5
1144<!DOCTYPE html>
<html>
<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" >
<input name="yfkg" type="submit" value="yfkg" onclick="submit"($yfkg=000)>
<input name="tfkg" type="submit" value="tfkg" onclick="submit"($tfkg=000)>
</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", $sql);
echo "
设计阈值成功
";
}
else{
echo "请重新输入
";
}
break;
case "tfkg":
if ($conn){
$sql = "UPDATE control SET yfkg=000";
$conn1= execute_sql($conn, "nongyedapeng", $sql);
echo "设计阈值成功
";
}
else{
echo "请重新输入
";
}
break;
default:
}
?> 
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你这个PHP代码完全不能执行，你链接数据那里都写得是错误，你有类？为什么要new mysqli ? 再说你链接数据库的函数也不对，人家是 mysqli_connect()
你应该是想要在表单中填入数据，然后将表单中的数据通过一个按钮提交到数据库中对吧？
根据你这个代码，你需要在这个html的目录下新建一个1.php，将你的php代码放入其中即可