Dynamic Dropdown menu - PHP - php

Question

First of all, I am new to mysqli and prepare statements so please let me know if you see any error. I have this static drop down menu :
HTML code:
<ul class="menu sgray fade" id="menu">
<li>Bike
<!-- start mega menu -->
<div class="cols3">
<div class="col1">
<ol>
<li>bikes</li>
<li>wheels</li>
<li>helmets</li>
<li>components</li>
</ol>
</div>
<div class="col1">
<ol>
<li>pedals</li>
<li>GPS</li>
<li>pumps</li>
<li>bike storage</li>
</ol>
</div>
<div class="col1">
<ol>
<li>power meters</li>
<li>hydratation system</li>
<li>shoes</li>
<li>saddles</li>
</ol>
</div>
</div>
<!-- end mega menu -->
</li>
I want to make a dynamic dropdown menu. I managed to show the $categoryName and the $SubCategoryName with this function:
function showMenuCategory(){
$db = db_connect();
$query = "SELECT * FROM Category";
$stmt = $db->prepare($query);
$stmt->execute();
$stmt->bind_result($id,$categoryName,$description,$pic,$active);
while($stmt->fetch()) {
echo'<li>'.$categoryName.'
<!-- start mega menu -->
<div class="cols3">
<div class="col1">
<ol>';
$dba = db_connect();
$Subquery = "SELECT * FROM Subcategory WHERE CategoryId = '".$id."'";
$Substmt = $dba->prepare($Subquery);
$Substmt->execute();
$Substmt->bind_result($Subid,$CatId,$SubCategoryName,$SubDescription);
while($Substmt->fetch()) {
echo'
<li>'.$SubCategoryName.'</li>';
}
echo'
</ol>
</div>
<!-- end mega menu -->
</li>';
}
}
The only problem is that it returns all the subcategories on the the same <div class="col1">:
what I would like to obtain is count the subcategories and if the result is more than 4 return the other items in the second and third column.
UPDATE***: thanks to the answer below now the menu looks like this:
thanks!

Answer 1

How about try this?
To explain further
What is happening is that for every subcategory fetched, I increment a counter. If that counter hits 4, it ends the <UL> and <DIV> and creates a new one which will represent the new column.
function showMenuCategory(){
$db = db_connect();
$query = "SELECT * FROM Category";
$stmt = $db->prepare($query);
$stmt->execute();
$stmt->bind_result($id,$categoryName,$description,$pic,$active);
while($stmt->fetch()) {
echo'<li>'.$categoryName.'
<!-- start mega menu -->
<div class="cols3">
<div class="col1">
<ol>';
$dba = db_connect();
$Subquery = "SELECT * FROM Subcategory WHERE CategoryId = '".$id."'";
$Substmt = $dba->prepare($Subquery);
$Substmt->execute();
$Substmt->bind_result($Subid,$CatId,$SubCategoryName,$SubDescription);
$count = 0;
while($Substmt->fetch()) {
echo'
<li>'.$SubCategoryName.'</li>';
$count+=1;
if ($count == 4) {
$count = 0;
echo '</ol></div><div class="col1"><ol>';
}
}
echo'
</ol>
</div>
<!-- end mega menu -->
</li>';
}
}
EDIT: Misunderstood the purpose of col1. They all should be col1 and should work now. If not, leave me a comment!

Answer 2

Try this:
function showMenuCategory(){
$db = db_connect();
$query = "SELECT * FROM Category";
$stmt = $db->prepare($query);
$stmt->execute();
$stmt->bind_result($id,$categoryName,$description,$pic,$active);
echo '<div class="cols3">';
while($stmt->fetch()) {
echo'<li>'.$categoryName.'
<!-- start mega menu -->
<div class="col1">
<ol>';
$dba = db_connect();
$Subquery = "SELECT * FROM Subcategory WHERE CategoryId = '".$id."'";
$Substmt = $dba->prepare($Subquery);
$Substmt->execute();
$Substmt->bind_result($Subid,$CatId,$SubCategoryName,$SubDescription);
while($Substmt->fetch()) {
echo'<li>'.$SubCategoryName.'</li>';
}
echo'</ol>';
}
echo '</div><!-- end mega menu --></li>';
}