我有這個方法可以作為其他方法的菜單：public static void menu(){ boolean works = true; System.out.println("\n1 - Register Car\t2 - Register Client\t3 - Rent car\n4 - Returning car\t5 - Show list\t6 - Exit\n"); System.out.print("Input: "); int operacao = scan.nextInt(); do { if (operacao == 1) { cadVeiculo(); } else if (operacao == 2) { cadCliente(); } else if (operacao == 3) { locar(); } else if (operacao == 4) { devolver(); } else if (operacao == 5) { listar(); } else if (operacao == 6){ works= false; } } while (works);}我不明白的是，當我運行程序并按“6”時，程序停止了，好的。但是當我按“1”注冊汽車然后按“6”退出程序時，程序一直要求我輸入另一個輸入。我所期望的是，當我運行該程序并按“6”退出時，它只會完全停止該程序。
1 回答
拉莫斯之舞
TA貢獻1820條經驗 獲得超10個贊
在do-while循環中移動println和行：scanner.nextInt()
這是一個僅打印出方法名稱的修改示例：
import java.util.Scanner;
class Main {
public static void main(String[] args) {
menu();
}
public static void menu() {
Scanner scan = new Scanner(System.in);
boolean works = true;
do {
System.out.println(
"\n1 - Register Car\t2 - Register Client\t3 - Rent car\n4 - Returning car\t5 - Show list\t6 - Exit\n");
System.out.print("Input: ");
int operacao = scan.nextInt();
if (operacao == 1) {
System.out.println("cadVeiculo();");
} else if (operacao == 2) {
System.out.println("cadCliente();");
} else if (operacao == 3) {
System.out.println("locar();");
} else if (operacao == 4) {
System.out.println("devolver();");
} else if (operacao == 5) {
System.out.println("listar();");
} else if (operacao == 6) {
System.out.println("Goodbye!");
works = false;
}
} while (works);
scan.close();
}
}
用法示例：
1 - Register Car 2 - Register Client 3 - Rent car
4 - Returning car 5 - Show list 6 - Exit
Input: 1
cadVeiculo();
1 - Register Car 2 - Register Client 3 - Rent car
4 - Returning car 5 - Show list 6 - Exit
Input: 3
locar();
1 - Register Car 2 - Register Client 3 - Rent car
4 - Returning car 5 - Show list 6 - Exit
Input: 4
devolver();
1 - Register Car 2 - Register Client 3 - Rent car
4 - Returning car 5 - Show list 6 - Exit
Input: 6
Goodbye!
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