在Java 8中，你可以这样做，将Iterable中的所有元素添加到Collection中并返回:

public static <T> Collection<T> iterableToCollection(Iterable<T> iterable) {
  Collection<T> collection = new ArrayList<>();
  iterable.forEach(collection::add);
  return collection;
}

灵感来自@Afreys的回答。

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();

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.

我经常使用fluentiatable .from(myIterable). tolist()。

我在试图获取项目列表时遇到了类似的情况，而不是在CrudRepository接口中声明的默认Iterable<T> findAll()。因此，在我的ProjectRepository接口(从CrudRepository扩展而来)中，我简单地声明了findAll()方法来返回一个List<Project>而不是Iterable<Project>。

package com.example.projectmanagement.dao;

import com.example.projectmanagement.entities.Project;
import org.springframework.data.repository.CrudRepository;
import java.util.List;

public interface ProjectRepository extends CrudRepository<Project, Long> {

    @Override
    List<Project> findAll();
}

我认为这是最简单的解决方案，不需要转换逻辑或使用外部库。

您可以使用Eclipse Collections工厂:

Iterable<String> iterable = Arrays.asList("1", "2", "3");

MutableList<String> list = Lists.mutable.withAll(iterable);
MutableSet<String> set = Sets.mutable.withAll(iterable);
MutableSortedSet<String> sortedSet = SortedSets.mutable.withAll(iterable);
MutableBag<String> bag = Bags.mutable.withAll(iterable);
MutableSortedBag<String> sortedBag = SortedBags.mutable.withAll(iterable);

您还可以将Iterable转换为LazyIterable，并使用转换器方法或任何其他可用的api。

Iterable<String> iterable = Arrays.asList("1", "2", "3");
LazyIterable<String> lazy = LazyIterate.adapt(iterable);

MutableList<String> list = lazy.toList();
MutableSet<String> set = lazy.toSet();
MutableSortedSet<String> sortedSet = lazy.toSortedSet();
MutableBag<String> bag = lazy.toBag();
MutableSortedBag<String> sortedBag = lazy.toSortedBag();

以上所有可变类型都扩展了java.util.Collection。

注意:我是Eclipse Collections的提交者。