能实现一个网页上面有多个按键，每个按键能提交不同的数据。现在的问题是不知道html如何通过post提交上来我想要的值？望大家帮帮忙
html代码
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
<script type="text/javascript" src="js/jquery-1.12.4.min.js"></script>
<style type="text/css">
body>div{height: 40px;}
label{
display: inline-block;
width: 50px; height: 22px; background:#dc3545; border-radius: 30px ; transition: background-color .6s linear;
}
/*在标签的前面插入一小圆形*/
label::after{
content: "关";width: 17px; height: 17px; background-color:#FFFFFF;color:#dc3545;display: inline-block;text-align: center;line-height: 17px;position: relative;left:3px;
transition: transform.6s linear; border-radius: 50%;
}
/*获得复选框后面的第一个兄弟元素label*/
input[type=checkbox]:checked+label{
background-color:#28a745;
}
input[type=checkbox]:checked+label::after{
content: "开";color:#28a745;transform: translate(27px);/*向右移动27px*/
}
</style>
</head>
<body>
<form action="1.php" method="post" >
<p align="center">
<input class="btn" name="yfkg" type="submit" value="yfkg" onclick="submit"($yfkg=000)>
<p align="center">
<input class="btn" name="tfkg" type="submit" value="tfkg" onclick="submit"($tfkg=000)>
<p align="center">
</form>
</html>
```php
<?php
header("Content-type: text/html; charset=utf-8");
//建立数据库连接*/
$servername = 'localhost';
$username ='root';
$password = '123';
$dbname = "nongyedapeng";
$conn = new mysqli($servername, $username, $password, $dbname);
function execute_sql($link, $database, $sql){
mysqli_select_db($link, $database)
or die("打开数据库失败: " . mysqli_error($link));
$result = mysqli_query($link, $sql);
return $result;
}
switch ($_POST["submit"]){
case "yfkg":
if ($conn){
$sql = "UPDATE control SET yfkg=111";
$conn1= execute_sql($conn, "nongyedapeng", $sql);
echo "<p align='center'>设计阈值成功</p>";
}
else{
echo "<p align='center'>请重新输入</p>";
}
break;
case "tfkg":
if ($conn){
$sql = "UPDATE control SET yfkg=000";
$conn1= execute_sql($conn, "nongyedapeng", $sql);
echo "<p align='center'>设计阈值成功</p>";
}
else{
echo "<p align='center'>请重新输入</p>";
}
break;
default:
}
?>
ajax调用php提供的接口
是不会发送请求吗？我看你代码中载入了 JQuery，你可以参考这里 https://www.w3school.com.cn/jquery/jquery_ref_ajax.asp