在 Java 8 流中按属性排序

Sorting by property in Java 8 stream

提问by Garret Wilson

Oh, those tricky Java 8 streams with lambdas. They are very powerful, yet the intricacies take a bit to wrap one's header around it all.

哦，那些带有 lambda 表达式的棘手 Java 8 流。它们非常强大，但复杂的东西需要一点时间才能将一个人的标题包裹起来。

Let's say I have a Usertype with a property User.getName(). Let's say I have a map of those users Map<String, User>associated with names (login usernames, for example). Let's further say I have an instance of a comparator UserNameComparator.INSTANCEto sort usernames (perhaps with fancy collators and such).

假设我有一个User带有属性的类型User.getName()。假设我有一张Map<String, User>与姓名（例如，登录用户名）相关联的用户的地图。让我们进一步说我有一个比较器的实例UserNameComparator.INSTANCE来对用户名进行排序（也许有花哨的校对者等）。

So how do I get a list of the users in the map, sorted by username? I can ignore the map keys and do this:

那么如何获取地图中的用户列表，按用户名排序？我可以忽略地图键并执行以下操作：

return userMap.values()
    .stream()
    .sorted((u1, u2) -> {
      return UserNameComparator.INSTANCE.compare(u1.getName(), u2.getName());
    })
    .collect(Collectors.toList());

But that line where I have to extract the name to use the UserNameComparator.INSTANCEseems like too much manual work. Is there any way I can simply supply User::getNameas some mapping function, just for the sorting, and still get the Userinstances back in the collected list?

但是我必须提取名称以使用的那一行UserNameComparator.INSTANCE似乎太多的手动工作。有什么方法可以简单地提供User::getName一些映射函数，仅用于排序，并且仍然将User实例返回到收集列表中？

Bonus: What if the thing I wanted to sort on were two levels deep, such as User.getProfile().getUsername()?

奖励：如果我想排序的东西有两个级别深，例如User.getProfile().getUsername()?

采纳答案by Misha

What you want is Comparator#comparing:

你想要的是Comparator#comparing：

userMap.values().stream()
    .sorted(Comparator.comparing(User::getName, UserNameComparator.INSTANCE))
    .collect(Collectors.toList());

For the second part of your question, you would just use

对于问题的第二部分，您只需使用

Comparator.comparing(
    u->u.getProfile().getUsername(), 
    UserNameComparator.INSTANCE
)