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php ajax无刷新上传图片实例代码

2022-2-5 17:01| 发布者: 菜鸟教程小白| 查看: 303| 评论: 0|原作者: [db:作者]|来自: [db:来源]

摘要: 将index.html和upload.php文件保存到支持php的空间的同级目录，测试运行即可

AJAX 客户端页面代码： index.html

<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! 