Simplify ifPresentOrElse chain

Question

Given the code:

  Optional<String> myOptional = getMyOptional();
  myOptional.ifPresentOrElse(
      s -> Optional.ofNullable(someMap.get(s))
      .ifPresentOrElse(g -> {
            doSomeStuff(); 
          },
          () -> doErrHandling()),
      () -> doErrHandling());

Now I am thinking about how to simplify the chain and remove the duplicate code line (() -> doErrHandling()).

Looks simple unless you want to use `g` in your actual code as `if (myOptional.isPresent() && someMap.containsKey(myOptional.get())) { doSomeStuff(); } else { doErrHandling(); }` — Naman, Jul 16 '19 at 12:06

score 9 · Accepted Answer · answered Jul 16 '19 at 12:04

9

Use map:

Optional<String> myOptional = getMyOptional() ;
myOptional.map(s -> someMap.get(s))
          .ifPresentOrElse(g -> doSomeStuff(), () -> doErrHandling());

map will return an Optional.empty() if the original Optional is empty, and will wrap the result of someMap.get(s) with an Optional otherwise.

answered Jul 16 '19 at 12:04

Eran

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This is the correct answer. May just use `someMap::get` if desired. – Ole V.V. Jul 16 '19 at 14:13

Simplify ifPresentOrElse chain

1 Answers1