php ajax无刷新上传图片实例代码-丞旭猿

AJAX 客户端页面代码： index.html

实现代码如下：

<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 + “rn”
datadata = data + “Content-Disposition: form-data; name=”photofile”; filename=”C:\a.txt”rn”
// datadata = data + “Content-Type: image/pjpeg” + vbCrLf
datadata = data + “Content-Type: text/plain” + “rn” + “rn”
text = “My name is Wilson Lin.”
postLength = text.length + data.length + 2 + spliter.length + 4
package = data + text + “rn” + spliter + “–rn”

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

实现代码如下：

<?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! 

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