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Author:??(---.7-200-80.adsl.skynet.be)
Date:???02-14-04 13:55

I try to get the results from this query to next page, using a pagination class, but I get this message and no results. How can I solve this ?

This is the code:

$total_results = mysql_query_test("SELECT COUNT(recept) AS Num FROM recepten WHERE ingredienten LIKE
'%$param1%' AND catID = ('$catID') ");
while($row = mysql_fetch_array($total_results));
$recept=$row['recept'];

echo "

Er zijn in het totaal ",mysql_result($total_results,0, 0),"
recepten welke beantwoorden aan uw zoekopdracht
and this is the error message:

Notice: Undefined index: ingredienten in /Library/WebServer/Documents/test4/results.php on line 311

Thanks

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foodstyling 02-14-04 13:55?
?Re: undifined index??new
bastien 02-16-04 17:56?
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foodstyling 02-16-04 22:43?
?Re: undifined index??new
mike... 02-22-04 07:11?
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??wrote: > > I try to get the results from this query to next page, using > a pagination class, but I get this message and no results. > How can I solve this ? > > This is the code: > > $total_results = mysql_query_test("SELECT COUNT(recept) AS > Num FROM recepten WHERE ingredienten LIKE > '%$param1%' AND catID = ('$catID') "); > while($row = mysql_fetch_array($total_results)); > $recept=$row['recept']; > > echo "

Er zijn in het totaal > ",mysql_result($total_results,0, 0)," > recepten welke beantwoorden aan uw zoekopdracht > and this is the error message: > > Notice: Undefined index: ingredienten in > /Library/WebServer/Documents/test4/results.php on line 311 > > Thanks ">??

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