[SOLVED] blocking injection

[serverman]

ok ive been reading the mysql_real_escape but i dont get it...

will some one please explane how to add it to my code and how it works (please and thank you)

oh and the code is a little sloppy i did it my self and im a noob (like 6 months of looking at php and 1 of working with it)

and if you have any tips on how to improve this please do share

<?php
//vars
$login = mysql_connect("---","---","---");
$firstname = $_POST['firstname'];
$comment = $_POST['comment'];
$ip = getenv('REMOTE_ADDR');
//test
if(!$firstname  || !$comment ) 
{
die("Fill the form properly!");
}
//connect
if (!$login)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("website_stuff", $login);$sql="INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('$_POST[firstname]','$_POST[lastname]','$_POST[email]','$_POST[comment]', '$ip')";
//query
if (!mysql_query($sql,$login))
  {
  die('Error: ' . mysql_error());
  }
// ending
echo "Thank you for leaving a comment."."<a href='../../home.PHP'>Back to Home</a>";
mysql_close($login)
?>

[LooieENG]

To use it, just do

 

<?php

$var = $_GET['var'];
$var = mysql_real_escape_string($var);

?>

 

And when you print the data

 

<?php

echo stripslashes($var);

?>

[DeanWhitehouse]

mysql_connect does not need to be a vairable, just add

or die

after it and then the script won't work if it can't connect

[MadTechie]

okay this is the line

$sql="INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('$_POST[firstname]','$_POST[lastname]','$_POST[email]','$_POST[comment]', '$ip')";

change to

$sql="INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('".mysql_real_escape($_POST['firstname'])."','".mysql_real_escape($_POST['lastname'])."','".mysql_real_escape($_POST['email'])."','".mysql_real_escape($_POST['comment'])."', '$ip')";

 

the reason, well read up on sql injection, but basically you are allowing anyone to control your whole database, that means anything you store in the database can be drop (removed) or updated (with anything of their choice)

 

put it this way, heres your insert

INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('$_POST[firstname]','$_POST[lastname]','$_POST[email]','$_POST[comment]', '$ip')"

now lets say the first name was mad last name techie, email [email protected], and comment was

nothing','0.0.0.0')--

looks weired i know but how will you code deal with it ?

basically your code will give the comment nothing and the ip 0.0.0.0

why ?

this is the resolved SQL statement

INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('mad','techie','[email protected]','nothing','0.0.0.0')--', '$ip')

the -- comments out the statement after it so you endup with

INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('mad','techie','[email protected]','nothing','0.0.0.0')

 

So yo add to your existing code

 

 

<?php
//vars
$login = mysql_connect("---","---","---");
$firstname = $_POST['firstname'];
$comment = $_POST['comment'];
$ip = getenv('REMOTE_ADDR');
//test
if(!$firstname  || !$comment ) 
{
die("Fill the form properly!");
}
//connect
if (!$login)
  {
  die('Could not connect: ' . mysql_error());
  }
/*--OLD
mysql_select_db("website_stuff", $login);$sql="INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('$_POST[firstname]','$_POST[lastname]','$_POST[email]','$_POST[comment]', '$ip')";
*/

mysql_select_db("website_stuff", $login);
$sql="INSERT INTO comment (FirstName, LastName, Email, Comment, Ip) VALUES('".mysql_real_escape($_POST['firstname'])."','".mysql_real_escape($_POST['lastname'])."','".mysql_real_escape($_POST['email'])."','".mysql_real_escape($_POST['comment'])."', '$ip')";

//query
if (!mysql_query($sql,$login))
  {
  die('Error: ' . mysql_error());
  }
// ending
echo "Thank you for leaving a comment."."<a href='../../home.PHP'>Back to Home</a>";
mysql_close($login)
?>

 

 

[serverman]

thanks dude i that was alot of help i give you 2 tumbs up for taking the time to explain and not just show me how to do it

thanks... php is sweet!

[MadTechie]

your welcome,

if this post is solved can you click solved, same other helpers having to read it all

 

as a side note the injection can be worse ie dropping the table or on a login

 

select * from users where username = '{$_POST['user']}' and password = '{$_POST['pass']}';

 

if the $_POST['user'] =

admin' --

 

OR

 

select * from users where password = '{$_POST['pass']}' and username = '{$_POST['user']}' ;

 

if the $_POST['user'] =

admin' OR username='admin'--

 

etc etc

 

but i assume you get the idea