How to filter Optional.empty in java 8

Question

I want to filter the values with 0 to print message invalid but I get Optional.empty as result

 Long l = 0L; Optional<String> l2 = Optional.of(l).filter(t -> t>0).map(t ->t.toString()); Optional<String>l3 =l2.filter(String::isEmpty).map(t ->"invalid"); System.out.println(l3);

Need return type to be Optional<String>

Expected output - Invalid
Actual Output - Optional.empty

Updated

Need return type to be String

Expected output - Invalid value 0
Actual Output - Optional.empty

Sorry but what you are doing looks like misuse of Optional. If you know that some value exists (regardless if it is correct or not, as long as it is not null) then Optional isn't right tool. Such cases should are better suited to be handled by simple if(validate(value)){..handle correct value..} else {..handle incorrect value..}. For now your question looks like XY problem — Pshemo
– Pshemo, Commented Dec 30, 2019 at 16:20
What if your input is not empty what is the expected result? — Youcef LAIDANI
– Youcef LAIDANI, Commented Dec 30, 2019 at 16:29

Youcef LAIDANI · Accepted Answer · 2019-12-30 16:37:04Z

6

No need to use Optional or filter, all you need is just :

String l3 = l > 0 ? l.toString() : "Invalid";

If you want to handle the null then :

String l3 = (l != null && l > 0) ? l.toString() : "Invalid";

If you want to show the number in the Invalid case, then :

String l3 = (l != null && l > 0) ? l.toString() : String.format("Invalid %d", l);

edited Dec 30, 2019 at 16:37

answered Dec 30, 2019 at 16:31

Youcef LAIDANI

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8 Comments

coder25 Over a year ago

did not get u , u mean to use it inside filter , String l3 = l > 0 ? l.toString() : "Invalid";

Youcef LAIDANI Over a year ago

@coder25 no, you don't need to use Optional no filter just like it is.

Pshemo Over a year ago

@coder25 but why is it there in the first place? Why involve Optional if your goal (based on how your question looks like now, not sure if it is your real goal) can be achieved in simpler way shown in above answer?

Youcef LAIDANI Over a year ago

@coder25 please show us the original problem, don't post an XY problem

Pshemo Over a year ago

@coder25 Sorry but I don't follow. It is like saying "it is there because it is there" which doesn't give us much. Usually things are in some place because someone made decision that it should be there, and I am asking for reasons behind that decision (why someone - author of code, which is probably you - though that Optional must be used in that place?).

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ernest_k · Accepted Answer · 2019-12-30 16:06:53Z

If you're on Java 9+, you can use or, which takes an alternative Optional if the present one is empty:

Optional<String> l3 = l2.filter(String::isEmpty) .or(() -> Optional.of("invalid"));

thanks. In case I do not need Optional need to print Invalid value +l2 , orElse wont work.Any idea how we can achieve

amer · Accepted Answer · 2019-12-30 15:49:29Z

1

How about not using filter.

String l3 = l2 .map(Object::toString) .orElse("invalid");

edited Dec 30, 2019 at 15:49

answered Dec 30, 2019 at 15:35

amer

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