我试图让当前代码从表单中获取on或off值,并使用对JS函数的函数调用将其放入SQL数据库中,该JS函数包含一个AJAX请求,然后将其发送到运行实际查询的PHP页面。'UPDATE onoff SET onoroff = $_POST["selectonoff"] WHERE ID = 1;';
$update = mysqli_query($db, $query">Toggle OFF</option>
* FROM `recipe` WHERE creator='$uid'"; $results = mysqli_query($conn,$query); $status = 0; //if OFF->0
$sql = "UPDATE `recipe` SET `status`=$status WHERE creator=$uid AND r