Question

我正在起草一个方案,以检查用户的年龄,然后在答复时向他们提供反馈。然而,我无法发现,如果用户输入的信息不是惯用,如何向用户发出错误的信息。我如何能够围绕这一点开展工作?

import javax.swing.JOptionPane;

public class ifStatements {

    public static void main(String[] args) {
        while (1 == 1) {
            int age = Integer.parseInt(JOptionPane.showInputDialog("Enter your Age: "));

            if (age >= 75) {
                JOptionPane.showMessageDialog(null, "You are old!");
                break;
            } else if (age >= 18) {
                JOptionPane.showMessageDialog(null, "You are an adult!");
                break;
            } else if (age > 0) {
                JOptionPane.showMessageDialog(null, "You are not an adult!");
                break;
            } else if (age <= 0) {
                JOptionPane.showMessageDialog(null, "Invalid response!");
            }
        }
    }
}

当用户投入任何价值而非分类时,即为错误信息。在这一具体情况下,I类“w”。我想能够发现所有类型的数据。也许超高技能,但我想学习如何处理所有例外,即便是这个小型方案所花东西。

Exception in thread "main" java.lang.NumberFormatException: For input string: "w"
    at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:67)
    at java.base/java.lang.Integer.parseInt(Integer.java:665)
    at java.base/java.lang.Integer.parseInt(Integer.java:781)
    at ifStatements.main(ifStatements.java:16)

Answer 1

喜欢这样做。仅仅重新提出不准确的意见。 Java公约规定,类别名称应从上层开始。

public class IfStatements {
    public static void main(String[] args) {
        int age = 0;
        while (age <= 0) {
            try {
                age = Integer.parseInt(
                        JOptionPane.showInputDialog("Enter your Age: "));
            } catch (NumberFormatException nfe) {
                displayError();
                continue;
            }
            if (age >= 75) {
                JOptionPane.showMessageDialog(null, "You are old!");
                break;
            } else if (age >= 18) {
                JOptionPane.showMessageDialog(null, "You are an adult!");
                break;
            } else if (age > 0) {
                JOptionPane.showMessageDialog(null, "You are not an adult!");
                break;
            }
            // must be <= 0 here so just display the message.
            displayError();
            
        }
    }
    private static void displayError() {
        JOptionPane.showMessageDialog(null,
                "Please enter a number > 0", "Invalid Input",
                JOptionPane.ERROR_MESSAGE);
    }
}

Answer 2

<<https://docs.oracle.com/en/java/javase/20/docs/api/java.base/java/lang/Integer.html#parseInt(java.lang.String)"rel=“nofollow noreferer”>Integer.parseInt > <

您可使用try-liver >,以捕获被推翻的例外情况。

The syntax is as follows.

try {

} catch (ExceptionType name) {

}

在try-block 栏内发言时,该栏目已卸下,并登录在fish>。

可通过 副渔获物/em>-block上公布的名称查阅Exception。

这里是贵国法典的一个实例,利用try- submissions验证输入值。

int age; while (1 == 1) { try { age = Integer.parseInt(JOptionPane.showInputDialog("Enter your Age: ")); if (age >= 75) { JOptionPane.showMessageDialog(null, "You are old!"); break; } else if (age >= 18) { JOptionPane.showMessageDialog(null, "You are an adult!"); break; } else if (age > 0) { JOptionPane.showMessageDialog(null, "You are not an adult!"); break; } else if (age <= 0) { JOptionPane.showMessageDialog(null, "Invalid response!"); } } catch (NumberFormatException exception) { JOptionPane.showMessageDialog(null, "Invalid age"); } }

下面是基本示范,根据您的法典改编,该守则使用除外>通过无效><>>>>>> /em> 输入。
The age method will attempt to parse the value returned by JOptionPane#showInputDialog.
If it is successful it will return the value, otherwise, if an exception is thrown, it s caught and a new IllegalArgumentException exception is thrown, with the input value as the message.

public static void main(String[] args) { int age; boolean exit; do { exit = true; try { age = age(); if (age >= 75) message("You are old!"); else if (age >= 18) message("You are an adult!"); else if (age > 0) message("You are not an adult!"); else { message("Invalid response!"); exit = false; } } catch (IllegalArgumentException exception) { String string = exception.getMessage(); if (string != null) { if (string.isBlank()) message("Input required"); else message("Invalid age, %s ".formatted(exception.getMessage())); exit = false; } } } while (!exit); } static int age() throws IllegalArgumentException { String string = null; try { string = JOptionPane.showInputDialog("Enter your Age: "); return Integer.parseInt(string); } catch (NumberFormatException exception) { throw new IllegalArgumentException(string); } } static void message(String string) { JOptionPane.showMessageDialog(null, string); }

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