上门的代码 通过AJAX的方式提交 到PHP 并记录到数据库 如何编写代码
时间: 2024-05-06 10:17:16 浏览: 109
以下是一个基本的示例代码,用于通过 AJAX 将数据提交到 PHP 并将其记录到数据库中:
HTML 代码:
```
<!DOCTYPE html>
<html>
<head>
<title>AJAX Submit Form Example</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
</head>
<body>
<form id="myForm">
<label for="name">Name:</label>
<input type="text" id="name" name="name"><br><br>
<label for="email">Email:</label>
<input type="email" id="email" name="email"><br><br>
<button type="submit">Submit</button>
</form>
<div id="result"></div>
<script type="text/javascript">
$(document).ready(function(){
$("#myForm").submit(function(event){
event.preventDefault();
var name = $("#name").val();
var email = $("#email").val();
$.ajax({
url: "submit.php",
type: "post",
data: {name: name, email: email},
success: function(response){
$("#result").html(response);
}
});
});
});
</script>
</body>
</html>
```
PHP 代码(submit.php):
```
<?php
// 连接到数据库
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
$conn = new mysqli($servername, $username, $password, $dbname);
// 检查连接
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// 获取提交的数据
$name = $_POST['name'];
$email = $_POST['email'];
// 将数据插入到数据库中
$sql = "INSERT INTO MyGuests (firstname, email) VALUES ('$name', '$email')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
```
请注意,此示例仅用于演示目的,并且可能需要进行适当的修改以适应您的具体要求。例如,您需要更改数据库连接详细信息,以及将表名和列名更改为您的实际数据库架构。