Dynamic menu

[NLCJ]

Hello,

I am creating a website, and I was trying to create a dynamic menu by using MySQL. So it would be easier to manage, through an Admin panel you can insert information in MySQL and the dynamic menu will take it out like this:

 

$d_getlinks = "SELECT * FROM paginas";
$d_links = mysql_fetch_array();

<div class="menu">
<a href="<?php echo $d_links['link']; ?>"><?php echo $d_links['menu']; ?></a>
</div>

 

This gives the result I wanted except that it only gives the first value, how can I get it to display ALL the links in that table?

 

Regards,

Chris

 

[PHP Monkeh]

This is more a PHP issue as your MySQL is fine.  You need to create a loop, like so:

 

<div class="menu">
<?php
$d_getlinks = "SELECT * FROM paginas";

while($d_links = mysql_fetch_array())
{
?>

<a href="<?php echo $d_links['link']; ?>"><?php echo $d_links['menu']; ?></a>

<?php
}
?>
</div>

[NLCJ]

Now it only shows the second value. 

 

Thank you though for answering, one step closer to the solution!

 

EDIT:

It shows all except the first one... :S

[PHP Monkeh]

I'm hopeless, sorry.  Try this:

 

<div class="menu">
<?php
$d_getlinks = "SELECT * FROM paginas";
$d_result = mysql_query($d_getlinks);

while($d_links = mysql_fetch_array($d_result))
{
?>

<a href="<?php echo $d_links['link']; ?>"><?php echo $d_links['menu']; ?></a>

<?php
}
?>
</div>

[NLCJ]

Thank you very much, it's working!