Php 付款详细信息未插入数据库中

在通过paypal付款后，我在数据库中存储付款详细信息时遇到问题。以下是支付成功的编码。它接受该值，但该值未存储在数据库中。我使用tmno作为用户id

<?php
session_start();
include 'con.php';

$cartID = $_SESSION['cartID'];
$amount = $_GET['amt'];
$currency = $_GET['cc'];
$transactionID = $_GET['tx'];
$payment_status = $_GET['st'];
$invoice = mt_rand();

$tmno = $_SESSION['tmno'];

$insert_payment = "INSERT INTO payment (paymentID,cartID) VALUES
('$transactionID','$cartID')";
$result_payment=mysqli_query($con,$insert_payment);

if($result_payment){
    $update_query = "UPDATE cart SET status=1 WHERE cartID='$cartID' ";
    $update_result = mysqli_query($con,$update_query);

    if($update_result){
        $sql1 = "SELECT * FROM cart WHERE tmno='$tmno' AND status=0 ";
        $result1 = mysqli_query($con,$sql1);

        if(mysqli_num_rows($result1)>0){
            $row1 = mysqli_fetch_array($result1);
            $cartID = $row1['cartID'];
        }

        else{ //if user doesn't have cart yet
            $sql2 = "INSERT INTO cart(tmno) VALUES ('$tmno') ";
            $result2 = mysqli_query($con,$sql2);
            $cartID = mysqli_insert_id($con);
        }

        $_SESSION['cartID'] = $cartID;

        echo "Your payment was successfully recorded!";
        echo "<a href='UserFee.php'><h3>Click here to go back to the homepage!</h3></a>";
    } else {
        echo "Oh no!";
    }
}else {
    echo "Failed!";
}

?>

---

那么会发生什么？你收到“哦，不”的留言了吗？您从mysql\u error（）

中获得了哪些详细信息？添加

ini\u集（'display\u errors'，1）；ini_集（“日志错误”，1）；错误报告（E_全部）；mysqli_报告（mysqli_报告错误| mysqli_报告严格）到脚本的顶部。这将强制任何
mysqli\uuuu
错误生成您可以在浏览器上看到的异常，其他错误也将在您的浏览器上看到。您的代码易受SQL注入攻击，您应该使用或帮助防止此类攻击。感谢您的帮助。我已经解决了。问题实际上来自数据库。它不接受cartID的值，因为其中有唯一的约束。
<?php
$con = mysqli_connect('DB_HOST', 'DB_USERNAME', 'DB_PASSWORD', 'DB_NAME') or die('Error Connect to DB');
mysqli_query($con,'utf-8');
?>