上传文件原生的ajax应改怎么写

Question

这是$ajax的写法

那类似这样的怎么写，网上实在是找不到了，那类似这样的怎么写，网上实在是找不到了

Answer 1

拿走不用客气

function uploadFile(url, file, onProgress, onSuccess, onError) {
  const xhr = new XMLHttpRequest();
  const formData = new FormData();

  formData.append('file', file);

  xhr.open('POST', url, true);

  xhr.upload.onprogress = function(event) {
    if (event.lengthComputable) {
      const percentComplete = event.loaded / event.total * 100;
      onProgress(percentComplete);
    }
  };

  xhr.onload = function() {
    if (xhr.status >= 200 && xhr.status < 300) {
      onSuccess(xhr.responseText);
    } else {
      onError(new Error('Upload failed with status ' + xhr.statusText));
    }
  };

  xhr.onerror = function() {
    onError(new Error('Upload failed with status ' + xhr.statusText));
  };

  xhr.send(formData);
}

// 使用方法
const fileInput = document.querySelector('input[type="file"]');
fileInput.addEventListener('change', function(event) {
  const file = event.target.files[0];
  uploadFile(
    '/upload', 
    file, 
    function(progress) {
      console.log('Upload progress: ' + progress + '%');
    },
    function(responseText) {
      console.log('Upload complete: ' + responseText);
    },
    function(error) {
      console.error('Upload failed: ' + error.message);
    }
  );
});