joomla ZipArchive生成无效的zip文件
问题描述:
Im使用Joomla 3.x作为我的网站内容管理器,并且还使用了NoNumber sourcerer扩展,它允许我在我的文章中的任何位置放置PHP代码。我使用ZipArchive压缩选定的文件并将其下载到.zip中。但是,当下载的zip文件被破坏时,因为在文本编辑器中查看时,您可以看到Joomla Framework中文章的所有HTML。有没有办法解决???我一直在努力,现在有一段时间了,任何帮助,将不胜感激。joomla ZipArchive生成无效的zip文件
这里是什么放在源码人员标签之间到底:
require_once("../php/Archive/Zip.php");
if((!empty($_POST['showdate']) || isset($_POST['showdate'])) && (!empty($_POST['showsheets']) || isset($_POST['showsheets'])))
{
// echo $_POST['showdate'];
// foreach($_POST['showsheets'] as $showsheet)
// {
// echo $showsheet;
// }
$files = $_POST['showsheets'];
unset($_POST['showsheets']);
$zippath = 'jdownloads/tmp/';
$zipname = $_POST['showdate'].'ShowSheets.zip';
unset($_POST['showdate']);
//$regex = "/^Jan | ^June | ^Sept/";
foreach(new DirectoryIterator($zippath) as $fileInfo){
if($fileInfo->isDot() || !$fileInfo->isFile() || $fileInfo->getFilename() == 'index.html')
continue;
// if (preg_match($regex, $fileInfo->getFilename())) {
unlink($fileInfo->getPathname());
// }
}
$zip = new ZipArchive;
$zip->open($zippath . $zipname, ZipArchive::CREATE);
foreach ($files as $file) {
$zip->addFile($file);
//$content = file_get_contents($file);
//$zip->addFromString(pathinfo ($file, PATHINFO_BASENAME), $content);
}
$zip->close();
// header('Content-Type: application/zip');
// header('Content-disposition: attachment; filename='.$zipname);
// header('Content-Length: ' . filesize($zippath . $zipname));
// readfile($zippath . $zipname);
// http headers for zip downloads
header("Pragma: public");
header("Expires: 0");
header("Cache-Control: must-revalidate, post-check=0, pre-check=0");
header("Cache-Control: public");
header("Content-Description: File Transfer");
header("Content-type: application/octet-stream");
header("Content-Disposition: attachment; filename=\"".$zipname."\"");
header("Content-Transfer-Encoding: binary");
header("Content-Length: ".filesize($zippath . $zipname));
//ob_end_flush();
readfile($zippath . $zipname);
}
?>
Windows资源管理器只是告诉我它的空,并试图toextract告诉我,它的无效和7zip的可以得到什么内部,但实际的.xlsx文件是腐败者,它不能对它们做任何事情。
这是它在文本编辑器中寻找什么时候从一个样本文件制作:
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-gb" lang="en-gb" dir="ltr">
<head>
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<base href="http://wheatbeltusa.com/index.php/testing-one" />
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<meta name="author" content="Super User" />
<meta name="description" content="Farm and Home Supplies" />
<meta name="generator" content="Joomla! - Open Source Content Management" />
<title>Testing One - Wheatbelt Inc.</title>
<link href="http://wheatbeltusa.com/index.php/testing-one" rel="canonical" />
<link href="/templates/protostar/favicon.ico" rel="shortcut icon" type="image/vnd.microsoft.icon" />
<link rel="stylesheet" href="/templates/protostar/css/template.css" type="text/css" />
<link rel="stylesheet" href="/media/system/css/frontediting.css" type="text/css" />
<script src="/media/jui/js/jquery.min.js" type="text/javascript"></script>
<script src="/media/jui/js/jquery-noconflict.js" type="text/javascript"></script>
<script src="/media/jui/js/jquery-migrate.min.js" type="text/javascript"></script>
<script src="/media/system/js/caption.js" type="text/javascript"></script>
<script src="/media/jui/js/bootstrap.min.js" type="text/javascript"></script>
<script src="/media/system/js/frontediting.js" type="text/javascript"></script>
<script type="text/javascript">
jQuery(window).on('load', function() {
new JCaption('img.caption');
});
jQuery(document).ready(function(){
jQuery('.hasTooltip').tooltip({"html": true,"container": "body"});
});
jQuery(document).ready(function()
等..然后是所有这就是我想要的,然后更多的Joomla HTML页脚和二进制(你的想法)
那么,如何做ppl在joomla zip文件,并包括他们的头被下载,而不是由此?
感谢所有帮助
答
OMG,
我只是通过剥离最终实现>并添加退出?;在读取文件之后立即转义它。如:
ob_clean();
flush();
readfile($filepath);
exit;
}