PHP Dynamic Footer Menu with MySQL

Question

I want to create a dynamic footer menu (not multilevel) with mysql.

My MySQL table is like below...

---------------------------------------------------
|  menu_id  |  menu_name  |  menu_url  | short_id  |
----------------------------------------------------
|  1        | Home        | index.php  | 1         |
----------------------------------------------------
|  2        | Contact Us  | contact.php| 3         |
----------------------------------------------------
|  3        | About Us    | abt.php    | 2         |
----------------------------------------------------

The HTML structure is like...

<div class="footer-menu">
 <ul>
    <li><a href="index.php">Home</a></li>
    <li><a href="abt.php">About Us</a></li>
    <li><a href="contact.php">Contact Us</a></li>
 </ul>
</div>

I have coded like...

mysql_select_db($db,$con);
$f_menu_qry = "SELECT * FROM footermenu ORDER BY short_id ASC";
$fm = mysql_query($f_menu_qry,$con);
$f_menu = mysql_fetch_assoc($fm);
$totrows = mysql_num_rows($fm);
$menu_name = $f_menu['menu_name'];
$menu_url = $f_menu['menu_url'];

    public function getFooterMenu(){
    global $fm, $f_menu, $menu_name, $menu_url, $totrows;
    $footer_menu = '';
    $cnt = 0;
     while ($clt_f_menu = mysql_fetch_assoc($fm)){
         $cnt++;
         $footer_menu = '<li><a href="'.$menu_url.'">'.$menu_name.'</a></li>';
         if($cnt == $totrows){
             return;
         }
     }
     echo $footer_menu;
    }

The output should be like Home / About Us / Contact Us

But the output is showing only Home.

I have tried different processes but those were showing errors(may be for my faulty code) but this time it is showing at least a link.

I am unable to do it.

What should I do ?

Raja · Accepted Answer · 2014-06-29 17:14:10Z

1

First of all , it's not a good practice to use old mysql extension. It's better to use PDO or MySQLi.

PDO : http://www.php.net/manual/en/book.pdo.php
MySQLi : http://www.php.net/manual/en/book.mysqli.php

Any way, I've modified your code a bit. You can try this one. Hope it'll work..

mysql_select_db($db,$con);
$f_menu_qry = "SELECT * FROM footermenu ORDER BY short_id ASC";   
$fm = mysql_query($f_menu_qry,$con);

$clt_f_menu = array();
while($rows = mysql_fetch_array($fm))
{
    $clt_f_menu[$rows['menu_id']] = array(
            'menu_name' => $rows['menu_name'],
            'menu_url' => $rows['menu_url']
        );
}

echo '<div class="footer-menu">';
    echo '<ul>';
    foreach ($clt_f_menu as $key => $value) {
        echo '<li><a href="'.$value['menu_url'].'">'.$value['menu_name'].'</a></li>';
    }
    echo '</ul>';
echo '</div>';

edited Jun 29, 2014 at 17:14

Raja

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answered Jun 29, 2014 at 13:19

Suresh

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2 Comments

Raja Over a year ago

I have tried your code. It is showing About Us / Contact Us but not showing Home

Raja Over a year ago

Got my answer. Thank you @mi6crazyheart

hellcode · Accepted Answer · 2014-06-29 13:33:05Z

0

There are two main aspects with

$footer_menu = '<li><a href="'.$menu_url.'">'.$menu_name.'</a></li>';

$footer_menu will have only the content from the last loop, because "=" will overwrite it in each loop and you are using global variables here which have the content from the first row only.

Try this:

mysql_select_db($db,$con);
echo getFooterMenu($con);

public function getFooterMenu($con){
  $sql = "SELECT * FROM footermenu ORDER BY short_id ASC";
  $mid = mysql_query($sql,$con);
  $footer_menu = '';
  while($rs = mysql_fetch_assoc($mid)) {
    $footer_menu.= '<li><a href="'.$rs["menu_url"].'">'.$rs["menu_name"].'</a></li>';
  }
  return $footer_menu;
}

edited Jun 29, 2014 at 13:33

answered Jun 29, 2014 at 13:18

hellcode

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