Warning: mysql_num_rows() expects parameter 1 to be resource, b
djamelpsy
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NHenry Messages postés 15235 Date d'inscription Statut Modérateur Dernière intervention -
NHenry Messages postés 15235 Date d'inscription Statut Modérateur Dernière intervention -
Bonjour, aider moi svp
j'ai pas compris cette erreur
(
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\EasyPHP-DevServer-14\data\localweb\projects\functions.php on line 22
pas de resultat trouvé...)
<?php
function results ($search)
{
$where = "";
$search = preg_split('/[\s\-]/',$search);
$count_keywords = count($search);
foreach($search as $key=>$searches)
{
$where .= "name LIKE '%$searches%'";
if($key != ($count_keywords-1))
{
$where .=" AND ";
}
}
21 $query = mysql_query("SELECT * FROM administrateur WHERE $where");
22 $rows = mysql_num_rows($query);
23 if($rows)
{
}else{
echo "pas de resultat trouvé...";
}
}
?>
j'ai pas compris cette erreur
(
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\EasyPHP-DevServer-14\data\localweb\projects\functions.php on line 22
pas de resultat trouvé...)
<?php
function results ($search)
{
$where = "";
$search = preg_split('/[\s\-]/',$search);
$count_keywords = count($search);
foreach($search as $key=>$searches)
{
$where .= "name LIKE '%$searches%'";
if($key != ($count_keywords-1))
{
$where .=" AND ";
}
}
21 $query = mysql_query("SELECT * FROM administrateur WHERE $where");
22 $rows = mysql_num_rows($query);
23 if($rows)
{
}else{
echo "pas de resultat trouvé...";
}
}
?>
