How to put current date in to my SQL datediff - php

Question

This is a query which I tried below.
$sql = "SELECT id,due, datediff(date('Y/m/d'), date) FROM sales
I am writing a sales program for identifying the due payment.
If the due payment is more than 7 error displayed payment pending what I tried so far I attached below of the full coding. when I tried the below code error ouput displayed
Notice: Undefined index: datediff(pay_date, date)
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$today = date("Y/m/d");
$sql = "SELECT id,due, datediff(date('Y/m/d'), date) FROM sales where due != '0' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$id = $row['id'];
$paydate = $row["datediff(pay_date, date)"] . " " . "Days" . "</br>";
if ($paydate >= 7) {
$status = "Payment pending";
$color = "#47f50b";
} elseif ($paydate >= 3) {
$status = "Payment pending please settle down the amount";
$color = "yellow";
} elseif ($paydate >= 1) {
$status = "kindly pay the payment";
$color = "#00a8ff";
} elseif ($paydate == 0) {
$status = "Disable your Order";
$color = "red";
}
?>

Answer 1

Assuming your due date column is called date, then your query should be
$sql = "SELECT id, due, DATEDIFF(date, CURDATE()) AS paydate FROM sales WHERE due != '0' ";
Note that we use CURDATE() to get the current date in MySQL, and that I have aliased the DATEDIFF result to paydate.
You also have an issue that you are trying to set
$paydate = $row ["datediff(pay_date, date)"]
however in your query your expression does not match datediff(pay_date, date). This is why it's better to use an alias as I did above, then you can change that line to simply
$paydate = $row['paydate'];
and if you change the formula in your query this part of the code will still work.
Finally you are appending " Days" to the value of $paydate. This will make comparisons with numbers e.g. in if($paydate >= 7) problematic and it would be better if you left off the " Days" until you output the value of $paydate.

Answer 2

Try using datediff(pay_date, date) as datedif in the SELECT and print using $row["datedif"]