我正在尝试构建一个简单的程序,该程序将继续执行,直到满足某个条件。在这种情况下,无论是真还是假。我已经玩了一段时间,但我仍然无法按照我的意愿去工作。
import java.util.Scanner;
public class oddeven {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter a number: ");
int num = scan.nextInt();
boolean play = true;
String restart = " ";
do {
if((num % 2) == 0) {
System.out.println("That number is even!");
} else {
System.out.println("That number is odd!");
}
System.out.println(
"Would you like to pick another number? (Type 'yes' or 'no')");
restart = scan.nextLine();
if(restart == "yes") {
System.out.println("Enter a number: ");
play = true;
} else {
System.out.println("Thanks for playing!");
play = false;
}
}
while(play == true);
}
}
答案 0 :(得分:0)
以下是您要执行的操作的代码。你一直在做3到4个错误,看看代码,你就会明白。
你还应该看到这个链接
What's the difference between next() and nextLine() methods from Scanner class?
import java.util.Scanner;
class test {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
boolean play = true;
do {
System.out.println("Enter a number: ");
int num = Integer.valueOf(scan.next());
String restart = " ";
if ((num % 2) == 0) {
System.out.println("That number is even!");
}
else {
System.out.println("That number is odd!");
}
System.out.println("Would you like to pick another number? (Type 'yes' or 'no')");
restart = scan.next();
if (restart.equalsIgnoreCase("yes")) {
play = true;
}
else {
System.out.println("Thanks for playing!");
play = false;
}
} while (play == true);
}
}