Is it possible to have the value of a form option set to being a variable like this??

 

//Find their teams and let them pick which team to use
$findteams = mysql_query("SELECT * FROM team WHERE owner='$id'") or die(mysql_error());
echo "Select which team to race with:<br><form action='regional.php' method=post> <select name=teamid value=teamid> Team:";
while($row = mysql_fetch_array( $findteams )) {
$teamid = $row['id'];
echo "<option name=teamid value={$row['id']}>{$row['teamname']}</option>";
}
echo '</select>';
echo "<bR><br>Is your team ready to go?<br><br><input type=submit name='stage1' value='Begin Race!'></form><br><br>";


if(isset($_POST['stage1']))
{ $_POST['teamid'] = $newteamid;
echo $newteamid;
$startrace = @mysql_query("INSERT INTO regionals (teamid, stage, ownerid) VALUE ('$newteamid', '1', '$id')") or die("Error: ".mysql_error());

Please let me know if I need to explain or elaborate at all, I'd appreciate any help.

Thanks guys!

[PravinS]

check your isset part, it should be like this

if (isset($_POST['stage1']))
{
	$newteamid = $_POST['teamid'];
	echo $newteamid;
	$startrace = @mysql_query("INSERT INTO regionals (teamid, stage, ownerid) VALUE ('$newteamid', '1', '$id')") or die("Error: ".mysql_error());
}

Edited August 6, 2013 by PravinS