make links for mysql/php menu

Question

Because I'm new to php I need help with some web page menu that is stored in MySQL database and then called to page with php. My PHP connection to database works well, and menu looks like it should but I don't know how to make links to pages from menu. Example: when menu is populated it would give me home, about, gallery the problem is how to make them links to specific pages like about.php? here is the code:

<?php
include('testconnect.php');
//calling menu from database
$query_sql = "SELECT * FROM meni";
$item_query = mysql_query($query_sql) or die(mysql_error());
$rsitem = mysql_fetch_assoc($item_query);
?>
<! DOCTYPE html>
<html>
<head><title></title>
<meta http-equiv="Content-Type" content="text/html;charset=UTF-8">
<link rel="stylesheet" href="style2.css" type="text/css" />

</head>
<body>
<nav>
<ul>
 <?php
 do {
 ?>
 <li><a href="menutest.php?meni_ID=<?php echo $rsitem['meni_ID']; ?>"><?php echo      $rsitem['item']; ?></a></li>
<?php    
}while ($rsitem = mysql_fetch_assoc($item_query));
?>
</ul>
</nav>
</body>
</html>

I put that menutest.php in link just to see if it works. MySql database has two fields meni_ID which is integer and item vhich is varchar. any kind of help would be good. or even example of other similar post. Thank you!

I,ll try to be more specific. Everything is good, menu is populated from database. I need to make them links to other pages. Example: populated menu is: home, about, gallery I need them to get me, when I click on them, to about.php... — youngster, Mar 31 '13 at 20:58

score 0 · Answer 1 · answered Mar 31 '13 at 20:51

<?php
include('testconnect.php');
//calling menu from database
$query_sql = "SELECT * FROM meni";
$item_query = mysql_query($query_sql) or die(mysql_error());
while($rsitem = mysql_fetch_array($item_query)) { ?>


<! DOCTYPE html>
<html>
<head><title></title>
<meta http-equiv="Content-Type" content="text/html;charset=UTF-8">
<link rel="stylesheet" href="style2.css" type="text/css" />

</head>
<body>
<nav>
<ul>

 <li><a href="menutest.php?meni_ID=<?php echo $rsitem['meni_ID']; ?>"><?php echo   $rsitem['item']; ?></a></li>


 <?php } ?>
</ul>
</nav>
</body>
</html>

That will produce invalid html... and won't change anything. — Hugo Dozois, Mar 31 '13 at 20:54

score 0 · Accepted Answer · edited May 23 '17 at 12:21

You can do it by passing variable from one page to another(about.php) with the help of GET variable. Refer this.

Also you can do it by passing an array (that consists of menu and page link) to another page(say about.php) by Session Variable.

<?php
// begin the session
session_start();
 // populate the array with menus or populate the array with the item that you want to pass to next(say about.php) page
$my_array=array('home','about','gallery');

// put the array in a session variable
$_SESSION['mymenu']=$my_array;

?>

In about.php page

<?php
// begin the session
session_start();

// loop through the session array with foreach
foreach($_SESSION['mymenu'] as $key=>$value)
    {
    // and print the menu
    echo $value.' <br />';
    }
?>

UPD: I have modified your code.

<?php
session_start();
include('testconnect.php');
//calling menu from database
$query_sql = "SELECT * FROM meni";
$item_query = mysql_query($query_sql) or die(mysql_error());
$rsitem = mysql_fetch_assoc($item_query);
?>
<! DOCTYPE html>
<html>
<head><title></title>
<meta http-equiv="Content-Type" content="text/html;charset=UTF-8">
<link rel="stylesheet" href="style2.css" type="text/css" />

</head>
<body>
<nav>
<ul>
 <?php
$idx=0;
 do {
 ?>
 <li><a href="menutest.php?meni_ID=<?php echo $rsitem['meni_ID']; ?>"><?php echo      $rsitem['item']; ?></a></li>
<?php  
$meni_ID_ARR[$idx]= $rsitem['meni_ID'];
$item_ARR[$idx]=$rsitem['item'];
$idx++;
}while ($rsitem = mysql_fetch_assoc($item_query));
// put the array in a session variable
$_SESSION['meni_ID_ARR_SESS']=$meni_ID_ARR;
$_SESSION['item_ARR_SESS']=$item_ARR;
?>
</ul>
</nav>
</body>
</html>

-

/*printing menu in about.php page*/
<?php
session_start();
?>
<?
    if(isset($_SESSION['meni_ID_ARR_SESS']) && isset($_SESSION['item_ARR_SESS']))
    {
        $meniIDS=$_SESSION['meni_ID_ARR_SESS'];
        $meniITEM=$_SESSION['item_ARR_SESS'];
        for($idx=0;$idx<count($_SESSION['meni_ID_ARR_SESS']);$idx++)
        {
            echo  '<li><a href="menutest.php?meni_ID='.$meniIDS[$idx].' ">' .$meniITEM[$idx].'</a></li>';
        }
    }
    else
    {
        echo "Session not set!";
        exit;
    }

?>

I will give it a try. Thanks. Now I'm reading posts in a link that you gave me. — youngster, Mar 31 '13 at 21:15
There is a dead loop in about.php and the error is: `for($idx=0;$idx — youngster, Mar 31 '13 at 21:57
Sorry that was a typo. I had fixed it.I haven't test the code.So there might be small Syntactic and Semantic Errors. — Ritesh Kumar Gupta, Mar 31 '13 at 21:59
@IgorStakic I have tested the code in my system just now.And its working perfectly. Please upvote and click on accept tick(sign) if it was helpful for you. Thankyou!! :) — Ritesh Kumar Gupta, Mar 31 '13 at 22:06
that is fine, I'm learning so it was good to test it and find the error. Now i will try to make the menu item to open pages. thank you for your time and help! — youngster, Mar 31 '13 at 22:08

make links for mysql/php menu

2 Answers2