Question

我在通过paypal付款后将付款详细信息存储在我的数据库中时遇到问题。这是付款成功的编码。它取值，但值不存储在数据库中。我使用tmno作为userID

<?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!";
}

?>

这是configure.php

<?php
$con = mysqli_connect('DB_HOST', 'DB_USERNAME', 'DB_PASSWORD', 'DB_NAME') or die('Error Connect to DB');
mysqli_query($con,'utf-8');
?>

显示的错误是

致命错误：未捕获的异常＆＃39; mysqli_sql_exception＆＃39;带有消息＆＃39;您的SQL语法中有错误;检查与您的MySQL服务器版本相对应的手册，以便在＆utff-8＆＃39;附近使用正确的语法。在第1行＆＃39;在/home/sptlabua/public_html/configure.php:3堆栈跟踪：＃0 /home/sptlabua/public_html/configure.php(3）：mysqli_query（对象（mysqli），＆＃39; utf-8＆＃39;）＃1 /home/sptlabua/public_html/payment_success.php(5）：include（＆＃39; / home / sptlabua /...'）＃home {home}抛出/ home / sptlabua / public_html / configure .php在第3行

付款详细信息未插入数据库

0 个答案: