扫描仪：扫描整数或字符串

Question

尝试使用 Scanner 获取 int 或 string.. 下面的代码是我所拥有的：

System.out.println("where would you like to send the following item: ");
System.out.println("1. groceries, 2. savings, 3. utilities");
System.out.print("Enter list name or list number to send: ");
String user = user_input.nextLine();
      if (user.toLowerCase().equals("groceries") || Integer.parseInt(user) == 1) {
             System.out.println("item was sent to the grocery list");

      }else if(user.toLowerCase().equals("savings") || Integer.parseInt(user) == 2) {
             System.out.println("item was sent to the savings list");

      }else if(user.toLowerCase().contains("utilities") || Integer.parseInt(user) == 3) {
             System.out.println("item was sent to the utilities list");

      }else{
             System.out.println("item is not in any category");

      }

我的问题是，每当我输入字符串时，都会收到 NumberFormatException 错误。键入 int 不会给我带来任何问题。我认为使用 String user = user_input.nextLine(); 获取输入并使用 Integer.parseInt(user) 在 if 语句中将数字字符串转换为 int 可以解决问题，但它仍然给我错误消息。

非常感谢任何帮助...

Answer 1

您可以尝试使用 user.equals("1") 而不是 Integer.parseInt(user) == 1。或者假设异常 java.lang.NumberFormatException 有效并捕获该异常并以正确的方式处理它，例如：

    System.out.println("where would you like to send the following item: ");
    System.out.println("1. groceries, 2. savings, 3. utilities");
    System.out.print("Enter list name or list number to send: ");
    String user = user_input.nextLine();
    try {
          if (user.toLowerCase().equals("groceries") || Integer.parseInt(user) == 1) {
                 System.out.println("item was sent to the grocery list");

          }else if(user.toLowerCase().equals("savings") || Integer.parseInt(user) == 2) {
                 System.out.println("item was sent to the savings list");

          }else if(user.toLowerCase().contains("utilities") || Integer.parseInt(user) == 3) {
                 System.out.println("item was sent to the utilities list");

          }else{
                 System.out.println("item is not in any category");

          }
    } catch (NumberFormatException nfe) {
        System.out.println("item is not in any category");
    }

更新：

    System.out.println("where would you like to send the following item: ");
    System.out.println("1. groceries, 2. savings, 3. utilities");
    System.out.print("Enter list name or list number to send: ");
    String user = user_input.nextLine();
    if (user.toLowerCase().equals("groceries") || user.equals("1")) {
        System.out.println("item was sent to the grocery list");

    }else if(user.toLowerCase().equals("savings") || user.equals("2")) {
        System.out.println("item was sent to the savings list");

    }else if(user.toLowerCase().contains("utilities") || user.equals("3")) {
        System.out.println("item was sent to the utilities list");

    }else{
        System.out.println("item is not in any category");

    }

Answer 2

我不是 100% 确定，但问题可能是当输入一个字符串时，它仍然尝试解析一个 int，导致它失败。例如，输入“2”是有效的，因为“2”是一个有效的字符串，而 2 是一个可以从字符串“2”解析的有效 int，而“groceries”是一个有效的字符串，但不包含可以被解析。

Answer 3

Odd822 是对的。问题是当您输入除“groceries”以外的任何字符串时，程序会尝试使用 Integer.parseint("stringwhichisnotaninteger")。 因此，它会引发异常。

这段代码应该可以工作：

if (user.toLowerCase().equals("groceries") || user.equals("1") ) {
        System.out.println("item was sent to the grocery list");

    }else if(user.toLowerCase().equals("savings") || user.equals("2")) {
        System.out.println("item was sent to the savings list");

    }else if(user.toLowerCase().contains("utilities") || user.equals("3")) {
        System.out.println("item was sent to the utilities list");

    }else{
        System.out.println("item is not in any category");

    }