function in php for displaying a menu
- by Stanislas Piotrowski
I met some trouble with a function I have written.
In fact I have an array in which I have different custom values.
I would like to display the result of that, so here is my function
function getAccess($var){
$data=mysql_fetch_array(mysql_query("SELECT * FROM `acces` WHERE `id`='.$var.'"));
return $data;
}
getAccess($_SESSION['id']);
$paramAcces = array(
'comptesfinanciers' => array(
'libelle' => 'COMPTES FINANCIERS',
'acces' => $data['comptesfinanciers'],
'lien' => 'financier',
'image' => 'images/finances.png'
)
);
I have done a var_dump of $paramAcces which return
array(1) { ["comptesfinanciers"]=> array(4) { ["libelle"]=> string(18) "COMPTES FINANCIERS" ["acces"]=> NULL ["lien"]=> string(9) "financier" ["image"]=> string(19) "images/finances.png" } } (that are the ecpected values).
Here is the function for displaying what is in the array
/**
* AFFICHAGE DE LA SECTION PARAMETRES SUR LA PAGE D'ACCUEIL
*/
function affichParam($paramAccees){
echo '<ul class="getcash-vmenu"><li><a href="index.php?p='.$paramAccees['lien'].'" class="active"><span class="t"><img src="'.$paramAccees['image'].'"> '.$paramAccees['libelle'].'</span></a></li></ul>';
}
The trouble is that actualy it return to me an empty line.
I really do not know what I'm wrong doin:
I call the function like that:
<?php
affichParam($paramAccees)
?>
In a second time I will add more value, so I think I will have to do a for each loop or something like that.
But actualy I just would like to display the first record.
Any kind of help will be much apprecitaed
