一个非常简单的待办事项列表，它要求输入，然后以ArrayList的形式打印出该列表，该列表分为多个部分（子列表）（我的视野很差，所以我必须使用大字体，当list太长了，问题是列表的末尾不在页面上。尽我所能使用home / end按钮快速查看页面，这不是最佳情况，我宁愿破坏ArrayList进入子列表，然后打印出子列表，每行一个，如下所示：
这是今天的待办事项清单： [醒来，Walk狗，吃早餐] [铺床，向后扫一下，学习Java]
import java.util.Scanner; 导入java.util.ArrayList;
/**
* @author Troy
*
*/
public class HelloWorld {
public static void main(String[] args) {
// I chose an ArrayList because the size does not have to be predetermined.
ArrayList<String> to_do = new<String>ArrayList();
System.out.println("What would you like to add to your to-do list?");
Scanner user_input = new Scanner(System.in);
//While the user_input still has entries, perform the following:
while (user_input.hasNextLine()) {
//Add next entry in the to-do list(user_input) to the ArrayList
String input = user_input.nextLine();
//If input = remove, remove the last item in the to_do list.(ArrayList)
if ("remove".equals(input)) {
if (to_do.size() > 0) {
to_do.remove(to_do.size() -1);
}}
/**If the user types in "exit", when prompted for the next item in their
* to_do list, close user_input, and print out...
*/
if ("exit".equals(input)) {
user_input.close();
System.out.println("Your to-do list is complete!");
ArrayList<String> sect1 = new ArrayList<String>(to_do.subList(0, to_do.size()));
if (to_do.size() <= 5) {
System.out.println(sect1 + "\n");
break;
}
ArrayList<String> sect2 = new ArrayList<String>(to_do.subList(6, to_do.size()));
if (to_do.size() > 5 && to_do.size() <=10) {
System.out.println(sect1 + "\n" + sect2);
break;
}
//If input does NOT equal "remove", add user_input to the to_do list.
if (!"remove".equals(input)) {
to_do.add(input);
}
System.out.println("\n");
/**Print the ArrayList called "to_do" split into sections AFTER writing,
* "Here is today's to-do list:"
* */
System.out.println("Here is today's to-do list: " + "\n");
if (to_do.size() <= 5) {
System.out.println(sect1 + "\n");
}
if (to_do.size() > 5 && to_do.size() <=10) {
System.out.println(sect1 + "\n" + sect2);
}
}
}
}}
答案 0 :(得分：0)
问题是这里的括号位置。行if (!"remove".equals(input)) {在if ("exit".equals(input)) {块内。我移动了if语句：
导入java.util.ArrayList;
导入java.util.Scanner;
public class HelloWorld {
public static void main(String[] args) {
// I chose an ArrayList because the size does not have to be predetermined.
ArrayList<String> to_do = new<String>ArrayList();
System.out.println("What would you like to add to your to-do list?");
Scanner user_input = new Scanner(System.in);
//While the user_input still has entries, perform the following:
while (user_input.hasNextLine()) {
//Add next entry in the to-do list(user_input) to the ArrayList
String input = user_input.nextLine();
//If input = remove, remove the last item in the to_do list.(ArrayList)
if ("remove".equals(input)) {
if (to_do.size() > 0) {
to_do.remove(to_do.size() -1);
}}
/**If the user types in "exit", when prompted for the next item in their
* to_do list, close user_input, and print out...
*/
if ("exit".equals(input)) {
user_input.close();
System.out.println("Your to-do list is complete!");
ArrayList<String> sect1 = new ArrayList<String>(to_do.subList(0, to_do.size()));
if (to_do.size() <= 5) {
System.out.println(sect1 + "\n");
break;
}
ArrayList<String> sect2 = new ArrayList<String>(to_do.subList(6, to_do.size()));
if (to_do.size() > 5 && to_do.size() <=10) {
System.out.println(sect1 + "\n" + sect2);
break;
}
//If input does NOT equal "remove", add user_input to the to_do list.
if (!"remove".equals(input) && !"exit".equals(input)) {
to_do.add(input);
}
System.out.println("\n");
/**Print the ArrayList called "to_do" split into sections AFTER writing,
* "Here is today's to-do list:"
* */
System.out.println("Here is today's to-do list: " + "\n");
if (to_do.size() <= 5) {
System.out.println(sect1 + "\n");
}
if (to_do.size() > 5 && to_do.size() <=10) {
System.out.println(sect1 + "\n" + sect2);
}
}
}
}}
答案 1 :(得分：0)
正如其他发布者所述，您的代码存在问题是if块的错误嵌套。这会使您的to_do.add位于if ("exit".equals(input))块中，因此您的列表仍然为空。我建议使用IDE并让其重新缩进（格式化）您的代码，然后这个问题将变得更加明显。
但是除此之外，您的代码中还有另一个问题：您的sect1占用了subList(0, to_do.size())，这是您的整个列表。这将导致它在一行上打印整个列表，这就是您所看到的。我建议您改为使用循环并将列表以相等的方式划分。由于subList已返回列表，因此您也不必将其包装在另一个ArrayList中，并且可以直接打印。
所以我将您的代码更正为：
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
/**
* @author Troy
*/
public class HelloWorld {
public static void main(String[] args) {
// I chose an ArrayList because the size does not have to be predetermined.
List<String> toDo = new ArrayList<String>();
System.out.println("What would you like to add to your to-do list?");
Scanner userInput = new Scanner(System.in);
// While the userInput still has entries, perform the following:
while (userInput.hasNextLine()) {
// Get the next line entered by the user
String input = userInput.nextLine();
//If input is "remove", remove the last item in the toDo list. (ArrayList)
if ("remove".equals(input)) {
if (toDo.size() > 0) {
toDo.remove(toDo.size() -1);
}
}
/*
* If the user types in "exit", when prompted for the next item in their
* toDo list, close userInput, and print out...
*/
else if ("exit".equals(input)) {
userInput.close();
System.out.println("Your to-do list is complete!");
System.out.println("Here is today's to-do list: ");
final int perLine = 3;
int i = 0;
while(i < toDo.size()) {
// Print from the start of our current chunk (i)
// to the end (i+3), or to the size of the list if our last chunk is smaller than "perLine".
System.out.println(
toDo.subList(i, Math.min(toDo.size(), i+perLine))
);
i+=perLine;
}
break;
}
/*
* If input is neither "remove" nor "exit", add input to the list
*/
else {
toDo.add(input);
}
}
}
}
我也将一些变量更改为camelCase而不是snake_case，就像Java中的惯例一样。