php ajax无刷新上传图片实例代码_AJAX相关_脚本之家--HTML5中文学习网

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 + "/r/n" 
datadata = data + "Content-Disposition: form-data; name=/"photofile/"; filename=/"C://a.txt/"/r/n" 
// datadata = data + "Content-Type: image/pjpeg" + vbCrLf 
datadata = data + "Content-Type: text/plain" + "/r/n" + "/r/n" 
text = "My name is Wilson Lin." 
postLength = text.length + data.length + 2 + spliter.length + 4 
package = data + text + "/r/n" + spliter + "--/r/n" 

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! 