Question

我试图使用此代码在mysql中的表中插入名称和密码。它会生成成功警报，但数据不会添加到表中。有人可以解决这个问题吗？

我的html文件 -

<html>
  <body>
    <table border="1">
      <tr>
        <td align="center">Test Form</td>
      </tr>
      <tr>
        <td>
          <table>
            
            <form onsubmit="return false" id="frm">
    
              <tr>
                <td>Name</td>
                <td><input type="text" id="name" name="name" size="50">
                </td>
              </tr>
              <tr>
                <td>Password</td>
                <td><input type="text" id="password" name="password" size="50">
                </td>
              </tr>
              <tr>
                <td></td>
                <td align="right">
                  <input type="submit" id="btnSubmit" name="submit" value="Send">
                </td>
              </tr>
            </form>
          </table>
        </td>
      </tr>
    </table>

    
  </body>
  

</html>

和javascript - 我想在不加载任何其他页面的情况下进行插入（保持在同一个html页面上）。此ajax部分会弹出警报，但数据不会添加到表中。

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>
    <script>
      $("#frm").submit(function() {
                    var name= $("#name").val();
                    var password= $("#password").val();

                    $.ajax({
                        type: "POST",
                        url: "db.php",
                        data: "name=" + name+ "&password=" + password,
                        success: function(data) {
                           alert("success!");
                        }
                    });


                });
      </script>

最后是PHP

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dynamic";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$order = "INSERT INTO userdata (name, password) VALUES ('$_POST[name]','$_POST[password]')";


$result = $conn->query($order);  
if($result){
    echo("Successfully added!");
} else{
   echo("Input failed!");
}
?>

Answer 1

试试这个 $ frm到#btnSubmit

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>
<script>
  $("#btnSubmit").click(function() {
         var data = {name: $("#name").val(), password: $("#password").val()'}
                $.ajax({
                    type: "POST",
                    url: "db.php",
                    data: data,
                    success: function(data) {
                       alert("success!");
                    }
                });


            });
  </script>

Answer 2

在HTML中更改代码：

<form onsubmit="return false" id="frm" method="POST">

</form>

添加表单方法POST

更改PHP代码：

$name = $_POST["name"];
$password = $_POST["password"];

$sql = "INSERT INTO userdata (name, password)
VALUES ($name, $password)";

并关闭您的连接

$conn->close();

使用AJAX和PHP将数据插入MySQL

2 个答案: