對於#3這基本上就是他們的網站。
我是一個.Net小夥子,所以我無法真正幫助您處理.php處理程序,您需要收到該文件,但我希望您覺得這很有用。
<html>
<head>
<link href="http://www.phpletter.com/css/general.css" rel="stylesheet" type="text/css" media="screen">
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="http://www.phpletter.com/contents/ajaxfileupload/ajaxfileupload.js"></script>
<script type="text/javascript">
function ajaxFileUpload()
{
$.ajaxFileUpload
(
{
//YOUR URL TO RECEIVE THE FILE
url: 'http://localhost/testing/postfile.php',
secureuri:false,
fileElementId:'fileToUpload',
dataType: 'json',
success: function (data, status)
{
if(typeof(data.error) != 'undefined')
{
if(data.error != '')
{
alert(data.error);
}else
{
alert(data.msg);
}
}
},
error: function (data, status, e)
{
alert(data.error);
alert(e);
}
}
)
return false;
}
</script>
</head>
<body>
<form name="form" action="" method="POST" enctype="multipart/form-data">
<table cellpadding="0" cellspacing="0" class="tableForm">
<thead>
<tr>
<th>Ajax File Upload</th>
</tr>
</thead>
<tbody>
<tr>
<td><input id="fileToUpload" type="file" size="45" name="fileToUpload" class="input"></td>
</tr>
<tr>
<td>Please select a file and click Upload button</td>
</tr>
</tbody>
<tfoot>
<tr>
<td><button class="button" id="buttonUpload" onclick="return ajaxFileUpload();">Upload</button></td>
</tr>
</tfoot>
</table>
</form>
</body>
</html>
第一個就足夠了。我這樣做。 – Zlatev 2010-09-07 17:57:03
但是它說什麼PHP文件應該處理文件? – Steven 2010-09-07 18:23:51