我可以用 Java 中的 List 以哪种方式写出 for 循环？

Question

是否可以使用 Lambda 表达式或更好的方法来写下 for 循环？

public TaskDTO convertToDTO(Task task) {

    for (int i = 0; i < task.getPrecedingTasks().size(); i++)

        this.precedingTasks.add(task.getPrecedingTasks().get(i).getName());

}

Answer 1

您的解决方案很好：task.getPrecedingTasks().stream().map(Task::getName).forEach(this.precedingTasks::add);但由于您只是检索 , 的一部分Task，map然后collect作为列表：this.precedingTasks&nbsp;=&nbsp;task.getPrecedingTasks().stream().map(Task::getName).collect(Collectors.toList());是不是更简单易懂？因为stream这里是做映射/转换然后收集。而且通过这种方式，您不需要对this.precedingTasksas进行初始化this.precedingTasks&nbsp;=&nbsp;new&nbsp;ArrayList<>();&nbsp;//&nbsp;to&nbsp;ensure&nbsp;it's&nbsp;not&nbsp;null;无论如何，这里只是个人喜好。

Answer 2

这是一个完整的例子，我把 System.out.println ... 你应该使用 this.precedingTasks.addAll( ...import java.util.ArrayList;import java.util.Arrays;import java.util.List;import java.util.stream.Collectors;public class Main {&nbsp; &nbsp; public static void main(String[] args) {&nbsp; &nbsp; &nbsp; &nbsp; Task t1 = new Task("myTask", Arrays.asList(new Task("innerTask1"), new Task("innerTask2")));&nbsp; &nbsp; &nbsp; &nbsp; System.out.println(t1.precedingTasks.stream().map(Task::getName).collect(Collectors.toList()));&nbsp; &nbsp; }&nbsp; &nbsp; static class Task {&nbsp; &nbsp; &nbsp; &nbsp; private String name;&nbsp; &nbsp; &nbsp; &nbsp; private List<Task> precedingTasks = new ArrayList<>();&nbsp; &nbsp; &nbsp; &nbsp; public Task(String name) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; this.name = name;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; public Task(String name, List<Task> precedingTasks) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; this.name = name;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; this.precedingTasks = precedingTasks;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; public String getName() {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return name;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; public List<Task> getPrecedingTasks() {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return precedingTasks;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}输出是[innerTask1, innerTask2]

Answer 3

我在我的案例中找到了正确的解决方案：task.getPrecedingTasks().stream().map(Task::getName).forEach(this.precedingTasks::add);谢谢，提示:)