I’m webdesigner and I touch NOTHING in development, so, I really need your help (exam in school). So, indeed, I’ve made a search form in the same page. There a field to search by town (ville) and the results should be appareate in a div. Sometimes it works, but I really need to fix that.
The error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\foodsurfing\index.php on line 55
There is the code
<?php
require_once('Connections/foodsurfing.php');
if(isset($_GET['recherche']))
$requete="SELECT * FROM fiche_membres WHERE ville LIKE '%".$_GET['recherche']."%'
LIMIT 0 , 1 ";
else
$requete="SELECT * FROM fiche_membres
LIMIT 1 , 1 ";
mysql_query("SET NAMES UTF8");
$resultat=mysql_query($requete);
if (false === $resultat) {
echo mysql_error();
}
?>
and the result’s div
<div id="gmap_result_container">
<div class="gmap_result">
<?php while($fiche_membres=mysql_fetch_array($resultat)) {?> <p>
<a class="prenom"><?php echo($fiche_membres['prenom']); ?></a>
<a> , </a>
<a class="ville"><?php echo($fiche_membres['ville']); ?></a></p>
<img src="images/avatar_unknown.jpg">
<p><?php echo($fiche_membres['experience_culinaire']); ?></p>
<p><a href="inscription.php" onMouseOver="roll.src='images/ensavoirplus_hover.png'" onMouseOut="roll.src='images/ensavoirplus.png'"><img src="images/ensavoirplus.png" name="roll" border="0" alt="En savoir plus" align="middle" class="ensavoirplus"></a></p>
<?php }?>
</div>
</div>
Thank you so much for your help !!
Glad to be into the community I just discover as beginner !
Warning says that the resource is not valid which is parameter of
mysql_fetch_array()method. You have to verify the reference before retrieving rows.