将Java流过滤为1，并且只有1个元素

我试图使用Java 8流在LinkedList中查找元素。但是，我想保证与筛选条件有且只有一个匹配。

以这段代码为例:

public static void main(String[] args) {

    LinkedList<User> users = new LinkedList<>();
    users.add(new User(1, "User1"));
    users.add(new User(2, "User2"));
    users.add(new User(3, "User3"));

    User match = users.stream().filter((user) -> user.getId() == 1).findAny().get();
    System.out.println(match.toString());
}

static class User {

    @Override
    public String toString() {
        return id + " - " + username;
    }

    int id;
    String username;

    public User() {
    }

    public User(int id, String username) {
        this.id = id;
        this.username = username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getUsername() {
        return username;
    }

    public int getId() {
        return id;
    }
}

这段代码根据用户的ID查找用户。但是不能保证有多少用户匹配过滤器。

更改过滤器行为:

User match = users.stream().filter((user) -> user.getId() < 0).findAny().get();

将抛出一个NoSuchElementException(很好!)

但是，如果有多个匹配，我希望它抛出一个错误。有办法做到这一点吗?

当前回答

Guava提供了MoreCollectors.onlyElement()，它在这里做正确的事情。但如果你必须自己做，你可以为这个创建自己的Collector:

<E> Collector<E, ?, Optional<E>> getOnly() {
  return Collector.of(
    AtomicReference::new,
    (ref, e) -> {
      if (!ref.compareAndSet(null, e)) {
         throw new IllegalArgumentException("Multiple values");
      }
    },
    (ref1, ref2) -> {
      if (ref1.get() == null) {
        return ref2;
      } else if (ref2.get() != null) {
        throw new IllegalArgumentException("Multiple values");
      } else {
        return ref1;
      }
    },
    ref -> Optional.ofNullable(ref.get()),
    Collector.Characteristics.UNORDERED);
}

…或者使用你自己的Holder类型而不是AtomicReference。您可以尽可能多地重用收集器。

2014-03-27 17:51:16

其他回答

public List<state> getAllActiveState() {
    List<Master> master = masterRepository.getActiveExamMasters();
    Master activeMaster = new Master();
    try {
        activeMaster = master.stream().filter(status -> status.getStatus() == true).reduce((u, v) -> {
            throw new IllegalStateException();
        }).get();
        return stateRepository.getAllStateActiveId(activeMaster.getId());
    } catch (IllegalStateException e) {
        logger.info(":More than one status found TRUE in Master");
        return null;
    }
}

In this above code, As per the condition if its find more than one true in the list then it will through the exception. When it through the error will showing custom message because it easy maintain the logs on server side. From Nth number of element present in list just want only one element have true condition if in list there are more than one elements having true status at that moment it will through an exception. after getting all the this we using get(); to taking that one element from list and stored it into another object. If you want you added optional like Optional<activeMaster > = master.stream().filter(status -> status.getStatus() == true).reduce((u, v) -> {throw new IllegalStateException();}).get();

2022-02-11 05:09:33

我们可以使用RxJava(非常强大的响应式扩展库)

LinkedList<User> users = new LinkedList<>();
users.add(new User(1, "User1"));
users.add(new User(2, "User2"));
users.add(new User(3, "User3"));

User userFound =  Observable.from(users)
                  .filter((user) -> user.getId() == 1)
                  .single().toBlocking().first();

如果没有找到用户或找到多个用户，单个操作符将抛出异常。

2016-01-07 21:40:10

我正在使用这两个收集器:

public static <T> Collector<T, ?, Optional<T>> zeroOrOne() {
    return Collectors.reducing((a, b) -> {
        throw new IllegalStateException("More than one value was returned");
    });
}

public static <T> Collector<T, ?, T> onlyOne() {
    return Collectors.collectingAndThen(zeroOrOne(), Optional::get);
}

2017-05-16 07:15:35

User match = users.stream().filter((user) -> user.getId()== 1).findAny().orElseThrow(()-> new IllegalArgumentException());

2020-03-28 11:38:46

如果你不使用Guava或Kotlin，这里有一个基于@skiwi和@Neuron答案的解决方案。

users.stream().collect(single(user -> user.getId() == 1));

users.stream().collect(optional(user -> user.getId() == 1));

其中single和optional是返回相应收集器的静态导入函数。

我认为，如果将过滤逻辑移到收集器内部，看起来会更简洁。同样，如果您碰巧用.filter删除字符串，代码中也不会中断任何内容。

代码https://gist.github.com/overpas/ccc39b75f17a1c65682c071045c1a079的要点

2020-10-14 12:54:03

将Java流过滤为1，并且只有1个元素

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