在java中获取整数输入

时间:2010-05-31 09:44:43

标签: java parsing input

我实际上是java编程的新手,我发现很难取整数输入并将其存储在变量中...如果有人可以告诉我如何操作或者提供一个例子,例如添加两个数字,我希望它由用户..

4 个答案:

答案 0 :(得分:5)

这是我的条目,包括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.");
            }
        }
    }
}

答案 1 :(得分:5)

java.util.Scanner是完成此任务的最佳选择。

来自文档:

  

例如,此代码允许用户从System.in中读取数字:

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

您需要阅读两行int。但是,不要低估Scanner的强大程度。例如,以下代码将一直提示输入一个数字:

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);

这就是你要写的全部内容,感谢hasNextInt(),你根本无需担心任何Integer.parseIntNumberFormatException

另见

相关问题


其他例子

Scanner可以使用java.io.File或普通String作为其来源。

以下是使用ScannerString进行标记并一次解析为数字的示例:

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"

使用正则表达式,这是一个稍微高级的用法:

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,Scanner非常强大!你应该更喜欢它StringTokenizer,现在是遗产类。

另见

相关问题

答案 2 :(得分: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

答案 3 :(得分:0)

如果您从控制台输入或任何其他String参数谈论这些参数,请使用静态Integer#parseInt()方法将其转换为Integer