hello i'm just training myself on web development so i wanted to create a drop down menu wich i have post before about it!
so i get the html code and css and the javascrip code of the bootstrap header menu and i want to customize it for my website menu, but the problem is in my data base i have 3 data but my meny displays only 2 and in the first drop down there is an error that is not in second and i can't figure out why the above pictures and my code explain lite more cleary !
<head>
<?php
include('./db/cnntdb.php');
$sqlrecup= "SELECT * FROM main";
$sqldonee= mysqli_query($cnndb, $sqlrecup) or die('could not get data');
$row=mysqli_fetch_array($sqldonee, MYSQLI_ASSOC);
?>
</head>
//and here is the menu
<ul class="nav navbar-nav">
<li class="active"><a href="#">HOME<span class="sr-only">(current)</span></a></li>
<?php do { ?>
<li class="dropdown"><a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false"> <?php echo $row['main'];
?>
<span class="caret"></span></a>
<ul class="dropdown-menu">
<?php
$mainsub=$row['idmain'];
$sqlsub = "SELECT * FROM `sub` WHERE `main` = '$mainsub' ";
$sqlsubdonee= mysqli_query($cnndb, $sqlsub) or die('could not get data');
$sub=mysqli_fetch_array($sqldonee, MYSQLI_ASSOC);?>
<?php do{ ?>
<li role="separator" class="divider"></li>
<li><a href="#"><?php echo $sub['sub'];?></a></li>
<?php } while ($sub=mysqli_fetch_assoc($sqlsubdonee));?>
<li role="separator" class="divider"></li>
</ul>
</li>
<?php } while ($row=mysqli_fetch_assoc($sqldonee));?>
</ul>
