Question

我有收集客户信息并将其推送到数据库的表格，以使查询无法正常工作

我尝试输入用户输入，它们向我显示并提示查询不成功的错误，因此我可以找出问题，即将数据推入数据库的查询。在我的sql语句中，我使用了最佳实践来对后杠，单引号和双引号进行处理。

<?php
/**
 * Created by PhpStorm.
 * User: Administrator
 * Date: 3/28/2019
 * Time: 12:05 PM
 */

require_once "config.php";
session_start();


$user_name = $_SESSION['username'];
$cardholder_name= $card_number= $security_code = $deliver_address='';
$cardholder_name_err = $card_number_err= $security_code_err =$deliver_address_err ='';

// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST")
{
    if(empty(trim($_POST["cardholder_name"]))){
        $cardholder_name_err = "Please enter the card holder name.";
    } else{
        $cardholder_name = trim($_POST["cardholder_name"]);
        if (!preg_match("/^[a-zA-Z ]*$/",$cardholder_name)) {
            $cardholder_name_err = "Only letters and white space allowed";
        }
    }

    if(empty(trim($_POST["card_number"]))) {
        $card_number_err = "Please enter the card number.";
    }elseif (strlen(trim($_POST['card_number'])) > 12 || strlen(trim($_POST['card_number'])) < 12 )
    {
        $card_number_err = "Please enter only 12 digits indicated on your card";
    }elseif(!preg_match('/^\d*$/',$_POST['card_number']))
    {
        $card_number_err = "Please enter numbers only";
    }else
    {
        $card_number = $_POST["card_number"];
    }

    if(empty(trim($_POST["security_code"]))) {
        $security_code_err = "Please enter the security code.";
    }elseif (strlen(trim($_POST['security_code'])) > 3 || strlen(trim($_POST['security_code'])) < 3 )
    {
        $security_code_err = "Please enter only 3 digits indicated at the back of your card";
    }elseif(!preg_match('/^\d*$/',$_POST['security_code']))
    {
        $security_code_err = "Please enter numbers only";
    }else
    {
        $security_code = $_POST["security_code"];
    }

    if (isset($_POST['radiosGroup'])){
        switch ($_POST['radiosGroup'])
        {
            case '1':
                if(empty(trim($_POST["new_address"]))){
                    $deliver_address_err = "Please enter your delivery address.";
                } else{
                    $deliver_address = trim($_POST["new_address"]);

                }
                break;
            case '2':
                //Collect the old address
                $sql = "SELECT address FROM users WHERE username = '$user_name'";
                $result = mysqli_query($link, $sql);
                if (mysqli_num_rows($result) > 0)
                {
                    while($row = mysqli_fetch_assoc($result))
                    {
                        $deliver_address = $row['address'];
                    }
                }

                break;
        }
    }

    if ($deliver_address_err == '' && $cardholder_name_err == '' && $card_number_err == '' && $security_code_err == '')
    {

        $sql1 = "INSERT INTO `orders`(`deliver_address`,`cardholder_name`, `card_number`, `security_code`) VALUES ('$deliver_address', '$cardholder_name', '$card_number', '$security_code')";
        $conn = mysqli_query($link, $sql1);
        if(!$conn){
           die("ERROR: Could not connect. " . mysqli_connect_error());
        }else{
            header ("Location: checkout.php");
        }
    }
}


?>

<?php include "header_member.php"; ?>

<div class="wrapper">
    <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">

        <div class="form-group" <?php echo (!empty($cardholder_name_err)) ? 'has-error' : ''; ?>>
            <label>Card holder name:</label>
            <input type="text" name="cardholder_name" class="form-control">
            <span class="help-block"><?php echo $cardholder_name_err; ?></span>
        </div>

        <div class="form-group" <?php echo (!empty($card_number_err)) ? 'has-error' : ''; ?>>
            <label>Card number:</label>
            <input type="text" name="card_number" class="form-control">
            <span class="help-block"><?php echo $card_number_err; ?></span>
        </div>

        <div class="form-group" <?php echo (!empty($security_code_err)) ? 'has-error' : ''; ?>>
            <label>Security code:</label>
            <input type="text" name="security_code" class="form-control">
            <span class="help-block"><?php echo $security_code_err; ?></span>
        </div>

        <!-- Default unchecked -->
        <div class="custom-control custom-radio">
            <input type="radio" class="custom-control-input" id="defaultUnchecked" name="radiosGroup"  value="2" onClick = yesnoCheck()>
            <label class="custom-control-label" for="defaultUnchecked">Deliver to my registered address</label>
        </div>

        <!-- Default checked -->
        <div class="custom-control custom-radio">
            <input type="radio" class="custom-control-input" id="defaultChecked" name="radiosGroup" value="1" onClick = yesnoCheck() checked>
            <label class="custom-control-label" for="defaultChecked">Deliver to another address</label>
        </div>


        <div class="form-group" <?php echo (!empty($deliver_address_err)) ? 'has-error' : ''; ?> id="noId">
            <label>New address:</label>
            <input type="text" name="new_address" class="form-control">
            <span class="help-block"><?php echo $deliver_address_err; ?></span>
        </div>

        <div class="form-group">
            <input type="submit" name= "action" class="btn btn-primary" value="Pay now">
        </div>
    </form>
    <script>
        function yesnoCheck() {
            if (document.getElementById('defaultChecked').checked) {
                document.getElementById('noId').style.visibility = 'visible';
            }
            else document.getElementById('noId').style.visibility = 'hidden';

        }
    </script>
</div>

<?php include "footer.php"?>

点击“立即付款”后。用户信息应该已经插入到数据库中，并将我定向到另一个页面。

编辑：我刚刚发现我还有一个索引，这是我的外键，当我们设置表之间的关系时，我认为它是从其他表中自动收集的。

即使我输入了有效的表名和列名，我的查询也不起作用

0 个答案: