Mysql Question(inserting Number From A Textfield)

Feelay

Feb 19 2008, 05:53 PM

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:

CODE

$expcomp=$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:

CODE

<?php
session_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");
?>

Comment/Reply (w/o sign-up)

TavoxPeru

Feb 20 2008, 04:45 AM

QUOTE(Feelay @ Feb 19 2008, 12:53 PM)

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:

CODE

$expcomp=$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:

CODE

<?php
session_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:

CODE

<?php
session_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,

Comment/Reply (w/o sign-up)

Feelay

Feb 20 2008, 03:58 PM

"(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.

Comment/Reply (w/o sign-up)

TavoxPeru

Mar 1 2008, 03:51 AM

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:

CODE

<?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:

CODE

<?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:

Best regards,

Comment/Reply (w/o sign-up)