Can't display my result in a table - PHP - php

Question

I am trying to display my query results on page. However, whenever I run the code although the query is correct it does not display anything.
Can anyone help? Would be muchly appreciated
<?php
session_start();
require_once("../config1.php");
if(isset($_POST["DailySales"])) {
$linkid = mysqli_connect(DB_DATA_SOURCE, DB_USERNAME, DB_PASSWORD, DB_DATABASE) or die("Could not connect:" . connect_error());
$sql = "SELECT Retdatetime AS date , sum(rentalrate + overduecharge) AS mny
FROM frs_FilmRental
WHERE shopid='2'
Order BY retdatetime DESC ";
$result = mysqli_query($linkid, $sql);
if (!$result)) {
printf("Errormessage: %s\n", mysqli_error($linkid));
}
echo "<table border = '1' align='center'>";
echo "<th> Shop ID 2</th></tr>";
while ($row = mysqli_fetch_assoc($result)) {
echo "<h2><center>Shop ID 2 daily sales : </center></h2>";
echo "<tr><td>";
echo $row['mny'];
echo "</td><td>";
echo $row ['date'];
echo "</td></tr>";
}
}
?>

Answer 1

replace
echo $row ['date'];
by
echo $row ['Retdatetime'];

Answer 2

To understand query errors you should use mysqli_error() in your code. If there are no errors for executed query, then you can run while loop for it.
<?php
session_start();
require_once("../config1.php");
if(isset($_POST["DailySales"])) {
$linkid = mysqli_connect(DB_DATA_SOURCE, DB_USERNAME, DB_PASSWORD, DB_DATABASE) or die("Could not connect:" . connect_error());
$sql = "SELECT Retdatetime , sum(rentalrate + overduecharge) AS mny
FROM frs_FilmRental
WHERE shopid='2'
Order BY retdatetime DESC ";
$result = mysqli_query($linkid, $sql);
if (!$result)) {
printf("Errormessage: %s\n", mysqli_error($linkid));
} else {
echo "<table border = '1' align='center'>";
echo "<tr><th> Shop ID 2</th></tr>";
while ($row = mysqli_fetch_assoc($result)) {
echo "<h2><center>Shop ID 2 daily sales : </center></h2>";
echo "<tr><td>";
echo $row['mny'];
echo "</td><td>";
echo $row ['date'];
echo "</td></tr>";
}
echo "</table>";
}
}
?>
fixes given in comments also applied
Please copy/paste the error, given by mysqli_error()