HOW TO SOLVE Notice: Undefined index: IN PHP

[KeyraRa]

hello guys.. i have a problem..can anyone tell me how to solve this:

Notice: Undefined index: BMK81A in C:\
2) Incorrect integer value: '' for column 'BMK81A' at row 1
3) Notice: Undefined index: BMK81 in C:
4) Notice: Undefined index: DL3 in C: and so on


here my code:

$IC2= $_SESSION['IC2'];
        $BMK81A = $_POST['BMK81A'];
        $BMK81 = $_POST['BMK81'];
        $DL3 = $_POST['DL3'];
        $DL2 = $_POST['DL2'];
        $DL1 = $_POST['DL1'];
        $S1 = $_POST['S1'];
        $S2 = $_POST['S2'];
        $S3 = $_POST['S3'];
        $S4 = $_POST['S4'];
        $S5 = $_POST['S5'];
        $S6 = $_POST['S6'];
        $S7 = $_POST['S7'];
        $D1 = $_POST['D1'];
        $A1 = $_POST['A1'];
        $RE1 = $_POST['RE1'];
        $LU2 = $_POST['LU2'];
        $NPT = $_POST['NPT'];
        $SRP = $_POST['SRP'];
        $KTDP = $_POST['KTDP'];
        $KDP = $_POST['KDP'];
        $USPD = $_POST['USPD'];
$query = " UPDATE  doku SET BMK81A='$BMK81A', BMK81='$BMK81', DL3='$DL3', DL2='$DL2', DL1='$DL1', S1='$S1', S2='$S2', S3='$S3', S4='$S4', S5='$S5', S6='$S6', S7='$S7', D1='$D1', A1='$A1', RE1='$RE1', LU2='$LU2', NPT='$NPT', SRP='$SRP', KTDP='$KTDP', KDP='$KDP', USPD='$USPD' WHERE NO_KPT2='$IC2' ";
mysql_query($query) or die(mysql_error());

- i'm doing a checkboxes.. if i select a checkboxex, then it will insert into database. then it should update data in the database if have any changes made..
- in database, i'm doing a default value..if selected it will be become 1, if not selected will become 0. so from this, i want to update in the database..how to make it?

[Luuk]

use function isset() for every array like ( i.e. line#2):

if (issset($_POST['BMK81A']) { 
  $BMK81A = $_POST['BMK81A'];
} else {
  $BMK81A = "no value given for this one";
}

[amnos]

//First assign default value for all variables
$BMK81A = 0;
$DL3=0;
//Check the value is posted by post method if it is assign it to the corresponding value
 if (isset($_POST['BMK81A']) { 
  $BMK81A = $_POST['BMK81A'];
 }
   
 if(isset($_POST['DL3'])){
 	$DL3=$_POST['DL3'];
 }

Like wise do this for all the variables for which you are getting notice error.