我有一个由两种方法组成的简单程序。第一个是菜单,询问玩家是想提出建议还是退出游戏。如果他们选择提出建议,则调用第二种方法。第二种方法是打印表单供用户填写。所以假设看起来像这样:
Person:
Weapon:
Room: Library
已为用户填写了库。其余的可以填写如下:
Person: Professor Plum
Weapon: Revolver
Room: Library
问题在于打印出的内容看起来不像我想要的。打印出来的内容如下:
Person: Weapon:
人和武器印在同一条线上,用户只能填写武器。此外,在Weapons填满之前,Room不打印。无论如何,我可以让它看起来像我在上一个例子中列出的方式吗?
import java.util.Scanner;
public class Main {
private String[] suggestions = new String[3];
private Scanner sc = new Scanner(System.in);
private String room = "Library";
public void menuSelection() {
System.out.println("Please make a selection...\n");
System.out.println("1. Make a suggestion");
System.out.println("2. Exit game");
int selection = sc.nextInt();
if (selection == 1)
makeSuggestion();
else
System.exit(0);
}
public void makeSuggestion() {
System.out.print("Person: ");
suggestions[0] = sc.nextLine();
System.out.print("Weapon: ");
suggestions[1] = sc.nextLine();
System.out.println("Room: " + room);
suggestions[2] = room;
}
public static void main(String[] args) {
Main main = new Main();
main.menuSelection();
}
}
答案 0 :(得分:1)
您没有打印值。使用System.out.println()
打印这些值。
System.out.print("Person: ");
suggestions[0] = sc.nextLine();
System.out.println(suggestions[0]);
System.out.print("Weapon: ");
suggestions[1] = sc.nextLine();
System.out.println(suggestions[1]);
System.out.print("Room: " + room);
suggestions[2] = room;
System.out.println(suggestions[2]);
<强>输出:强>
Person: Professor Plum
Weapon: Revolver
Room: Library
答案 1 :(得分:1)
在sc.nextLine();
int selection = sc.nextInt();
这将给出欲望输出
答案 2 :(得分:0)
System.out.println将在System.out.print将打印的新行上打印并且在同一行。您应该使用System.out.println()而不是System.out.print()
答案 3 :(得分:0)
这就是造成这个问题的原因:
int selection = sc.nextInt();
例如用sc.nextLine()替换它。