个人中心
gpt4 book ai didi

javascript - 必须按下按钮才能激活 php 代码吗？

转载 作者：行者123 更新时间：2023-11-28 10:37:11 25 4
gpt4 key购买 nike

所以这个网站的想法是，当您按下按钮时，它会从数据库中删除相应的行。然而，我的代码的问题是，第一次按下按钮后，填充 id 变量，然后在下一次按下时执行 php。我怎样才能避免这种情况并让 php 适应并在第一次按下按钮时执行 php？

其中包含嵌入 php 的 HTML 页面:

<!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>

<?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"];
$sql = "INSERT INTO freezerinventory (item)
VALUES ('$itemvar')";
    $add =  mysqli_query($conn, $sql);
}
else {
    $sql = "";
}




mysqli_close($conn);


?>

<form action="<?=$_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);
} 

$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()'  id = 'delete' type='submit'>Remove</button>

                <input type='hidden' name='id' value=".$row['id']." />

            </div></div>

</form></td></tr>";



}
    echo "</table>";
?>

 <script>

function deleteitem (e) {

            e.preventDefault();

    <?php 



    $id = $_POST['id'];

    $delete = "DELETE FROM freezerinventory WHERE id=$id";
   $del =  mysqli_query($conn, $delete);

        ?>
        }
        else {
    }
    </script> 

<script>

function myFunction() {
  confirm("I am an alert box!");
}

</script>

<script>

(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>

最佳答案

您可以创建单独的处理脚本并将表单提交到按钮的 URL 路径“onClick”。不能用JS代码直接调用PHP函数。

关于javascript - 必须按下按钮才能激活 php 代码吗？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/59778321/

25 4 0
文章推荐： json - 使用 SwiftyJSON 解析具有相似元素的 JSON
文章推荐： html - 如何为与文本相关的 Angular Material 的 mdRadioButton 指令设置垂直对齐？
文章推荐： javascript - 如何以 Angular react 形式执行最小和最大验证？
文章推荐： java - JSF 数据表不显示任何数据
行者123
个人简介

我是一名优秀的程序员,十分优秀！

滴滴打车优惠券免费领取
滴滴打车优惠券
全站热门文章
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com