必须按下按钮才能激活php代码？

我已经纠正和美化了所有代码，最小化了它，现在应该是它：<?php$servername = "localhost";$username = "root";$password = "";$dbname = "myDB";$sql = "CREATE DATABASE myDB";$conn = new mysqli($servername, $username, $password, $dbname);if ($conn->connect_error) {    die("Connection failed: " . $conn->connect_error);} $sql = "CREATE TABLE freezerinventory (id INT AUTO_INCREMENT PRIMARY KEY, item VARCHAR(30) NOT NULL,reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP)";if(isset($_POST['itemx'])) {  $itemvar = $_POST["itemx"];  $stmt = $conn->prepare("INSERT INTO freezerinventory (item) VALUES ('?')");  $stmt->bind_param("s", $itemvar);  $stmt->execute();} else {    //the 'itemx' post parameter isn't set, make an alert or something}mysqli_close($conn);?><!doctype html><html>  <head>    <meta charset="UTF-8">    <meta name="viewport" content="width=device-width, initial-scale=1">    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.4.1/css/bootstrap.min.css">    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>    <script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.16.0/umd/popper.min.js"></script>    <script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.4.1/js/bootstrap.min.js"></script>    <title>Untitled Document</title>    <style>      table {        font-family: arial, sans-serif;        border-collapse: collapse;        width: 100%;      }      td, th {        border: 1px solid #dddddd;        text-align: left;        padding: 8px;      }      tr:nth-child(even) {        background-color: #dddddd;      }    </style>  </head>  <body>    <div class="container-fluid">    <h2 style"text-align:center";>Please enter the item you want to add to the list below </h2><form action="<?php echo $_SERVER["PHP_SELF"];?>" class="needs-validation" novalidate method="post">    <div class="form-group">      <label for="uname"></label>      <input type="text" class="form-control" id="itemx" placeholder="Enter an item for the freezer here" name="itemx" required>      <div class="valid-feedback">Valid.</div>      <div class="invalid-feedback">Please fill out this field.</div>    </div>    <button id = "SubmitButton" name = "SubmitButton" type="button submit" class="btn btn-primary">Add to list</button></form>    </div><?php$servername = "localhost";$username = "root";$password = "";$dbname = "myDB";$aVar = mysqli_connect('localhost','root','','myDB');$conn = new mysqli($servername, $username, $password, $dbname);if ($conn->connect_error) {    die("Connection failed: " . $conn->connect_error);} if(isset($_POST['id'])) {  $delete = $con->prepare("DELETE FROM freezerinventory WHERE id= ?");  $delete->bind_param("s", $_POST['id']);  $delete->execute();}$sql = "SELECT * FROM freezerinventory";$result = mysqli_query($aVar, $sql);echo "<table><tr>    <th>Item name</th>    <th>Date added</th>     <th>remove</th>  </tr>    "; while($row = mysqli_fetch_array($result))   {  echo "      <tr><td>" . $row['item'] . "</td><td>" . $row['reg_date'] . "</td><td><form action='' method='POST'>    <div class= 'input-group' > <div class='input-group-append'>        <button class='btn btn-danger'  onclick='deleteitem(".$row['id'].")' id='delete'>Remove</button>        <input type='hidden' name='id'/>    </div></div>    </form></td></tr>";}    echo "</table>";?><script>function deleteitem (id_data) {  $.post('<?php echo $_SERVER["PHP_SELF"];?>', {id: id_data}, function(data) {    console.log(data);  //callback data  });}function myFunction() {  confirm("I am an alert box!");}(function() {  'use strict';  window.addEventListener('load', function() {    // Get the forms we want to add validation styles to    var forms = document.getElementsByClassName('needs-validation');    // Loop over them and prevent submission    var validation = Array.prototype.filter.call(forms, function(form) {      form.addEventListener('submit', function(event) {        if (form.checkValidity() === false) {          event.preventDefault();          event.stopPropagation();        }        form.classList.add('was-validated');      }, false);    });  }, false);})();</script></body></html>它可能仍然需要一些修复，我在那里留下了一些未使用的html元素，但从理论上讲，这应该有效。它利用将数据发送到。您不必使用，但从理论上讲，这应该可以解决您的问题。如果您遇到任何问题，请在我的答案下发表评论。ajaxphpajax注意：只是为了澄清，我没有尝试这样做，因此可能存在一些错误。

必须按下按钮才能激活php代码？

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