How can I display other elements from a table when I choose an element from a select menu? I have tried to do something, as you can see in the link below, but I don't see anything :(
http://faous.net/test/profile/page1.php
<html>
<header>
<title>test PHP</title>
</header>
<body>
<h1>Formulaire</h1>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="get">
<?php
echo "<select name='sub1'>";
while($row=mysql_fetch_array($result)){
echo "<option value='" . $row['nom'] . "'>" . $row['nom'] . "</option>";
}
echo "</select>";
echo "<br />";
$result2=mysql_query($sql);
$base=mysql_fetch_array($result2);
//$nnnn= $_GET["nom"];
//$aaaa=$_GET["base"];
//$bbbb=$_GET["hauteur"];
//$cccc=$_GET["rayon"];
$nnnn = $_GET[$row];
$aaaa = $_GET["base"];
$bbbb = $_GET["hauteur"];
$cccc = $_GET["rayon"];
echo "Nom: $nnnn <br /> Base: $aaaa <br /> Hauteur: $bbbb<br /> Rayon: $cccc";
echo "<br /> ";
?>
</form>
</body>
</html>
<?php
mysql_close($link);
?>
You have to submit your form on change of select like this,
<select onchange="this.form.submit()">
In your case its,
echo "<select name='sub1' onchange='this.form.submit()'>";
mysql_* are deprecated, use mysqli_* or PDO instead
These lines are completely wrong
$result2=mysql_query($sql);// what is $sql here??
$base=mysql_fetch_array($result2);
$nnnn = $_GET[$row];// what is $row??
$aaaa = $_GET["base"]; // where is the form element with name base??
$bbbb = $_GET["hauteur"]; // where is the form element with name hauteur??
$cccc = $_GET["rayon"]; // where is the form element with name rayon??