在java中获取整数输入

Question

我实际上是java编程的新手，发现很难接受整数输入并将其存储在变量中...如果有人能告诉我如何做到这一点或提供一个示例，例如将用户给出的两个数字相加，我会很高兴..

Answer 1

这是我的条目，包含相当强大的错误处理和资源管理：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Simple demonstration of a reader
 *
 * @author jasonmp85
 *
 */
public class ReaderClass {

    /**
     * Reads two integers from standard in and prints their sum
     *
     * @param args
     *            unused
     */
    public static void main(String[] args) {
        // System.in is standard in. It's an InputStream, which means
        // the methods on it all deal with reading bytes. We want
        // to read characters, so we'll wrap it in an
        // InputStreamReader, which can read characters into a buffer
        InputStreamReader isReader = new InputStreamReader(System.in);

        // but even that's not good enough. BufferedReader will
        // buffer the input so we can read line-by-line, freeing
        // us from manually getting each character and having
        // to deal with things like backspace, etc.
        // It wraps our InputStreamReader
        BufferedReader reader = new BufferedReader(isReader);
        try {
            System.out.println("Please enter a number:");
            int firstInt = readInt(reader);

            System.out.println("Please enter a second number:");
            int secondInt = readInt(reader);

            // printf uses a format string to print values
            System.out.printf("%d + %d = %d",
                              firstInt, secondInt, firstInt + secondInt);
        } catch (IOException ioe) {
            // IOException is thrown if a reader error occurs
            System.err.println("An error occurred reading from the reader, "
                               + ioe);

            // exit with a non-zero status to signal failure
            System.exit(-1);
        } finally {
            try {
                // the finally block gives us a place to ensure that
                // we clean up all our resources, namely our reader
                reader.close();
            } catch (IOException ioe) {
                // but even that might throw an error
                System.err.println("An error occurred closing the reader, "
                                   + ioe);
                System.exit(-1);
            }
        }

    }

    private static int readInt(BufferedReader reader) throws IOException {
        while (true) {
            try {
                // Integer.parseInt turns a string into an int
                return Integer.parseInt(reader.readLine());
            } catch (NumberFormatException nfe) {
                // but it throws an exception if the String doesn't look
                // like any integer it recognizes
                System.out.println("That's not a number! Try again.");
            }
        }
    }
}

Answer 2

java.util.Scanner< /a> 是完成此任务的最佳选择。

从文档中：

例如，此代码允许用户从 System.in 读取数字：

 扫描仪 sc = new Scanner(System.in);
 int i = sc.nextInt();

读取 int 只需两行。不过，请不要低估 Scanner 的强大功能。例如，以下代码将不断提示输入数字，直到给出一个数字：

Scanner sc = new Scanner(System.in);
System.out.println("Please enter a number: ");
while (!sc.hasNextInt()) {
    System.out.println("A number, please?");
    sc.next(); // discard next token, which isn't a valid int
}
int num = sc.nextInt();
System.out.println("Thank you! I received " + num);

这就是您所需要编写的全部内容，并且感谢 hasNextInt() 你不必担心任何 Integer.parseInt< /code> 和 NumberFormatException 根本没有。

另请参阅

• Java 教程/基本类/基本 I/O/扫描 和格式

相关问题

• 当输入错误的类型时，如何防止扫描仪抛出异常？ (java)
• 如何使用扫描器仅接受有效的 int 作为输入

---

其他示例

扫描器 可以使用 java.io.File，或普通字符串。

下面是使用 Scanner 标记 String 并一次性解析为数字的示例：

Scanner sc = new Scanner("1,2,3,4").useDelimiter(",");
int sum = 0;
while (sc.hasNextInt()) {
    sum += sc.nextInt();
}
System.out.println("Sum is " + sum); // prints "Sum is 10"

这是使用正则表达式的稍微更高级的用法：

Scanner sc = new Scanner("OhMyGoodnessHowAreYou?").useDelimiter("(?=[A-Z])");
while (sc.hasNext()) {
    System.out.println(sc.next());
} // prints "Oh", "My", "Goodness", "How", "Are", "You?"

如您所见，扫描仪相当强大！您应该更喜欢它 StringTokenizer ，现在是一个遗留类。

另请参阅

• Java 教程/基本类/正则表达式
• < a href="http://www.regular-expressions.info/tutorial.html" rel="noreferrer">regular-expressions.info/Tutorial

相关问题

• 扫描仪、StringTokenizer 与 String.Split

Answer 3

你的意思是来自用户的输入

   Scanner s = new Scanner(System.in);

    System.out.print("Enter a number: ");

    int number = s.nextInt();

//process the number

Answer 4

如果您正在谈论来自控制台输入的这些参数或任何其他 String 参数，请使用 static Integer#parseInt() 方法将它们转换为 整数。