Javascript Perl CGI代码错误
你好,这段代码中有漏洞吗?攻击者能否绕过此选项并上载其他文件或webshell? 或者是安全的Javascript Perl CGI代码错误,javascript,perl,cgi,Javascript,Perl,Cgi,你好,这段代码中有漏洞吗?攻击者能否绕过此选项并上载其他文件或webshell? 或者是安全的 #!/usr/bin/perl use CGI; $q = new CGI; if (defined($q->param('Head'))) { print $q->header(-type=>'image/bmp'); open(HEAD,"head".$q->param('Head')); open(HAIR,"hair".$q->param('Hai
#!/usr/bin/perl
use CGI;
$q = new CGI;
if (defined($q->param('Head'))) {
print $q->header(-type=>'image/bmp');
open(HEAD,"head".$q->param('Head'));
open(HAIR,"hair".$q->param('Hair'));
open(NOSE,"nose".$q->param('Nose'));
open(MOUTH,"mouth".$q->param('Mouth'));
open(EYES,"eyes".$q->param('Eyes'));
while (read(HEAD,$headb,1)) {
read(HAIR,$hairb,1);
read(NOSE,$noseb,1);
read(MOUTH,$mouthb,1);
read(EYES,$eyesb,1);
print (chr (ord($headb)&ord($hairb)&ord($noseb)&ord($mouthb)&ord($eyesb)));
}
}
else {
print $q->header;
print $q->start_html(-title=>"Hello wolrd test",-script=>{'src'=>'/js.js'},-style=>{'src'=>'/css.css'});
print $q->div(
$q->h1("Avatar Generator"),
$q->p("make your face !"),
"<video><source src='https://zippy.gfycat.com/DesertedEasygoingArabianwildcat.webm'></source></video><canvas></canvas>",
$q->start_form(-id=>"frm",-method=>"POOP",-action=>"#",-onchange=>"loadImage()"),
$q->br(),
$q->table(
$q->Tr($q->td([$q->b("Head"),$q->input({-name=>"Head",-type=>'range',-min=>1,-max=>4})])),
$q->Tr($q->td([$q->b("Hair"),$q->input({-name=>"Hair",-type=>'range',-min=>0,-max=>2})])),
$q->Tr($q->td([$q->b("Nose"),$q->input({-name=>"Nose",-type=>'range',-min=>1,-max=>3})])),
$q->Tr($q->td([$q->b("Mouth"),$q->input({-name=>"Mouth",-type=>'range',-min=>1,-max=>3})])),
$q->Tr($q->td([$q->b("Eyes"),$q->input({-name=>"Eyes",-type=>'range',-min=>1,-max=>3})]))
),
$q->end_form
);
open SELF, "index.cgi";
print $q->comment("DEBUG SOURCE\n".do { local $/; <SELF> });
print $q->end_html();
}
还有JS代码
任何帮助都将非常感谢不要接受来自进程所有者以外的用户的任意文件名
通过为所有五个字段传递相同的文件名,客户端可以获取服务器可以访问的任何文件
但与为其中一个字段传递以下内容相比,这算不了什么:
rm -rf / |
嗯,好的,还有别的吗?不会。处理用户输入的整个代码都被破坏了。实际上不可能比这更坏。