php ajax无刷新上传图片实例代码

AJAX 客户端页面代码： index.html
复制代码代码如下:
<html>
<body>
<h1>Ajax file upload sample</h1><br/><input id="uplaod" name="btn_send" type="button" value="上传测试"/>
<div id=result></div>
<PRE class=js name="code"><SCRIPT LANGUAGE=JavaScript>
// 上传函数
function btn_send.onclick（） {
data = ""
spliter = "-------7d8d733180846"
datadata = data + spliter + " "
datadata = data + "Content-Disposition: form-data; name="photofile"; filename="C:\a.txt" "
// datadata = data + "Content-Type: image/pjpeg" + vbCrLf
datadata = data + "Content-Type: text/plain" + " " + " "
text = "My name is Wilson Lin."
postLength = text.length + data.length + 2 + spliter.length + 4
package = data + text + " " + spliter + "-- "

alert（package）
// 把XML文档发送到Web服务器
var xmlhttp = new ActiveXObject（"Microsoft.XMLHTTP"）;
xmlhttp.open（"POST","./upload.php",false）;
xmlhttp.setRequestHeader（"Content-Type", "multipart/form-data; boundary=-----7d8d733180846"）;
xmlhttp.setRequestHeader（"Content-Length", postLength）;
xmlhttp.send（package）;
// 显示服务器返回的信息
result.innerHTML=xmlhttp.ResponseText;
}
</SCRIPT>
</PRE>
</body>
</html>

PHP服务器端代码： upload.php
复制代码代码如下:
<？php
// $_FILES["photofile"]:是获得上传图片的数组
// $uploadfile：存放地址
$uploadfile = "D:/".$_FILES["photofile"]["name"];
copy（ $_FILES["photofile"]["tmp_name"], $uploadfile ）;
echo "URL: <a href="http://localhost/".$_FILES["photofile"]["name"]."" target="_blank">".$_FILES["photofile"]["name"]."</a><br/>";
？>
Upload successed！