在使用AJAX提交后，如何将验证错误回显到我的表单？

document.getElementById('myForm').addEventListener('submit', function(event) {
  event.preventDefault(); // 阻止表单的默认提交行为
  // 其他操作
});

var formData = new FormData(document.getElementById('myForm')); // 获取表单数据
var xhr = new XMLHttpRequest();
xhr.open('POST', '/validate', true);
xhr.setRequestHeader('Content-Type', 'application/json');
xhr.onreadystatechange = function() {
  if (xhr.readyState === 4 && xhr.status === 200) {
    var response = JSON.parse(xhr.responseText);
    // 处理验证结果
  }
};
xhr.send(JSON.stringify({ data: formData }));

@app.route('/validate', methods=['POST'])
def validate():
    data = request.get_json()
    # 进行表单验证逻辑
    if valid:
        return jsonify({'success': True})
    else:
        return jsonify({'success': False, 'errors': {'field1': 'Error message 1', 'field2': 'Error message 2'}})

var form = document.getElementById('myForm');
var errorMessages = response.errors;
for (var field in errorMessages) {
  if (errorMessages.hasOwnProperty(field)) {
    var errorMessage = errorMessages[field];
    var errorElement = document.createElement('span');
    errorElement.className = 'error-message';
    errorElement.innerHTML = errorMessage;
    form.querySelector('[name="' + field + '"]').insertAdjacentElement('afterend', errorElement);
  }
}