我已经创建了一些PHP代码来创建表，并且它可以工作-注意，ID是自动递增的。
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "TrainingDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// sql to create table
$sql = "CREATE TABLE Courses (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
course VARCHAR(40) NOT NULL
)";
if ($conn->query($sql) === TRUE) {
echo "Table Courses created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close();
?>
我创建了一些php代码来插入一条记录，并且它可以正常工作。注意，我只需要插入课程而不是ID。
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "TrainingDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO Courses (course)
VALUES ('Course1')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
最后，我创建了一些php代码，以便用户可以在表单上输入数据并将其保存在表中。我在使用自动递增键时遇到问题。我收到错误消息警告：mysqli_prepare（）恰好需要2个参数，给定1个参数我已经试过$ sql =“ INSERT INTO Courses（id，course）VALUES（？，？）”；我也不确定“ sss”。
<!DOCTYPE html>
<html lang="en">
<head>
<title>Course Example</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.16.0/umd/popper.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/js/bootstrap.min.js"></script>
</head>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "TrainingDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['save'])){
echo "save<br>";
$sql = "INSERT INTO Courses (course)
VALUES (?)";
$stmt = mysqli_prepare($sql);
$stmt->bind_param("sss", $_POST['course_name']);
$stmt->execute();
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
<div class="container">
<h2>Course</h2>
<form course="/course_page.php" method="post">
<label for="course_name">Course Name:</label><br>
<input type="text" id="course_name" name="course_name" size="40" maxlength="40"><br><br>
<button type="submit" class="btn btn-primary" name="save" value="save">Save</button>
</form>
</div>
</body>
</html>
您需要
$stmt = mysqli_prepare($conn, $sql);
建立连接。
O.Jones说了什么，加上那里您只需要一个“ s”：bind_param("sss", $_POST['course_name']);