为 mysqli 插入查询似乎不起作用，我不知道为什么

Question

对于学校的一个项目，我必须为一家公司创建一个网站。当我偶然发现这个问题时，我正在研究注册系统。由于某种原因，我的查询不适用于我的数据库。我试过在phpmyadmin中输入查询，效果很好，但是如果我想在php中使用查询，它就不起作用了。我使用的数据库是预制的，它不是自动递增的，所以当我想插入新数据时，我必须添加所有行的所有值。

这是到我正在使用的数据库的连接 我使用标题消息来指示注册是否成功：

function dbConnectionRoot () {

        $dbServername = "localhost";
        $dbUsername = "root";
        $dbPassword = "";
        $dbName = "wideworldimporters";

        $connection = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);

        return $connection;
    }

这是我正在使用的插入脚本：

$sql = "INSERT INTO customers (
                    CustomerID, CustomerName, CustomerCategoryID, 
                    PhoneNumber, FaxNumber, EmailAddress, 
                    HashedPassword, BillToCustomerID, BuyingGroupID, 
                    PrimaryContactPersonId, AlternateContactPersonID, DeliveryMethodID, 
                    DeliveryCityID, PostalCityId, AccountOpenedDate, 
                    StandardDiscountPercentage, IsStatementSent, IsOnCreditHold, 
                    PaymentDays, WebsiteURL, DeliveryAddressLine1, 
                    DeliveryPostalCode, PostalAddressLine1, PostalPostalCode, 
                    LastEditedBy, ValidFrom, ValidTo) 
                    VALUE (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?);";
                $statement = dbConnectionRoot()->prepare($sql);
                mysqli_stmt_init(dbConnectionRoot());

                if (dbConnectionRoot()->connect_errno) {
                    header("Location: signup:php?error=connection");
                    exit();
                } elseif(!dbConnectionRoot()->query($sql)) {
                    header("Location: signup.php?error=queryerror");
                    exit();
                } else {
                    //First we has the password. This is because if a hacker were to hack into the database, it could only see the hashed passwords.
                    // We use this hashing method(bcrypt) because it is always updated when there is a security breach.
                    $hashedPassword = password_hash($password, PASSWORD_DEFAULT);
                    $maxCustomerID++;
                    //Again, we are making a prepared statement for security. This time we use 27 s'es because we want 27 different variables
                    $statement->bind_param("sssssssssssssssssssssssssss", 
                    $maxCustomerID, $fullName, $customerCategory, $phoneNumber, $faxNumber, $email, $hashedPassword, $billToCustomerID, 
                    $buyingGroupID, $primaryContactPersonId, $alternateContactPersonID, $deliveryMethodID, $deliveryCityID, $postalCityId, $accountOpenedDate, 
                    $standardDiscountPercentage, $isStatementsent, $isOnCreditHold, $paymentDays, $websiteURL, $deliveryAddressLine1, $deliveryPostalCode, 
                    $postalAddressLine1, $postalPostalCode, $lastEditedBy, $validFrom, $validTo);
                    $statement->execute();
                    return($statement);
                    //A message that the signup was succesfull
                    header("Location: signup.php?signup=success");
                    exit();

这是sn-p的注册表

<form action="signupfunctions.php" method="post">
    <label>Volledige naam: </label><input type="text" name="FullName"><br>
    <label>E-mail adres: </label><input type="text" name="EmailAddress"><br>
    <label>Wachtwoord: </label><input type="password" name="Password"><br>
    <label>Herhaal uw wachtwoord: </label><input type="password" name="PasswordRepeat"><br>
    <label>Telefoon nummer: </label><input type="text" name="PhoneNumber"><br>
    <label>Fax nummer: </label><input type="text" name="FaxNumber"><br>
    <button type="submit" name="signupbutton" >Registreer</button>
</form>

Answer 1

您的 SQL 语句中有语法错误。应该是：INSERT INTO ... VALUES，而不是 VALUE。