Best solution to this issue

rn14

Hi,

I currently have an application that provides users with 4 different menus on four different pages with each menu containing 7 options. All menus are stored in database which enables content to be changed by site administrator. However a new menu is added every week. This should be added as the first menu with the next menu then becoming the previous first menu and the next menu becoming the previous second and the next menu becoming the previous third. This is the solution I want to achieve but do not currently have. The current process requires the site admin to update each meal in every menu every week meaning they have to change 28 different options which isn't very efficent when it should be only one. In the database at the moment I just have four unrelated tables and simply let the site admin directly update these. Hope I have explained this well. Any suggestions as to how to put together an efficent solution would be a fantastic help. I can provide my current code if its a help.

Thanks

Pikachu2000

rn14;10889664 wrote:
Hi,

I currently have an application that provides users with 4 different menus on four different pages with each menu containing 7 options. All menus are stored in database which enables content to be changed by site administrator. However a new menu is added every week. This should be added as the first menu with the next menu then becoming the previous first menu and the next menu becoming the previous second and the next menu becoming the previous third. This is the solution I want to achieve but do not currently have. The current process requires the site admin to update each meal in every menu every week meaning they have to change 28 different options which isn't very efficent when it should be only one. In the database at the moment I just have four unrelated tables and simply let the site admin directly update these. Hope I have explained this well. Any suggestions as to how to put together an efficent solution would be a fantastic help. I can provide my current code if its a help.

Thanks

A little more detail on how each menu is utilized would be helpful, or if it's a live site, a link would be better. I understand what you're saying, but just can't picture it in action. At the moment, I'm thinking along the lines of using a LIMIT() statement in the query.

rn14

The site address is http://www.rossnoonan.com/mffy/our_menus.php . Would be great if you could have a look. I am new to this so would obviously like to keep solution as clear and straight forward as possible but i suppose that may no be possible.

Thanks for your help

Pikachu2000

rn14;10889695 wrote:
The site address is http://www.rossnoonan.com/mffy/our_menus.php . Would be great if you could have a look. I am new to this so would obviously like to keep solution as clear and straight forward as possible but i suppose that may no be possible.

Thanks for your help

I think the best solution would be to put it all in one table with an autoincrement index key. In that manner, it would only be necessary to add the current week's menu instead of making 28 entries in four separate tables. Then you could query the database with a limit statement to pull each week's menu. You could either leave the previous entries in there as an archive, or run a script in a cron job to clean them out periodically.

Alternately, if there are a limited number of menu choices that get recycled periodically, they could all go into a table and then use an admin interface to assign each item to the appropriate week.

rn14

<?
include('config.php');
$link=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD)or die("Could not connect: " .
mysql_error());
mysql_select_db("");
$result = mysql_query("SELECT * FROM Menu LIMIT 0,5");
while($row = mysql_fetch_array($result))
{
$image = $row['Picture'];
$title = $row['Title'];
$description = $row['Description'];
}
mysql_close($link);
?>


 <div class="dishes">
  <span>
   <?echo strip_tags($image, '<img>, <span>'); ?>
  </span>
  <h2><?echo strip_tags($title, '<span>');?></h2>
  <p><?echo strip_tags($description, '<span>');?></p>
  <div class="clear"></div>
 </div>

How would i go about printing each row that is returned from the select statement in this div class??

Thanks in advance