

(1)如何获取当前文件的数据流呢?
答:通过FormData()实例化的对象,将文件数据append在一个变量里面
(2)如何获取数据?
答:在type为file的input表单中,自带一个files属性。
HTML页面发送文件上传请求:
	<input type="file" name="upload_img" id='upload_img'/>
	<img src="" id='myfile_img' alt='' title='' width='300'/> 
	<script type="text/javascript">
	
	var uploadImg = document.getElementById('upload_img');
	var myfileImg = document.getElementById('myfile_img');
	uploadImg.onchange = function()
	{
	var imgName = this.files[0].name;
	//let reader = new FileReader();
	var fordata = new FormData();
	fordata.append('my_file',this.files[0]);
	
	//向服务器发送文件数据
	ajaxPost(fordata,function(obj){
	var content = JSON.parse(obj.response);
	console.log(content);
	if(content.status == 'sucess'){
	myfileImg.src = './images/'+imgName;
	}
	});
	}
	function ajaxPost(data,fn)
	{
	var xhr = new XMLHttpRequest();
	xhr.open('post','./upload.php','true');
	xhr.send(data);
	xhr.onload = function()
	{
	fn(this);
	} 
	}
	</script>服务器处理文件数据,生成上传的文件:
$success = array('status' => 'sucess', 'code' => '1');
$error = array('status' => 'error', 'code' => '0');
if (!empty($_FILES)) {
 $file = $_FILES['my_file'];
 $new_file_dir = dirname(__FILE__) . '/images/' . $file['name'];
 @move_uploaded_file($file['tmp_name'], $new_file_dir);
 exit(json_encode($success));
} else {
 exit(json_encode($error));
}