我正在尝试制作一个简单的偶数或奇数程序。我希望它一直运行,直到用户进入' q'。但我无法接受' q'作为一个字符串。
import java.util.Scanner;
class EvenOrOdd {
public static void main(String[] args) {
Scanner myScanner = new Scanner(System.in);
System.out.println("Welcome to my program that checks if a number is even or odd.");
while (true) {
System.out.println();
System.out.print("Please type number in a number ['q' to quit]: ");
int number;
String quit;
try {
number = myScanner.nextInt();
} finally {
quit = myScanner.nextLine();
}
if (quit.equals("q")) {
break;
} else if (number % 2 == 0) {
System.out.println(number + " is Even.");
} else {
System.out.println(number + " is Odd.");
}
}
}
}
当我输入数字时,程序运行正常,但当我输入' q'时,控制台会抛出错误:
Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at EvenOrOdd.main(EvenOrOdd.java:19)
我知道这对你们很多人来说可能很容易,但我刚拿起一本java书并试图完成任务。任何帮助将不胜感激!
答案 0 :(得分:1)
从扫描仪中取出一个字符串,检查是否为“q”,如果不是,则将其转换为int,然后检查偶数或奇数。
public static void main(String[] args) {
Scanner myScanner = new Scanner(System.in);
System.out.println("Welcome to my program that checks if a number is even or odd.");
while (true) {
System.out.println();
System.out.print("Please type number in a number ['q' to quit]: ");
String inText = myScanner.next();
if (inText.equals("q")){
break;
}
int number = Integer.valueOf(inText);
if (number % 2 == 0) {
System.out.println(number + " is Even.");
} else {
System.out.println(number + " is Odd.");
}
}
}
答案 1 :(得分:1)
你可以做这样的事情,我发现在这种情况下布尔值更适合循环而不是while(true)和break:
public class EvenOrOdd {
public static void main(String[] args) {
Scanner myScanner = new Scanner(System.in);
System.out
.println("Welcome to my program that checks if a number is even or odd.");
boolean enterLoop = true;
while (enterLoop) {
System.out.println();
System.out.print("Please type number in a number ['q' to quit]: ");
String scannerinput = myScanner.nextLine();
if (scannerinput.equals("q")) {
enterLoop = false;
} else {
checkNumber(scannerinput);
}
}
}
private static void checkNumber(String scannerinput) {
try {
int number = Integer.parseInt(scannerinput);
if (number % 2 == 0) {
System.out.println(number + " is Even.");
} else {
System.out.println(number + " is Odd.");
}
} catch (Exception e) {
System.out.println("No Number!");
}
}
}
答案 2 :(得分:1)
import java.util.Scanner;
class EvenOrOdd {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Welcome to my program that checks if a number is even or odd.");
while (true) {
System.out.print("\nPlease type number in a number ['q' to quit]: ");
String input = scanner.next();
if (input.equals("q")) {
break;
} else {
int number = Integer.parseInt(input);
System.out.print(number + " is ");
System.out.print(number%2 == 0 ? "Even." : "Odd.");
}
}
}
}
那样做。 :)
答案 3 :(得分:0)
不要使用myScanner.nextInt()
,只需使用myScanner.next()
即可获得String
。然后,如果它不是" q",请使用Integer.valueOf(inputString)
获取int
并检查偶数/奇数。
while (true) {
String input = myScanner.next();
if ("q".equals(input)) {
break;
} else {
int number = Integer.valueOf(input);
if (number % 2 == 0) {
System.out.println(number + " is Even.");
} else {
System.out.println(number + " is Odd.");
}
}
}
答案 4 :(得分:0)
这是因为程序期望输入int类型,基本上是程序输出:请在数字中键入数字[' q'退出],之后它会到达myScanner.nextInt();
行并等待输入,因为" q"不是整数,它会引发异常。
快速解决方案是使用myScanner.nextLine()
,然后将字符串转换为整数,除非它等于' q'。像这样:
import java.util.Scanner;
public class EvenOrOdd {
public static void main(String[] args) {
Scanner myScanner = new Scanner(System.in);
System.out.println("Welcome to my program that checks if a number is even or odd.");
while (true) {
System.out.println();
System.out.print("Please type number in a number ['q' to quit]: ");
String string = myScanner.nextLine();
int number = 0;
if (string.equals("q")) {
myScanner.close(); // Close the scanner.
break;
} else if ((number = toInteger(string)) == -1){ // Is the string a number, less than Integer.MAX_VALUE and greater than Integer.MIN_VALUE?
System.out.printf("%s is not a valid integer!%n",string);
} else if (number % 2 == 0) {
System.out.println(number + " is Even.");
} else {
System.out.println(number + " is Odd.");
}
}
}
private static int toInteger(String str){
try{
return Math.abs(Integer.parseInt(str));
}catch(NumberFormatException e){
return -1;
}
}
}
顺便说一句,请务必关闭扫描仪,否则可能会发生资源泄漏。
答案 5 :(得分:0)
尝试这种方法。查看代码注释以获取更多详细信息。
import java.util.Scanner;
class EvenOrOdd {
public static void main(String[] args) {
Scanner myScanner = new Scanner(System.in);
System.out.println("Welcome to my program that checks if a number is even or odd.");
String input=null;
int number;
boolean flag=true; // loop flag
do {
System.out.println();
System.out.print("Please type number in a number ['q' to quit]: ");
// Take user input as String
input=myScanner.nextLine();
try
{
// convert the string value to integer value
number = Integer.parseInt(input);
if (number % 2 == 0)
{
System.out.println(number + " is Even.");
}
else
{
System.out.println(number + " is Odd.");
}
}
catch (NumberFormatException nfe)
{
// control goes here if input is not integer value
if(input.equals("q")) // exist option
flag=false;
else // invalid input
System.out.println("Invalid input, Please enter integer value or (q) to exist");
}
} while (flag);
}
}
答案 6 :(得分:0)
你应该更新你的方法流,你实际上是在尝试传递和整数字符串的验证,所以首先把输入作为String
并检查它是否是你的退出字符 q 然后如果不是尝试用int
将字符串解析为Integer.parseInt(input)
原语并测试它是奇数还是偶数。
如果过程失败,假设它既不是 q 也不是数字(任何其他字符),则会提示用户告知用户使用有效号码或" q"退出:
import java.util.Scanner;
class EvenOrOdd {
public static void main(String[] args) {
Scanner myScanner = new Scanner(System.in);
System.out.println("Welcome to my program that checks if a number is even or odd.");
while (true) {
System.out.println();
System.out.print("Please type number in a number ['q' to quit]: ");
int number;
String input = myScanner.next();
if (input.equals("q")) {
break;
} else {
try {
number = Integer.parseInt(input);
if (number % 2 == 0) {
System.out.println(number + " is Even.");
} else {
System.out.println(number + " is Odd.");
}
} catch (NumberFormatException nfe) {
System.out.println("Enter valid number or \"q\" to quit!");
}
}
}
}
}