如何将可选的 < T > 转换为流 < T > ？

我想用一个可选项预置一个流。因为 Stream.concat只能连接流，所以我有这样一个问题:

如何将可选的 < T > 转换为流 < T > ？

例如:

Optional<String> optional = Optional.of("Hello");
Stream<String> texts = optional.stream(); // not working

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If restricted with Java-8, you can do this:

Stream<String> texts = optional.map(Stream::of).orElseGet(Stream::empty);

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You can do:

Stream<String> texts = optional.isPresent() ? Stream.of(optional.get()) : Stream.empty();

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最佳答案

In Java-9 the missing stream() method is added, so this code works:

Stream<String> texts = optional.stream();

See JDK-8050820. Download Java-9 here.

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I can recommend Guava's Streams.stream(optional) method if you are not on Java 9. A simple example:

Streams.stream(Optional.of("Hello"))

Also possible to static import Streams.stream, so you can just write

stream(Optional.of("Hello"))

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If you're on an older version of Java (lookin' at you, Android) and are using the aNNiMON Lightweight Stream API, you can do something along the lines of the following:

    final List<String> flintstones = new ArrayList<String>()\{\{
add("Fred");
add("Wilma");
add("Pebbles");
}};


final List<String> another = Optional.ofNullable(flintstones)
.map(Stream::of)
.orElseGet(Stream::empty)
.toList();

This example just makes a copy of the list.

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A nice library from one of my ex collegues is Streamify. A lot of collectors, creating streams from practicly everything.

https://github.com/sourcy/streamify

Creating a stream form an optional in streamify:

Streamify.stream(optional)

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It can depend of how you want to convert empty optional to stream element. If you want to interpret it as "nothing" (or "no element"):

Stream<String> texts = optional.stream(); // since JDK 9
Stream<String> texts = optional.map(Stream::of).orElseGet(Stream::empty); // JDK 8

But if you want to interpret it as null:

Stream<String> texts = Stream.of(optional.oreElse(null));