Java流迭代列表最有效的方法

Question

给定一个

 List<User> listUser; //Userobject contains fields userName,userId, forName, last Name

是否可以不用写 5 行来输出用户字段值

listUser.get(0).getUserName

listUser.get(0).getUserId

listUser.get(0).getforName and so on 

所以输出应该是

userName -> Hans

userId -> 1

forName -> foo

lastName -> bar

只需一行代码而无需编写 5 行？即通过流或其他方式？

listUser.stream().map(s->s.getUserName) //same as above?

Answer 1

一种解决方案是toString为具有五行代码的类编写一个方法：public String toString() {&nbsp; &nbsp; return "User name: " + getUserName() + "\n" +&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;"User ID: " + getUserID() + "\n" +&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;"Form name: " + getFormName() + "\n" +&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;//...&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;;}然后Stream您可以简单地执行以下操作：listUser.forEach(System.out::println);