问题描述
如果有列表< Person> ，是否有可能获得所有的列表person.getName（）出于那个？
是否有准备好的电话，或者我必须写一个foreach循环，如：
When there is an List<Person>, is there a possibility of getting List of all person.getName() out of that?Is there an prepared call for that, or do I have to write an foreach loop like:
List<Person> personList = new ArrayList<Person>();
List<String> namesList = new ArrayList<String>();
for(Person person : personList){
namesList.add(personList.getName());
}
推荐答案
Java 8和上面：
List<String> namesList = personList.stream()
.map(Person::getName)
.collect(Collectors.toList());
如果您需要确保获得 ArrayList 因此，您必须将最后一行更改为：
If you need to make sure you get an ArrayList as a result, you have to change the last line to:
...
.collect(Collectors.toCollection(ArrayList::new));
Java 7及以下版本：
Java 8之前的标准集合API不支持此类转换。你必须编写一个循环（或将它包装在你自己的某个map函数中），除非你转向一些更高级的集合API /扩展。
The standard collection API prior to Java 8 has no support for such transformation. You'll have to write a loop (or wrap it in some "map" function of your own), unless you turn to some fancier collection API / extension.
（ Java代码段中的行正好是我要使用的行。）
(The lines in your Java snippet are exactly the lines I would use.)
在Apache Commons中，您可以使用和
In Apache Commons, you could use CollectionUtils.collect and a Transformer
在Guava中，您可以使用方法。
In Guava, you could use the Lists.transform method.
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