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php ajax无刷新上传图片实例代码

  • 作者: 姐就是姐_别侮辱姐
  • 来源: 51数据库
  • 2021-09-21
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'], $uploadfile );
echo "url: <a .$_files['photofile']['name']."' target='_blank'>".$_files['photofile']['name']."</a><br/>";
?>
upload successed!
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