所以我试图让用户输入他们的答案作为布尔值,然后进行错误检查,但这似乎并没有起作用。我是在正确的轨道上还是有更好的方法来做到这一点?我曾经考虑过在输入是一个字符串的地方做它,并且只是将它与我指定的其他两个字符串进行检查" true"并且" false,"但我的while循环:
while (!continueGame.equalsIgnoreCase(answerTrue) && !continueGame.equalsIgnoreCase(answerFalse)) {
System.out.println("Would you like to try again? (true/false)");
continueGame = keyboard.nextBoolean();
System.out.println();
}
也没有工作。我相当肯定它与我的不相关,但我不确定为什么。无论如何,下面是我使用布尔值而不是字符串进行错误检查的方法。字符串版本基本上是相同的方式,但只是为字符串修改。
public static boolean continueGame() {
boolean continueGame;
System.out.println("Would you like to try again? (true/false)\n");
continueGame = keyboard.nextBoolean();
System.out.println();
while (continueGame != true && continueGame != false) {
System.out.println("Would you like to try again? (true/false)");
continueGame = keyboard.nextBoolean();
System.out.println();
}
if (continueGame) {
return true;
}
else {
System.out.println("We accept your surrender.");
return false;
}
} //End continueGame method
答案 0 :(得分:0)
基本上我只是交换循环的工作方式,因此可以达到破坏条件。
import java.*;
import java.util.Scanner;
public class Loop
{
public static void main(String[] args)
{
boolean continueGame=true;
Scanner keyboard = new Scanner(System.in);
System.out.println("Would you like to try again? (true/false)\n");
continueGame = keyboard.nextBoolean();
while (continueGame) {
System.out.println("Would you like to try again? (true/false)");
continueGame = keyboard.nextBoolean();
if(!continueGame)
{
System.out.println("We accept your surrender.");
}
}
}
}
答案 1 :(得分:0)
while(true) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Would you like to try again? (true/false)");
if(keyboard.hasNextBoolean()){
continueGame = keyboard.nextBoolean();
System.out.println();
break;
}
}
这将无限期地提示用户输入true / false,直到他们输入有效的布尔值。然后它将突破循环,您可以继续使用其余的代码。