undifined index

  • foodstyling
  • Novice
  • Novice
  • foodstyling
  • Posts: 16
  • Loc: Antwerp

Post 3+ Months Ago

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

echo "<h1>Er zijn in het totaal<font color = #042C89> ",mysql_result($total_results,0, 0),"</font>
recepten welke beantwoorden aan uw zoekopdracht </h1\n";
  1. $total_results = mysql_query_test("SELECT COUNT(recept) AS Num FROM recepten WHERE ingredienten LIKE
  2. '%$param1%' AND catID = ('$catID') ");
  3. while($row = mysql_fetch_array($total_results));
  4. $recept=$row['recept'];
  5. echo "<h1>Er zijn in het totaal<font color = #042C89> ",mysql_result($total_results,0, 0),"</font>
  6. recepten welke beantwoorden aan uw zoekopdracht </h1\n";


and this is the error message:

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

Thanks
  • Anonymous
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  • Posts: ?
  • Loc: Ozzuland
  • Status: Online

Post 3+ Months Ago

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