我正在开发一个用户可以上传zip文件的Web服务器。上传后,应该解压缩文件。我尝试了unzip系统命令以及Archive::Extract
模块。我仍然得到错误:
Insecure dependency in eval while running with -T switch at /usr/lib/perl5/5.10.0/Module/Load/Conditional.pm line 332, <GEN3> line 34
请帮忙。
答案 0 :(得分:0)
我猜你正在做类似
的事情system "unzip $value_from_browser";
所以整个事情是邀请shell代码注入(浏览器中的文件名“bla.zip; rm -fr *”)
使用multiargument语法,避免使用shell:
system 'unzip', $value_from_browser;
答案 1 :(得分:0)
更安全的方法可能是使用Archive::zip
,这是我的一些工作代码,它应该给你如何做的要点。 Archive::zip
模块也有一些信息。
sub unpack_zipfiles
{
my $zipfile = shift; # Name of uploaded zip file
my $zip = Archive::Zip->new();
my $status = $zip->read($zipfile);
if ($status == AZ_OK )
{
my @members = $zip->memberNames();
my $n = 1;
foreach my $f (@members)
{
my @f2 = split(/\//,$f);
# Part 1 - get a filename, and save the file
my $zf = getNextFile($f2[$#f2]);
my $unzipped = qq{$unpack_dir/$zf};
$zip->extractMemberWithoutPaths($f,$unzipped);
# Part 2 - check if it's new or the same, and adjust the filename if necessary
$zf = check_or_revert($unpack_dir,$zf,$f2[$#f2]);
$n++;
push @msgs,qq{Unzipped file: $zf};
# Add file to list of unpacked file (for diagnostic purposes)
push @{$data{unpacked}},$unzipped;
# Unzipped files are plonked back in the file queue, because there is a loop that deals with them.
push @{$data{fileq}},$unzipped;
}
}
else
{
die "Error: $zipfile is not a valid zip file, Zip status=$status";
}
}