只要你调用contains, containsAll, equals, hashCode, remove, retainAll, size或toArray，你就必须遍历元素。

如果您偶尔只调用isEmpty或clear等方法，我认为您最好是惰性地创建集合。例如，你可以有一个支持数组列表来存储先前迭代的元素。

我不知道任何一个库中有这样的类，但是编写它应该是一个相当简单的练习。

从CollectionUtils:

List<T> targetCollection = new ArrayList<T>();
CollectionUtils.addAll(targetCollection, iterable.iterator())

下面是这个实用方法的完整源代码:

public static <T> void addAll(Collection<T> collection, Iterator<T> iterator) {
    while (iterator.hasNext()) {
        collection.add(iterator.next());
    }
}

在JDK 8+中，不使用任何额外的库:

Iterator<T> source = ...;
List<T> target = new ArrayList<>();
source.forEachRemaining(target::add);

编辑:上面的是迭代器。如果你在处理Iterable，

iterable.forEach(target::add);

两个评论

There is no need to convert Iterable to Collection to use foreach loop - Iterable may be used in such loop directly, there is no syntactical difference, so I hardly understand why the original question was asked at all. Suggested way to convert Iterable to Collection is unsafe (the same relates to CollectionUtils) - there is no guarantee that subsequent calls to the next() method return different object instances. Moreover, this concern is not pure theoretical. E.g. Iterable implementation used to pass values to a reduce method of Hadoop Reducer always returns the same value instance, just with different field values. So if you apply makeCollection from above (or CollectionUtils.addAll(Iterator)) you will end up with a collection with all identical elements.

如果您可以更新到Spring Data 3，那么这个问题已经解决了。有一个新的界面ListCrudRepository，它做的正是你想要的。

这是来自https://spring.io/blog/2022/02/22/announcing-listcrudrepository-friends-for-spring-data-3-0:的界面

public interface ListCrudRepository<T, ID> extends CrudRepository<T, ID> {
    <S extends T> List<S> saveAll(Iterable<S> entities);
    List<T> findAll();
    List<T> findAllById(Iterable<ID> ids);
}

注意，在版本3中必须实现两个接口

所以在版本2中

public interface PersonRepository<Person, Long> extends 
   PagingAndSortingRepository<Person, Long> {}

在版本3中应改为:

public interface PersonRepository<Person, Long> extends
    PagingAndSortingRepository<Person, Long>,ListCrudRepository<Person, Long> {}

其他变化见https://spring.io/blog/2022/02/22/announcing-listcrudrepository-friends-for-spring-data-3-0

因为RxJava是一个锤子，而这个看起来像钉子，你可以这样做

Observable.from(iterable).toList().toBlocking().single();