Jeg har følgende kode der via linket
</pre>
pcoq.php?siteid=101-1125-5750S
</pre>
skal kunen finde nogle oplysninger og bruge dem det virker ikke!
Hvorfor?
<pre>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
<title>Nyt dokument</title>
</head>
<body>
<?
$database = mysql_connect("localhost","*****************","**************");
mysql_select_db("test",$database);
$foresp = mysql_query("SELECT * FROM pcoquality WHERE siteid='$siteid' ");
$row = mysql_fetch_array($foresp);
$siteplaz = $row["sitepla"];
$siten = $row["siten"];
$sitedesc = $row["sitedesc"];
$siteqi = $row["siteqi"];
$siteqt = $row["siteqt"];
?>
<table width="100%" height="100%" border="1">
<tr>
<td width="100%" height="20%"><?echo("The site $siten have get a Quality price for $sitedesc as the quality type $siteqt<img src='$siteqi' height='120' width='143' > ");?></td>
</tr>
<tr>
<td width="100%" height="*"><? include ("$siteplaz"); ?></td>
</tr>
</table>
</body>
</html>
</pre>
Håber der er nogle der kan hjælpe mig!
-Thomas
Ændre det til
<pre>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
<title>Nyt dokument</title>
</head>
<body>
<?
$siteid = $_GET["siteid"];
$database = mysql_connect("localhost","*****************","**************");
mysql_select_db("test",$database);
$foresp = mysql_query("SELECT * FROM pcoquality WHERE siteid='$siteid' ");
$row = mysql_fetch_array($foresp);
$siteplaz = $row["sitepla"];
$siten = $row["siten"];
$sitedesc = $row["sitedesc"];
$siteqi = $row["siteqi"];
$siteqt = $row["siteqt"];
?>
<table width="100%" height="100%" border="1">
<tr>
<td width="100%" height="20%"><?echo("The site $siten have get a Quality price for $sitedesc as the quality type $siteqt<img src='$siteqi' height='120' width='143' > ");?></td>
</tr>
<tr>
<td width="100%" height="*"><? include ("$siteplaz"); ?></td>
</tr>
</table>
</body>
</html>
</pre>
Så burde det virke.
MH.
The-Freak
Livet er for kort til at kede sig.