为什么我的 java 程序跳过我的输入并挂在 IF 语句上？

Question

除了这个单一方法，我的程序中的其他所有东西都可以工作，它跳过 sc.next() 然后挂在我的 if 上。

public static String getChoice() {
    boolean choiceSelector = false;

    // construct a Scanner object for getting input
    Scanner sc = new Scanner(System.in);

    System.out.println("Do you wish to continue? Y/N");
    String choice = " ";

    do{
    choice = sc.next();

    if(choice.equalsIgnoreCase("Y")){
        choice="Y";
        choiceSelector = true;
    }
    if(choice.equalsIgnoreCase("N")){
        choice="N";
        choiceSelector = true;
    } else
        choice = sc.next("Please enter a correct value.");

    } while(choiceSelector == false);

    return choice;
}//end getChoice

Answer 1

可能放另一个else ，因为即使输入最初也是Y ，您仍然会要求他们提供正确的值：

if(choice.equalsIgnoreCase("Y")){
    choice="Y";
    choiceSelector = true;
} else if(choice.equalsIgnoreCase("N"))...

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和休息

choice = sc.next("Please enter a correct value."); // not a valid pattern provided as an input here

作为

System.out.println("Please enter a correct value.");
choice = sc.next();

---

和

return choiceSelector; // the boolean value