<?php
$liendb = mysql_connect ("localhost", "grandmo_test", "123") or die ('I cannot connect to the database.');
mysql_select_db ("grandmo_test");
$sql = "CREATE TABLE eleve (
ideleve int(10) unsigned NOT NULL auto_increment,
nom varchar(64) NOT NULL default '',
prenom varchar(64) NOT NULL default '',
PRIMARY KEY (ideleve),
KEY nom (nom)
) TYPE MyISAM COMMENT='Premier'";
$sgl = "INSERT INTO eleve (nom,prenom) VALUES ('Savon','Pierre')";
mysql_query ($sql);
$sgl = "SELECT nom, prenom FROM eleve WHERE ideleve = '1'";
$resultat = mysql_query ($sql);
$eleve = mysql_fetch_array ($resultat);
$nom = $eleve ['nom'];
$prenom = $eleve ['prenom'];
echo "eleve [1], nom = $nom, prenom = $prenom";
mysql_close($liendb);
?>
$eleve = mysql_fetch_array ($resultat);
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/grandmo/public_html/liendb.php on line 24
eleve [1], nom = , prenom =
😕
I'm not sure if the warning come from this simple code or from the hosting server. If someone can find mistake in this code, I will appreciate.
Alain
🙂
