java - 扫描仪 nextLine() 在 nextInt() 之后被跳过

Question

我正在尝试制作一个循环的计算器程序，用户可以在其中输入一个数字，然后输入他希望对该数字执行的操作，直到他输入“=”作为运算符。包含结果的变量在类中初始化为零，并且应用的默认运算符是“+”。

import java.util.Scanner;

public class main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        Calculator c = new Calculator();
        boolean flow = true;

        while(flow) {
            System.out.println("Number :");
            int userEntry = scan.nextInt();
            System.out.println("Operator :");
            String operation = scan.nextLine();

            switch(operation) {
            case "+":
                c.setOperation(Calculator.ADDITION);
                break;
            case "-":
                c.setOperation(Calculator.SOUSTRACTION);
                break;
            case "*":
                c.setOperation(Calculator.MULTIPLICATION);
                break;
            case "=":
                c.getResult();
                return;
            default:
                System.out.println("Please enter a valid operator.");
            }
            c.apply(userEntry);
            c.getResult();
        }
    }
}

但是每次我尝试运行该程序时，我都会得到这个结果

Number :
4
Operator :
Please enter a valid operator.
Number :
67
Operator :
Please enter a valid operator.
Number :

程序不允许我将输入放入运算符中并直接跳到下一个 int 输入。我一直在尝试各种方法来编写此内容，例如将该部分从循环中取出，但它仍然是相同的错误，我看不出导致此问题的原因。任何帮助将不胜感激。

Answer 1

/** Scanning problems */
class Scanning {
    public static void main(String[] args) {
        int num;
        String txt;
        Scanner scanner = new Scanner(System.in);

        // Problem: nextLine() will read previous nextInt()
        num = scanner.nextInt();
        txt = scanner.nextLine();

        // solution #1: read full line and convert it to integer
        num = Integer.parseInt(scanner.nextLine());
        txt = scanner.nextLine();

        // solution #2: consume newline left-over
        num = scanner.nextInt();
        scanner.nextLine();
        txt = scanner.nextLine();
    }
}