Question

我制作了一个程序，要求用户输入一个数字，然后检查它是否等于，小于或大于随机数。一个siple猜谜游戏。它工作得很好，但我想在输入的同一行显示输出，但我无法弄清楚如何做到这一点。

Ok, I've picked a number between 1 and 100. What is your first guess?
Guess 1: 36
<
Guess 2: 50
>
Guess 3: 46
That is correct. The hidden number is 46.

这就是我希望它的样子：

Guess 1: 36 <
Guess 2: 50 >
Guess 3: 46 That is correct. The hidden number is 46.

我该怎么做？当我使用Scanner.nextInt();

时，光标在输入后向下移动

这是我目前的代码：

while (!isCorrect) {
        System.out.print("Guess " + guessCount + ":  ");
        currentGuess = input.nextInt();

        if (currentGuess == randomNumber) {
            isCorrect = true;
            System.out.println("That is correct. The hidden number was " + randomNumber);
        }
        else if (currentGuess > randomNumber) {
            System.out.println(" >");
            ++guessCount;
        }
        else if (currentGuess < randomNumber) {
            System.out.println(" <");
            ++guessCount;
        }

    }

Answer 1

这不是一贯可能的。您可以尝试回显\b退格符，但这可能不起作用。一旦用户键入新行，就无法将其删除。

Answer 2

你能够完成的最好的事情就是重新回应输入。

while (!isCorrect) {
    System.out.print("Guess " + guessCount + ":  ");
    currentGuess = input.nextInt();

    if (currentGuess == randomNumber) {
        isCorrect = true;
        System.out.println("That is correct. The hidden number was " + randomNumber);
    }
    else if (currentGuess > randomNumber) {
        System.out.println(currentGuess  + " >");
        ++guessCount;
    }
    else if (currentGuess < randomNumber) {
        System.out.println(currentGuess  + " <");
        ++guessCount;
    }

}

输出：

Guess 1: 36 
36 <
Guess 2: 50
50 >
Guess 3: 46 
That is correct. The hidden number is 46.

Answer 3

如果您确实需要格式化输出，可以尝试将输入保存到arraylist，清除屏幕，然后再将所有先前输入打印到屏幕。

List<Integer> foo = new ArrayList<Integer>();

while (!isCorrect) {
    System.out.print("Guess " + guessCount + ": ");
    guessCount++;
    currentGuess = input.nextInt();
    foo.add(currentGuess);
    for(int a = 0; a < 50; a++) {
        System.out.println('\n');
    }
    for(int i = 0; i < foo.size(); i++) {
        currentGuess = foo.get(i);
        System.out.print("Guess " + (i+1) + ": " + currentGuess);
        if (currentGuess == randomNumber) {
            isCorrect = true;
            System.out.println(" That is correct. The hidden number was " + randomNumber);
        }
        else if (currentGuess > randomNumber) {
            System.out.println(" >");
            }
        else if (currentGuess < randomNumber) {
            System.out.println(" <");
        }
    }
}

要清除屏幕，我刚打印出50个空行。

此处提出了清除控制台的解决方案： Java: Clear the console 但是，它对我或克里斯不起作用。

输入后显示输出

3 个答案: