Mysql Question(inserting Number From A Textfield)

Hey!

I am trying to do a "Admin give EXP script".
But I can't make it work.
The value is not updating, but the update query is correct.( I think:P)
I think the fault is here:

$expcomp=$givexpp['exp'] += $givexp;

The $givexp is the variable for the amount of Xp the admin wants to give.
the $givexpp is the variable for the user info (in this case, the experince he already have).

The datatype for the XP in the database is INT. So I have no idea if it can take data from a normal textfield.
If you need to see all the code, here you go:

<?phpsession_start();include "database.php";$givexpto= mysql_real_escape_string($_POST['givexpto']);$givexp= mysql_real_escape_string($_POST['givexp']);$queryxp =("SELECT * FROM characters WHERE user='$givexpto'")or die(mysql_error());					if(mysql_num_rows(mysql_query($queryxp)) == 1)	{		$givexpp = mysql_fetch_assoc(mysql_query($queryxp));		}else{ die("User Don't Excist. Please Go Back And Try Again");}$expcomp=$givexpp['exp'] += $givexp;mysql_query("UPDATE characters SET exp='$expcomp' WHERE user='$givexpto'");header("location:index.php");?>

Hey!
I am trying to do a "Admin give EXP script".
But I can't make it work.
The value is not updating, but the update query is correct.( I think:P)
I think the fault is here:

$expcomp=$givexpp['exp'] += $givexp;

The $givexp is the variable for the amount of Xp the admin wants to give.
the $givexpp is the variable for the user info (in this case, the experince he already have).

The datatype for the XP in the database is INT. So I have no idea if it can take data from a normal textfield.
If you need to see all the code, here you go:

<?phpsession_start();include "database.php";$givexpto= mysql_real_escape_string($_POST['givexpto']);$givexp= mysql_real_escape_string($_POST['givexp']);$queryxp =("SELECT * FROM characters WHERE user='$givexpto'")or die(mysql_error());					if(mysql_num_rows(mysql_query($queryxp)) == 1)	{		$givexpp = mysql_fetch_assoc(mysql_query($queryxp));		}else{ die("User Don't Excist. Please Go Back And Try Again");}$expcomp=$givexpp['exp'] += $givexp;mysql_query("UPDATE characters SET exp='$expcomp' WHERE user='$givexpto'");header("location:index.php");?>

Always debug your variables to verify if its values are the expected ones, you can do it simply by echoing it.

Try this:

<?phpsession_start();include "database.php";$givexpto= mysql_real_escape_string($_POST['givexpto']);$givexp= (int) mysql_real_escape_string($_POST['givexp']);echo $givexp;$queryxp="SELECT * FROM characters WHERE user='$givexpto'";	if(mysql_num_rows(mysql_query($queryxp))>0) {	$givexpp = mysql_fetch_assoc(mysql_query($queryxp));} else { die("User Don't Excist. Please Go Back And Try Again"); }$expcomp=$givexpp['exp'] + $givexp;mysql_query("UPDATE characters SET exp=$expcomp WHERE user='$givexpto'");header("location:index.php");exit;?>

Best regards,

"(int)" don't work. I've tryed that before, and now I tryed it with your code.It is still showing value "0" where "$givexp" is.

Which method do you use on your form to submit your data??? If you are using the GET method simply replace the $_POST's variables with $_GET's like this:

<?php$givexpto= mysql_real_escape_string($_GET['givexpto']);$givexp= (int) mysql_real_escape_string($_GET['givexp']);?>

And, I'm not pretty sure about this but just in case, try this code instead:

<?php$givexpto= mysql_real_escape_string($_POST['givexpto']);$givexp= (int) $_POST['givexp'];/* USE THIS IF YOU USE THE GET METHOD  $givexpto= mysql_real_escape_string($_GET['givexpto']);$givexp= (int) $_GET['givexp'];*/?>

Also, take a look to these topics:

Php String To Int Typecasting

Checking To See If Something Is Not An Integer

Best regards,