Question

任务是询问一些随机卡片的定义。引入卡并访问此方法后，对应于密钥的值存在，并且仍返回null。

pair.get（a）始终显示为空

  public static void ask() {
        System.out.println("How many times to ask?");
        int ask = scan.nextInt();
        scan.nextLine();

        Random random = new Random();
        Object[] values = pair.keySet().toArray();
        int retur = random.nextInt(values.length);
        int i = 0;
        for (String iterator : pair.keySet()) {
            if (i <= ask) {
                System.out.println("Print the definition of \"" + values[retur] + "\":");
                String a = scan.nextLine();
                System.out.println(a.equals(pair.get(values[retur])) ? "Correct answer." :
                        "Wrong answer. The correct one is \"" + pair.get(values[retur]) +
                                "\", you've just written the definition of \"" + pair.get(a)  + "\".");
            }else
                break;
        }

Answer 1

如果我正确理解了您的代码，那么这里的问题是，您正在尝试使用另一个值pair.get(a)来通过a检索一个值（由于依赖于用户输入，该值可能甚至不存在！）。

假设您仍想实现此功能，则需要遵循以下原则。

// Get the key referenced by a (if exists)
var aKey = pair.entrySet()
               .stream()
               .filter(entry -> a.equals(entry.getValue()))
               .map(Map.Entry::getKey)
               .findFirst();

// If the key for value a does not exist, print incorrect input (you can handle this however you like), otherwise print original statement
if (aKey.isEmpty()) {
    System.out.println("Incorrect input!");
} else {
    System.out.println(a.equals(pair.get(values[retur])) ? "Correct answer." :
                        "Wrong answer. The correct one is \"" + pair.get(values[retur]) +
                                "\", you've just written the definition of \"" + pair.get(aKey.get())  + "\".");
}

LinkedHashMap get（key）即使存在该值也返回null

1 个答案: