如何在PHP中重置ajax成功后的图像文件预览？

<input type="file" id="imageUpload" accept="image/*">
<div id="imagePreview"></div>

document.getElementById('imageUpload').addEventListener('change', function() {
  var file = this.files[0];
  var formData = new FormData();
  formData.append('image', file);

  // 发送ajax请求
  var xhr = new XMLHttpRequest();
  xhr.open('POST', 'upload.php', true);
  xhr.onload = function() {
    if (xhr.status === 200) {
      var fileName = xhr.responseText;
      var imageURL = 'path/to/your/image/directory/' + fileName;
      document.getElementById('imagePreview').innerHTML = '<img src="' + imageURL + '">';
    }
  };
  xhr.send(formData);
});

<?php
if ($_FILES['image']['error'] === UPLOAD_ERR_OK) {
  $tempName = $_FILES['image']['tmp_name'];
  $fileName = uniqid() . '_' . $_FILES['image']['name'];
  $destination = 'path/to/your/image/directory/' . $fileName;
  move_uploaded_file($tempName, $destination);
  echo $fileName;
}
?>