没有为准备好的语句中的参数提供数据-错误php-MySQLi

Question

我对php非常陌生。所以，我试着制作一个简单的表格来点三明治，但当我点击提交按钮时，我收到了这个错误"没有为准备语句"中的参数提供数据；。

顺便说一句，我从YouTube视频中复制了大部分代码，我不知道代码的某些部分实际上是做什么的

这是我的代码：

<?php
if (isset($_POST['submit'])) {
if (isset($_POST['nombre']) && isset($_POST['apellido']) &&
isset($_POST['bocadillo']) && isset($_POST['extra']) &&
isset($_POST['comentario']) && isset($_POST['comentario'])) {

$nombre = $_POST['nombre'];
$apellido = $_POST['apellido'];
$bocadillo = $_POST['bocadillo'];
$extra = $_POST['extra'];
$comentario = $_POST['comentario'];
$host = "localhost";
$dbUsername = "------";
$dbpassword = "------";
$dbName = "------";
$conn = new mysqli($host, $dbUsername, $dbpassword, $dbName);
if ($conn->connect_error) {
die('Could not connect to the database.');
}
else {
$Select = "SELECT extra FROM pedidos WHERE extra = ? LIMIT 1";
$Insert = "INSERT INTO pedidos(nombre, apellido, bocadillo, extra, comentario) values(?, ?, ?, ?, ?)";
$stmt = $conn->prepare($Select);
$stmt->execute();
$stmt->bind_result($resultemail);
$stmt->store_result();
$stmt->fetch();
$rnum = $stmt->num_rows;
if ($rnum == 0) {
$stmt->close();
$stmt = $conn->prepare($Insert);
if ($stmt->execute()) {
echo "New record inserted sucessfully.";
}
else {
echo $stmt->error;
}
}
else {
echo "Someone already registers using this email.";
}
$stmt->close();
$conn->close();
}
}
else {
echo "All field are required.";
die();
}
}
else {
echo "Submit button is not set";
}
?>

Answer 1

两个查询都缺少bind_param语句

$Select = "SELECT extra FROM pedidos WHERE extra = ? LIMIT 1";
$stmt = $conn->prepare($Select);
$stmt->bind_param("s", $extra);
$stmt->execute();

然后在插入中

$Insert = "INSERT INTO pedidos(nombre, apellido, bocadillo, extra, comentario) values(?, ?, ?, ?, ?)";
$stmt = $conn->prepare($Select);
$stmt->bind_param("sssss", $nombre, $apellido, $bocadillo, $extra, $comentario);
$stmt->execute();

https://www.w3schools.com/php/php_mysql_prepared_statements.asp