Passing multiple checkbox values to different columns of database - php

Question

I am pretty new to PHP, but have tried searching for other questions similar to mine and been unable to find anything that is close enough to my situation to help me solve this.
I am trying to code a web page that allows users to select as many or as few items as they would like to order. The item values are identical to their Primary Key in the Item table.
Once submitted, each different item value should be input into the same row of a database table based on the date{pk}. Within that row, there are numerous columns: Item1ID, Item2ID, Item3ID, etc.
So far, the value of each item selected is assigned to a new array. However, I cannot simply input the array values into a column -- I need each array index to be placed into a sequential column. The code is below:
$date = new DateTime();
$td = $date->format('Y-m-d');
$x = 1;
$checkedItems = $_POST['Item'];
$count = count($checkedItems);
echo $count;
$foodID = "Item".$x."ID";
While($x<=$count){
if(isset($_POST['Item'])){
if (is_array($_POST['Item'])) {
foreach($_POST['Item'] as $values){
$selectedFoods = substr($values,0,4);
$addFoodOrderQuery= sprintf("UPDATE WeeklyBasketFoodOrder SET '%s' = %s WHERE `foodOrderDate` = '%s'",
$foodID, $selectedFoods, $td);
$result= mysqli_query($db, $addFoodOrderQuery);
}
}
} else {
$values = $_POST['Item'];
echo "You have not selected any items to order.";
}
$x++;
}
If you need any further clarification, please let me know. After submitting the code, the database item#ID tables are different, but they are now empty instead of "NULL."