Form form, after clicking submit, if the submission is successful, a pop-up window will prompt that the submission is successful. After clicking, it will jump to a new page.

You can use JavaScript to achieve this functionality. The specific steps are as follows:

1. Add a submit button to the form and add an event listener to it.
2. Use the XMLHttpRequest object in the event listener to submit form data to the server.
3. Check the status code returned by the server in the onreadystatechange event handler of the XMLHttpRequest object. If it is 200, the submission is successful, and a prompt box will pop up.
4. Add a "Confirm" button in the prompt box. When the user clicks the Confirm button, use the window.location.href property to jump to a new page.

Here is sample code:

<form id="myForm">
  <!-- 在表单中添加需要提交的表单元素 -->
  <input type="text" name="name">
  <input type="email" name="email">
  <button type="submit" id="submitBtn">提交</button>
</form>

<script>
  const form = document.getElementById('myForm');
  const submitBtn = document.getElementById('submitBtn');

  submitBtn.addEventListener('click', function(event) {
    event.preventDefault(); // 防止表单自动提交

    const xhr = new XMLHttpRequest();
    xhr.open('POST', '/submit');
    xhr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

    xhr.onreadystatechange = function() {
      if (xhr.readyState === XMLHttpRequest.DONE && xhr.status === 200) {
        alert('提交成功！');
        window.location.href = '/newpage'; // 跳转到新页面
      }
    };

    xhr.send(new FormData(form));
  });
</script>

Please note that the above sample code is only used to demonstrate how to implement the described functions, and you need to adjust it according to the actual situation.