我想使這個程序輸出一個錯誤,當有人進入一個字母作爲菜單選擇輸出,當有人進入一個字母爲菜單選擇了錯誤
package main;
import java.util.Scanner;
public class Main {
Scanner input = new Scanner(System.in);
public static void main(String[] args) {
int UserIn = 0;
double numberInputA;
double numberInputB;
double Answer;
Scanner input = new Scanner(System.in);
while (UserIn != 5) {
System.out.println("What type of calculation would you like to perform?");
System.out.println("1. Add\n2. Subtract\n" +
"3. Multiply\n4. Divide\n5. Exit");
UserIn = input.nextInt();
if (UserIn > 5) {
System.out.println("ERROR choose the numbers on the menu");
} else if (UserIn < 1) {
System.out.println("ERROR choose the numbers on the menu");
}
if (UserIn < 5) {
System.out.println("Enter first number:");
numberInputA = input.nextDouble();
System.out.println("Enter second number");
numberInputB = input.nextDouble();
if (UserIn == 1) {
Answer = numberInputA + numberInputB;
System.out.println("Number 1 is:" + numberInputA);
System.out.println("Number 2 is:" + numberInputB);
System.out.println("Result is : " + Answer);
System.out.println("");
}
if (UserIn == 2) {
Answer = numberInputA - numberInputB;
System.out.println("Number 1:" + numberInputA);
System.out.println("Number 2:" + numberInputB);
System.out.println("Result is :" + Answer);
System.out.println("");
}
if (UserIn == 3) {
Answer = numberInputA * numberInputB;
System.out.println("Number 1:" + numberInputA);
System.out.println("Number 2:" + numberInputB);
System.out.println("Result is :" + Answer);
System.out.println("");
}
if (UserIn == 4) {
Answer = numberInputA/numberInputB;
System.out.println("Number 1:" + numberInputA);
System.out.println("Number 2:" + numberInputB);
System.out.println("Result is :" + Answer);
System.out.println("");
}
}
if (UserIn == 5) {
System.out.println("Program exited");
}
}
}
}
對不起,我們不「請爲我做X」的事情做。只要知道你可以使用(例如)RegEx來檢查一個字符串是否包含一個有效的數字,你可以使用'Integer.parseInt(String)'方法從'String'中獲得一個'int'。 –
處理每個選擇的大部分代碼都完全相同;我會考慮稍微調整一下,以便於閱讀和修改。還要注意,例如,-10 <5,但不是有效的菜單選項。請請始終縮進代碼 - 這有助於大大提高可讀性。 –