如何通过Javascript访问php代码中的输入名称？

var data = {
  name: "John",
  age: 25
};

// 使用Ajax发送POST请求
var xhr = new XMLHttpRequest();
xhr.open("POST", "your_php_script.php", true);
xhr.setRequestHeader("Content-Type", "application/json");

xhr.onreadystatechange = function() {
  if (xhr.readyState === 4 && xhr.status === 200) {
    // 请求成功后的处理逻辑
    var response = JSON.parse(xhr.responseText);
    console.log(response);
  }
};

xhr.send(JSON.stringify(data));

<?php
$name = $_POST['name'];
$age = $_POST['age'];

// 在这里可以对获取到的数据进行处理或响应

$response = array(
  "message" => "Data received successfully",
  "name" => $name,
  "age" => $age
);

echo json_encode($response);
?>

<form id="myForm" action="your_php_script.php" method="post">
  <input type="text" name="name" id="nameInput" placeholder="Name" required>
  <input type="number" name="age" id="ageInput" placeholder="Age" required>
  <button type="submit">Submit</button>
</form>

document.getElementById("myForm").addEventListener("submit", function(event) {
  event.preventDefault(); // 阻止表单默认提交行为

  var form = event.target;
  var formData = new FormData(form);

  // 使用Fetch API发送POST请求
  fetch(form.action, {
    method: "POST",
    body: formData
  })
  .then(function(response) {
    return response.json();
  })
  .then(function(data) {
    console.log(data);
  })
  .catch(function(error) {
    console.error(error);
  });
});

<?php
$name = $_POST['name'];
$age = $_POST['age'];

// 在这里可以对获取到的数据进行处理或响应

$response = array(
  "message" => "Data received successfully",
  "name" => $name,
  "age" => $age
);

echo json_encode($response);
?>