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如何运用PHPAjax实现图片的无刷新上传_PHP教程

时间：2021-07-01 10:21:17 帮助过：5人阅读

作为一个PHP Ajax客户端页面代码： index.html

<html>   
<body>   
<h1>Ajax file upload sampleh1><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"   
datadatadata = data + spliter + "rn"   
datadatadata = data + "Content-Disposition: form-data; name="photofile"; filename="C:\a.txt"rn"   
// datadatadata = data + "Content-Type: image/pjpeg" + vbCrLf   
datadatadata = 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 Ajax服务器端代码： upload.php

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/>";   
?>

以上所写代码就是基本的PHP Ajax实现无刷新图片上传的具体解决办法。

http://www.bkjia.com/PHPjc/446386.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/446386.htmlTechArticle作为一个 PHP Ajax客户端页面代码： index.html html body h1 Ajaxfileuploadsample / h1 br / input id = "uplaod" name = "btn_send" type = "button" value = "上传测试" /...

如何运用PHPAjax实现图片的无刷新上传_PHP教程

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