Java '；如果'；语句始终返回true

每当我运行程序第69行的

else if（room==6）

语句时，无论room的值是多少，它总是返回

true

import java.util.Scanner;

public class AdventureTwo {
    public static void main(String[] args) {
        Scanner keyboard = new Scanner(System.in);
        String choice = "";
        int room = 1;

        while (room != 0) {
            if (room == 1) {
                System.out.println("You find yourself in a room. There's a \"blue\" door and a \"red\" door.\nWhich do you choose?");
                choice = keyboard.next();

                if (choice.equalsIgnoreCase("red")) {
                    room = 2;
                } else if (choice.equalsIgnoreCase("blue")) {
                    room = 3;
                }
            } else if (room == 2) {
                System.out.println("Going through the red door reveals the \"family\" room.\nYou can also go \"back\". What is your choice?");
                choice = keyboard.next();
                if (choice.equalsIgnoreCase("family")) {
                    room = 4;
                } else if (choice.equalsIgnoreCase("back")) {
                    room = 1;
                }
            } else if (room == 3) {
                System.out.println("You find two new rooms, the \"kitchen\" and the \"bathroom\".\nYou can also go \"back\", What is your choice?");
                choice = keyboard.next();
                if (choice.equalsIgnoreCase("kitchen")) {
                    room = 5;
                } else if (choice.equalsIgnoreCase("bathroom")) {
                    room = 6;
                } else if (choice.equalsIgnoreCase("back")) {
                    room = 1;
                }
            } else if (room == 4) {
                System.out.println("In the family room you see a tv \"remote\" do you pick it up and turn on the tv, or go \"back\"?");
                choice = keyboard.next();
                if (choice.equalsIgnoreCase("remote")) {
                    System.out.println("You pick up the remote and turn on the tv. You are immediately engrossed in the rv program.");
                    System.out.println("In fact you are so interested you stay there watching tv until you die of dehydration");
                    room = 0;
                } else if (choice.equalsIgnoreCase("back")) {
                    room = 2;
                }
            } else if (room == 5) {
                System.out.println("In the kitchen you find some \"crackers\", you are pretty hungry and they don't look that old.\nYou can also go \"back\".");
                if (choice.equalsIgnoreCase("crackers")) {
                    System.out.println("After eating one cracker, you feel the need to have another one, and another , and another, this goes on until you\nhave eaten so much your stomach bursts and you die.");
                    room = 0;
                } else if (choice.equalsIgnoreCase("back")) {
                    room = 3;
                }
            } else if (room == 6) { // line 69
                System.out.println("In the bathroom there is an open \"window\", you can also go \"back\". What is your choice?");
                choice = keyboard.next();
                if (choice.equalsIgnoreCase("window")) {
                    System.out.println("You climb out the window and drop to the ground. You are free!");
                    room = 0;
                } else if (choice.equalsIgnoreCase("back")) {
                    room = 3;
                }
            }
        }
        System.out.print("Game Over.");
    }
}

---

我看到您正在更改某些其他if中的房间值。这是否会影响后续的if块？尝试使用switch…case语句

例如：

while (room != 0) { switch(room) { case 1: { System.out.println("You find yourself in a room. There's a \"blue\" door and a \"red\" door.\nWhich do you choose?"); choice = keyboard.next(); if (choice.equalsIgnoreCase("red")) { room = 2; } else if (choice.equalsIgnoreCase("blue")) { room = 3; } break; } case 2: { //room==2 code break; } case 3: { //etc... break; //don't forget to put the "break" before each case's close bracket } } } while（房间！=0）{ 开关（室）{ 案例1：{ System.out.println（“您发现自己在一个房间里，有一个“蓝色”门和一个“红色”门。\n您选择哪一个？”）； 选择=键盘。下一步（）； if（选择等信号情况（“红色”））{ 房间=2间； }else if（选择.相等信号情况（“蓝色”））{ 房间=3间； } 打破 } 案例2：{ //房间==2码 打破 } 案例3：{ //等等。。。 break；//不要忘记在每个案例的右括号前加上“break” } } }

---

hth:-）

您可以创建单独的方法来处理每个房间中发生的事情，然后创建这些方法的数组或ArrayList（例如，称为“roomMethod”）。每个方法都返回用户选择的下一个房间。那么主要代码是：

  while (room != 0 ) {
    room = roomMethod[room]();
  } 

---

第69行是哪一行？它的

if/else if

hell…哇，饼干消费导致的死亡。太残酷了。你错过了

choice=keyboard.next（）在房间5。@Joormatt作为旁注，我建议使用
switch/case
而不是这么多
if/else if
语句（请参阅）。我还没有谈到数组，但当我谈到数组时，我会回顾一下，thanks@Joormatt这在java中是行不通的。FredK很可能习惯于用javascript编写代码。就这样做了，它帮助简化了代码，并解决了intellij的问题，尽管它的
case n:
不是
case:n