I just trying to show a result of a consult in MySQL (PHP). The code is:
$example = mysql_query("SELECT count(*) as text FROM table WHERE name = '$name'");
$qtd = mysql_num_rows($example);
while($data = mysql_fetch_array($qtd)){
$count = $data["text"];
}
echo "<h3>($count)</h3>";
Error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\Site\index.php on line 9