我想使按钮像这样，所以我编写了这段代码。当我单击保存按钮时，我也收到了“数据已插入”消息，同时mysql在数据库中创建了新的id，但数据库中的name为空。名称输入不保存。
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<title>Insert</title>
</head>
<body>
<label>Like</label>
<input type="text" id="name">
<button type="submit" id="button">SAVE</button>
<script>
$(document).ready(function(){
$("#button").click(function(){
var name=$("#name").val();
$.ajax({
url:'insert.php',
method:'POST',
data:{
name:name,
},
success:function(data){
alert(data);
}
});
});
});
</script>
</body>
</html>
insert.php：
<?php
$conn = new mysqli('localhost', 'root', 'mypassword', 'loveme');
$name=$_POST['name'];
$sql="INSERT INTO `data` (`id`, `name`) VALUES (NULL, '$name')";
if ($conn->query($sql) === TRUE) {
echo "data inserted";
}
else
{
echo "failed";
}
?>
此代码可能会对您有所帮助
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', '[email protected]')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
如果连接失败，请通过捕获连接来确保已成功设置连接。还尝试登录“名称”，并确保其实际发送到您的php代码。