【发布时间】:2012-01-07 19:14:51
【问题描述】:
现在我正在使用以下方法将文件上传到 PHP
<form enctype="multipart/form-data" action="http://sserver.com/fileupload.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="30000000" />
<input type="hidden" name="filename" value="file_uploaded.gif" />
<input type="hidden" name="username" value="foobar"/>
Please choose a file:
<input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>
我在php中阅读了$_POST和$_FILE来完成这样的上传。
$target = $_SERVER['DOCUMENT_ROOT']."/test/upload/";
$target = $target . basename( $_FILES['uploaded']['name']) ;
echo $target;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['uploaded']['name']). " has been uploaded";
}
else {
echo "Sorry, there was a problem uploading your file.";
}
我的问题是,我能否将上述代码 (HTML) 更改为 Ajax XMLHttpRequest 而无需更改 PHP。
【问题讨论】:
标签: php ajax upload xmlhttprequest