comparing string with null in perl

[Shrutisinha]

if (int(substr($re cord,1,8)) > int(0)||
(substr($record ,11,16))> int(0))
{
if (int(substr($re cord,1,9)) >= int(100000000)) {
$Ban = trim(substr($re cord,1,9));
}

if ((substr($recor d,48,8)) ne null) {

$prd_typ = trim(substr($re cord,48,8));

}

this don't work as it doesn't recognise the null value, somebody please help me in this problem
what i am trying to do isas soon as my prd_typ hit null value i want to print prev value what is in the prd_typ ..

[KevinADC]

there is no "null" in perl, try like this:

if ( substr($record, 48,8) ) {

that will return true if the substr function is able to parse $record with the arguments 48,8 so the expresion in the following block will be processed.

[Shrutisinha]

there is no "null" in perl, try like this:

if ( substr($record, 48,8) ) {

that will return true if the substr function is able to parse $record with the arguments 48,8 so the expresion in the following block will be processed.

hi thanks kevin that helped me
i changed the code to this and it works now
############### ############### ############
## Extracting Ban And Product Type records
############### ############### ############

if (int(substr($re cord,1,8)) > int(0)||
(substr($record ,11,16))> int(0))
{
if (int(substr($re cord,1,9)) >= int(100000000)) {
$Ban = trim(substr($re cord,1,9));
print OutPutFile $prd_typ;
}

if ((substr($recor d,48,8)) ne " "){

$prd_typ = trim(substr($re cord,48,8));

}