如何捕获form.submit的响应

const form = document.querySelector('form');
form.addEventListener('submit', (event) => {
  event.preventDefault();

  const xhr = new XMLHttpRequest();
  xhr.open('POST', form.action);
  xhr.onload = () => {
    if (xhr.status === 200) {
      const response = xhr.responseText;
      console.log('响应内容：', response);
    } else {
      console.error('请求失败，状态码：', xhr.status);
    }
  };
  xhr.send(new FormData(form));
});

const form = document.querySelector('form');
form.addEventListener('submit', async (event) => {
  event.preventDefault();

  try {
    const response = await fetch(form.action, {
      method: 'POST',
      body: new FormData(form),
    });

    if (response.ok) {
      const responseText = await response.text();
      console.log('响应内容：', responseText);
    } else {
      console.error('请求失败，状态码：', response.status);
    }
  } catch (error) {
    console.error('请求失败，错误信息：', error);
  }
});