> From: "Voisine" <[EMAIL PROTECTED]>
> Sent: Sunday, September 29, 2002 7:52 PM
> Subject: [PHP] Undefined constant error
> I'm learning PHP from a book "PHP for newbie writen in French" and I
> have an error on one of the exemple. Undifined constant 'compteur' on
> line 15 which is :
> if (compteur == 1) {
>
> What I'm doing wrong? This is the script
>
> <?php
> include("config.inc.php");
> $query = "SELECT * FROM Type ORDER BY animalType";
> $result = mysql_query($query) or die ("Exécution de la sélection
> impossible");
>
> echo "<h1 align='center'>Catalogue</h1><p><h3>Quel type d'animal
> cherchez-vous ?</h3>\n";
> echo "<form action='montre_animaux.php' method='post'>\n";
> echo "<table cellpadding='5' border='1'>";
> $compteur = 1;
> while ($ligne = mysql_fetch_array($result)) {
> extract($ligne);
> echo "<tr><td valign='top' width='15%'>";
> echo "<input type='radio' name='interet' value='$animalType'>";
> if (compteur == 1) {
you're missing the $ in from of compteur. Should be:
if ($compteur == 1) {
make sure you have register_globals on in php.ini or your next question will
be 'why isn't $compteur set to the value posted?' Better would be to change
the line to the new magic globals, and learn the right way from scratch:
if ($_POST['compteur'] == 1) {
For the full explaination see:
http://www.php.net/manual/fr/language.variables.predefined.php
> echo "checked";
> }
Also, I think this code needs to place the 'checked' within the <input
type="radio"> tag
so you end up with:
echo "<input type='radio' name='interet' value='$animalType'";
if ($_POST['compteur'] == 1) {
echo "checked";
}
echo ">";
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
