java 如何使用java Stream检查Collection是否为空

How to check if Collection is not empty using java Stream

提问by Saurabh Kumar

I am new to Java 8. I am not able to understand what is wrong in the following piece of code. The idea is to sent Collection<User>if its not empty. But if the collection is empty than sent HttpStatus.NOT_FOUNDEntity response.

我是 Java 8 的新手。我无法理解以下代码中有什么问题。这个想法是Collection<User>如果它不为空就发送。但是，如果集合为空，则发送HttpStatus.NOT_FOUND实体响应。

@RequestMapping(value = "/find/pks", 
                method = RequestMethod.GET, 
                produces = MediaType.APPLICATION_JSON_VALUE)
public ResponseEntity<Collection<User>> getUsers(@RequestBody final Collection<String> pks)
{
    return StreamSupport.stream(userRepository.findAll(pks).spliterator(), false)
         .map(list -> new ResponseEntity<>(list , HttpStatus.OK))
         .orElse(new ResponseEntity<>(HttpStatus.NOT_FOUND));
}

Eclipse shows me error in the following point .orElse

Eclipse 显示以下错误 .orElse

The method orElse(new ResponseEntity<>(HttpStatus.NOT_FOUND))is undefined for the type Stream<ResponseEntity<User>>

该orElse(new ResponseEntity<>(HttpStatus.NOT_FOUND))类型的方法未定义Stream<ResponseEntity<User>>

My base interface method looks like following

我的基本接口方法如下所示

Iterable<T> findAll(Iterable<PK> pks);

采纳答案by Holger

You are mixing two things up. The first task is to convert the Iterableto a Collectionwhich you can indeed solve using the StreamAPI:

你把两件事混在一起了。第一个任务是将转换Iterable到Collection你的确可以解决使用StreamAPI：

Collection<User> list=
    StreamSupport.stream(userRepository.findAll(pks).spliterator(), false)
   .collect(Collectors.toList());

Note that this stream is a stream of Users, not a stream of lists. Therefore you can't map a listto something else with this stream. The mapoperation will map each elementof the stream to a new element.

请注意，此流是s流User，而不是列表流。因此，您无法list使用此流将 a 映射到其他内容。该map操作会将流的每个元素映射到一个新元素。

Then you can use this list to create the ResponseEntity

然后你可以使用这个列表来创建 ResponseEntity

return list.isEmpty()? new ResponseEntity<>(HttpStatus.NOT_FOUND):
                       new ResponseEntity<>(list, HttpStatus.OK);

You can combine these steps by creating a Collectorperforming this steps though this does not provide any advantage, it's only a matter of style:

您可以通过创建Collector执行这些步骤来组合这些步骤，尽管这不会提供任何优势，这只是风格问题：

ResponseEntity<User> responseEntity=
    StreamSupport.stream(userRepository.findAll(pks).spliterator(), false)
   .collect(Collectors.collectingAndThen(Collectors.toList(),
      list -> list.isEmpty()? new ResponseEntity<>(HttpStatus.NOT_FOUND):
                              new ResponseEntity<>(list, HttpStatus.OK) ));

回答by Bohemian

It's not necessary, and often a mistake, to cram everything into one line. In this case, you can't - there's no such API for your intention.

将所有内容都塞进一行是没有必要的，而且常常是错误的。在这种情况下，您不能 - 没有适合您的 API。

Keep it simple:

把事情简单化：

Collection<User> list = <your stream code that gets a list>;
if (list.isEmpty())
    return new ResponseEntity<>(HttpStatus.NOT_FOUND);
return new ResponseEntity<>(list, HttpStatus.OK);

but if you absolutely must:

但如果你绝对必须：

return <your code>.map(list -> new ResponseEntity<>(list, list.isEmpty() ? HttpStatus.NOT_FOUND : HttpStatus.OK));

回答by the8472

That depends on your terminal operation of the stream, remember that a stream can only be consumed once.

这取决于你对流的终端操作，记住一个流只能被消费一次。

• Is it a grouping by/statistics operation? Then you'll get a 0 count or an empty map of groups.
• If it collects into a list then it's an empty list.
• If it's one of the methods that returns an Optional(such as findAny) then you can use the optional's null-checking methods.

• 它是按/统计操作分组吗？然后你会得到一个 0 计数或一个空的组图。
• 如果它收集到一个列表中，那么它就是一个空列表。
• 如果它是返回一个Optional（例如findAny）的方法之一，那么您可以使用可选的空检查方法。