"No data supplied for parameters in prepared statement"

Question

所以我正在重写一个脚本，以包括准备好的语句。它以前工作得很好，但现在我得到"没有数据提供的参数在准备好的语句"，当脚本运行。这里的问题是什么?

<?php
require_once("models/config.php");

$firstname = htmlspecialchars(trim($_POST['firstname']));
$firstname = mysqli_real_escape_string($mysqli, $firstname);
$surname = htmlspecialchars(trim($_POST['surname']));
$surname = mysqli_real_escape_string($mysqli, $surname);
$address = htmlspecialchars(trim($_POST['address']));
$address = mysqli_real_escape_string($mysqli, $address);
$gender = htmlspecialchars(trim($_POST['gender']));
$gender = mysqli_real_escape_string($mysqli, $gender);
$city = htmlspecialchars(trim($_POST['city']));
$city = mysqli_real_escape_string($mysqli, $city);
$province = htmlspecialchars(trim($_POST['province']));
$province = mysqli_real_escape_string($mysqli, $province);
$phone = htmlspecialchars(trim($_POST['phone']));
$phone = mysqli_real_escape_string($mysqli, $phone);
$secondphone = htmlspecialchars(trim($_POST['secondphone']));
$secondphone = mysqli_real_escape_string($mysqli, $secondphone);
$postalcode = htmlspecialchars(trim($_POST['postalcode']));
$postalcode = mysqli_real_escape_string($mysqli, $postalcode);
$email = htmlspecialchars(trim($_POST['email']));
$email = mysqli_real_escape_string($mysqli, $email);
$organization = htmlspecialchars(trim($_POST['organization']));
$organization = mysqli_real_escape_string($mysqli, $organization);
$inriding = htmlspecialchars(trim($_POST['inriding']));
$inriding = mysqli_real_escape_string($mysqli, $inriding);
$ethnicity = htmlspecialchars(trim($_POST['ethnicity']));
$ethnicity = mysqli_real_escape_string($mysqli, $ethnicity);
$senior = htmlspecialchars(trim($_POST['senior']));
$senior = mysqli_real_escape_string($mysqli, $senior);
$student = htmlspecialchars(trim($_POST['student']));
$student = mysqli_real_escape_string($mysqli, $student);

$order= "INSERT INTO persons (firstname, surname, address, gender, city, province,  postalcode, phone, secondphone, email, organization, inriding, ethnicity, senior, student_id) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
$stmt = mysqli_prepare($mysqli, $order);
mysqli_stmt_bind_param($stmt, "sssd", $firstname, $surname, $address, $gender, $city, $province, $postalcode, $phone, $secondphone, $email, $organization, $inriding, $ethnicity, $senior, $student);
mysqli_stmt_execute($stmt); 
echo $stmt->error;
$result = mysqli_query($mysqli,$stmt);
if ($result === false) {
echo "Error entering data! <BR>";
echo mysqli_error($mysqli);
 } else {
echo "User $firstname added <BR>";
 }
?>

Answer 1

您仅通过控制字符串"sssd"绑定了四个参数，但是您有许多参数。当使用mysqli绑定变量时，每个参数需要一个字符，例如:

mysqli_stmt_bind_param($stmt, "sssdsssssssssdd", $firstname, $surname, $address, 
    $gender, $city, $province, $postalcode, $phone, $secondphone, $email, 
    $organization, $inriding, $ethnicity, $senior, $student);

(我假设senior和student是整数，并且需要"d"代码)

您不需要使用mysqli_real_escape_string()处理任何变量——这就是使用参数的意义所在。如果您也进行转义，您将在数据库中的数据中获得文字反斜杠字符。

并且在任何情况下都不需要使用htmlspecialchars()—您将在输出到HTML时使用它，而不是在插入到数据库时使用它。您将在数据库中的数据中获得像&这样的文字序列。

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你的下一个错误:

"可捕获的致命错误:mysqli_stmt类的对象无法转换为…中的字符串"

这是由以下原因引起的:

$result = mysqli_query($mysqli,$stmt);

该函数期望第二个参数是一个字符串，一个新的SQL查询。但是您已经准备好了这个查询，所以您需要以下内容:

$result = mysqli_stmt_execute($stmt);