Question

我试图将键盘输入变成整数，但是我的程序不断崩溃。当输入了诸如“ k”之类的字符时，它可以工作，但是当我输入“ 5”时，则崩溃。关于我在做什么错的任何想法吗？

// Getting an integer value.
public static int getInt() {
    int numberEntered = 0;
    String entry = "";
    Scanner keyboard = new Scanner(System.in);
    while (!keyboard.hasNextInt()) {
        entry = keyboard.next();
        System.out.println("That is not an integer.  " + "Please try again.");
    }
    numberEntered = Integer.parseInt(entry);
    System.out.print(numberEntered);
    return numberEntered;
}

输出：

Error given: k That is not an integer. 
Please try again. 
8 
Exception in thread "main" java.lang.NumberFormatException: For input string: "k" at 
  java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) 
    at java.base/java.lang.Integer.parseInt(Integer.java:652) 
    at java.base/java.lang.Integer.parseInt(Integer.java:770) 
    at Program2.getInt(Program2.java:56) 
    at Program2.problemSelectionMenu(Program2.java:40) 
    at Program2.main(Program2.java:14)

Answer 1

您检查以确保输入具有下一个int，但是一旦Scanner具有下一个int时，就永远不会将int解析为entry，因此它仍然是错误的输入。您需要将用户输入的int分配给entry。您只需拨打nextInt()：

即可轻松完成此操作

while (!keyboard.hasNextInt()) {
    entry = keyboard.next();
    System.out.println("That is not an integer.  " + "Please try again.");
}
numberEntered = keyboard.nextInt();
System.out.print(numberEntered);
return numberEntered;

Answer 2

当您提供字符作为输入时，虽然循环条件为true，所以它进入while循环并扫描字符和字符的打印值，但是当您提供整数时，in while循环条件变为false并且不会进入while循环。在while循环之外，您正在解析整数，您不需要这样做，因为您将整数作为输入。您所要做的就是在 integer.parseInt（entry），您必须扫描整数，即 int我= keyboard.nextInt（）; 因为在while循环条件下，您仅检查给定的输入是否为整数。但是如果是整数，您还没有扫描输入。

尝试一下！

int numberEntered = 0;
    String entry = "";
    Scanner keyboard = new Scanner(System.in);

    while (!keyboard.hasNextInt()) {
         entry = keyboard.next();

        System.out.println("That is not an integer.  " +"Please try again.");
    }
    numberEntered = keyboard.nextInt();
    System.out.print(numberEntered);

键盘输入到整数java

2 个答案: