ok tak så langt ... brugeren kender arrangement navn her tjekker jeg bla om der ledige pladser
skal selv til påske frokost så der er ingen respons før i aften
her er min kode den sidste query giver ingen data
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="
http://www.w3.org/1999/xhtml"><head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Oversigt over tilmeldte</title>
</head>
<body>
<?php
$aid = $_GET['aid'];
include 'connect.php';
$result = mysql_query("SELECT * FROM tilmdb WHERE arrid='$aid'")
//$result = mysql_query("SELECT arrdb.adag, tilmdb.* FROM arrdb JOIN tilmdb ON arrdb.anavn = tilmdb.arrid WHERE tilmdb.arrid = $aid.")
or die(mysql_error());
echo '<form id="form1" name="form1" method="post" action="tilm_list.php?aid='.$aid.'">
<table width="1000" border="1" cellpadding="0" cellspacing="0">
<tr>
<td width="172">Navn</td>
<td width="175">Adresse</td>
<td width="109">Telefon</td>
<td width="173">E-mail</td>
<td width="89">Antal voksne</td>
<td width="89">sted</td>
<td width="70">Slet</td>
</tr>';
while($row = mysql_fetch_array( $result )) {
$tid = $row['tid'];
$navn = $row['tnavn'];
$adr = $row['tadr'];
$by = $row['tby'];
$pnr = $row['tpnr'];
$mail = $row['tmail'];
$tel = $row['ttel'];
$numv = $row['numv'];
$numb = $row['numb'];
$anavn = $row['anavn'];
echo '<table width="1000" border="1" cellpadding="0" cellspacing="1">
<tr>
<td width="172">'.$navn.'</td>
<td width="175">'.$adr.' '.$by.', '.$pnr.'</td>
<td width="109">'.$tel.'</td>
<td width="173">'.$mail.'</td>
<td width="89">'.$numv.'</td>
<td width="89">'.$anavn.'</td>
<td width="70"><a href="delete.php?tid=' . $tid . '" onclick="return confirm(\'Er du sikker?\');">SLET</a>
</tr>';
}
?>
</body>
</html>