error in a text box to mysql when I use '

[computel]

Getting this error in my script

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 's incredible attention to detail, which will now serve as the new benchmark for '

 

I'm entering text into a text box and when I use ' in the text I get this error. example it's will cause the error.

 

Here is the script

 

<?
require_once("conn.php");
require_once("includes.php");
require_once("access.php");

if(isset($_POST[s1]))
{
if($_POST[content_type] == "cheat")
{
	$ItemID = $_POST[item_id_1];
	$ContentTitle = strip_trim($_POST[cheat_title]);
	$ContentText = strip_trim($_POST[cheat_text]);
}
else
{
	$ItemID = $_POST[item_id_2];
	$ContentTitle = strip_trim($_POST[review_title]);
	$ContentText = strip_trim($_POST[review_text]);
	$rating = $_POST[rating];
}

$q1 = "insert into games_content set 
							ContentType = '$_POST[content_type]',
							ItemID = '$ItemID',
							ContentTitle = '$ContentTitle',
							ContentText = '$ContentText',
							rating = '$rating',
							date_added = '$t',
							user_id = '$_SESSION[MemberID]' ";
mysql_query($q1) or die(mysql_error());

$last = mysql_insert_id();

header("location:view.php?cmd=$_POST[content_type]&id=$last&content_id=$ItemID");
exit();
}


if($_POST[content_type] == "cheat" || empty($_POST[content_type]))
{
$checked1 = "checked";
}
elseif($_POST[content_type] == "review")
{
$checked2 = "checked";
}

require_once("templates/HeaderTemplate.php");
require_once("templates/AddReviewTemplate.php");
require_once("templates/FooterTemplate.php");

?>

 

thanks in advance

[Pikachu2000]

Run any string type data coming from your form through mysql_real_escape_string() before using it in a database query. That should always be done to prevent SQL injection attacks anyhow.

[computel]

how do I do this I have almost 0 experience in PHP & mysql. I bought this script and it doesn't work.

 

Any help would be great.

 

Thanks

 

[Zane]

It's as easy as doing this

$variable = mysql_real_escape_string($_POST['variable']);

Just like any other function.

[computel]

OK That seems to work. Thanks everyone that replied.

 

 

but the line return doesn't work.

 

example

 

the dog ran fast

 

the boy ran faster.

 

here is how it is stored in the database and show on the script

 

the dog ran fast

the boy ran faster.

 

how do I make it that is allows the line breaks when stored in the database and when shown