How to determine if a long type is actually NULL?

Question

Our MySQL database has a null (long type) column (called referral). We use sleep mode for ORM.

I am trying to get the column value for a given member. Some of them are NULL, and if not, then its identifier points to another member, which is the referee.

The problem is the java code, which I am trying to determine if this member column is null, if not, to do something.

String referrerAffiliateId = Long.toString(member.getReferral());
if (referrerAffiliateId != null){
    //do something
}

member.getReferral () returns the value (long type) of the referral column. Some of these columns are null and some are not.

The above code compiles fine, but I get a nullPointerException when I call a method on a user whose referral column is null.

How to properly perform detection on this?

Thanks in advance!

Full answer:

Thanks @Marcelo for the best correct answer.

Here is the code in its final state:

Long referrerAffiliateId = member.getReferral();
if (referrerAffiliateId != null) {
    //...
}

Answer 1

Assuming that member.getReferral()returns a Long, use:

if (member.getReferral() != null)

In Hibernate, if you want to define the nullability value in a property, you should not use primitive types because they will always have a default value. 0 for lengths.

Answer 2

An exception is probably an Long.toString()attempt to check the value before converting to a string:

Long ref = member.getReferral();
if (ref == null) {
  // Do something...
} else {
  String referrerAffiliateId = Long.toString(ref);
  // ...
}

Answer 3

String referrerAffiliateId = Long.toString(member.getReferral());
if (referrerAffiliateId != null){
    //do something
}

To:

if (member.getReferral() != null){

    String referrerAffiliateId = Long.toString(member.getReferral());
    //do something
}

, NullPointerException, Long.toString() .

Answer 4

:

    Long ref = member.getReferral();
    String referrerAffiliateId = null;
    if(ref != null){
    referrerAffiliateId = Long.toString(ref); 
    }