在java中获取整数输入_Java_Parsing_Input

在java中获取整数输入

java parsing input

在java中获取整数输入,java,parsing,input,Java,Parsing,Input,实际上，我是java编程新手，我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，或者提供一个示例，比如添加用户给定的两个数字..如果您是从控制台输入中谈论这些参数，或者任何其他字符串参数，使用静态方法将它们转换为整数。如果您正在从控制台输入中谈论这些参数或任何其他字符串参数，请使用静态方法将它们转换为整数。以下是我的条目，包括相当健壮的错误处理和资源管理： import java.io.BufferedReader; import java.io.IOException; imp

实际上，我是java编程新手，我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，或者提供一个示例，比如添加用户给定的两个数字..

如果您是从控制台输入中谈论这些参数，或者任何其他字符串参数，使用静态方法将它们转换为整数。

如果您正在从控制台输入中谈论这些参数或任何其他字符串参数，请使用静态方法将它们转换为整数。

以下是我的条目，包括相当健壮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.");
            }
        }
    }
}

以下是我的文章，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.");
            }
        }
    }
}

你是说用户的输入

   Scanner s = new Scanner(System.in);

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

    int number = s.nextInt();

//process the number

你是说用户的输入

   Scanner s = new Scanner(System.in);

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

    int number = s.nextInt();

//process the number

是完成此任务的最佳选择

从文件中：

例如，此代码允许用户从System.in读取数字：

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

读一个int只需要两行，但不要低估扫描仪的强大功能。例如，以下代码将一直提示输入一个数字，直到给出一个为止：

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);

这就是您需要编写的全部内容，而且由于它，您根本不必担心Integer.parseInt和NumberFormatException

另见相关问题其他例子扫描器可以使用一个或一个平面作为其源

下面是一个使用Scanner标记字符串并同时解析为数字的示例：

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"

这里有一个更高级的用法，使用正则表达式：

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

正如你们所看到的，扫描仪是相当强大的！您应该更喜欢它，它现在是一个遗留类

另见相关问题是完成此任务的最佳选择