Kolla vad det beror på:
$SQL2 = mysql_query("select * from lank_ar where kat='".$kat['kat']."' ORDER by datum DESC") or die(mysql_error());
1 svar · 188 visningar · startad av jwradhe
ska göra ett länkarkiv nu men får felmeddelande :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/htdocs/users/jwradhe/lank/index.php on line 34
på raden :
while($lank=mysql_fetch_array($SQL2)){
Hela koden :
<?
$server="localhost";
$dbname="xxx";
$user="xxx";
$pwd="xxx";
$db = mysql_connect($server, $user, $pwd);
mysql_select_db($dbname, $db);
$SQL1 = mysql_query("select * from lank_kat ORDER by kat ASC ");
while($kat=mysql_fetch_array($SQL1)){
?>
<br>
<font face="Verdana, Arial, Helvetica, sans-serif" size="2"> <b>
<? echo $kat["kat"]; ?>
</b></font><br>
<table width="500" border="0" cellspacing="0" cellpadding="0">
<tr> <?
$SQL2 = mysql_query("select * from lank_ar where kat='".$kat['kat']."' ORDER by datum DESC ");
while($lank=mysql_fetch_array($SQL2)){
?>
<td width="324"> <font face="Verdana, Arial, Helvetica, sans-serif" size="1">
<b>
<? echo $lank["datum"]; ?>
</b> |
<? echo $lank["lank"]; ?>
</font> </td>
</tr><? } ?>
</table>
<br>
<?
}
mysql_close($db); ?>
Kolla vad det beror på:
$SQL2 = mysql_query("select * from lank_ar where kat='".$kat['kat']."' ORDER by datum DESC") or die(mysql_error());