Java 扫描仪输入数据类型

Scanner input data types

提问by Claude Rhay

I need to write a test class that will do the following:

我需要编写一个测试类来执行以下操作：

• a. Let the user input an integer and display it.
• b. Let the user input a float value and display it.
• c. Let the user input his/her name (no white spaces) and display the name as: “Hello <name>, welcome to Scanner!”
• d. Let the user input a character and display it.
• e. Let the user input any string (with white spaces) and display it.

• 一种。让用户输入一个整数并显示它。
• 湾 让用户输入一个浮点值并显示它。
• C。让用户输入他/她的名字（没有空格）并将名字显示为：“Hello <name>, welcome to Scanner!”
• d. 让用户输入一个字符并显示它。
• e. 让用户输入任何字符串（带空格）并显示它。

My questions is, how can I simply scan just a Characterand display it? And in number 2, How can I input a Stringwith white spaces and display it? (letters "d" and "e")

我的问题是，我怎样才能简单地扫描Character并显示它？在数字 2 中，如何输入String带有空格的 a 并显示它？（字母“d”和“e”）

I've searched around, but I cannot find the simplest solution (since I'm new to Java and programming).

我四处搜索，但找不到最简单的解决方案（因为我是 Java 和编程的新手）。

Here is my code so far:

到目前为止，这是我的代码：

package aw;

import java.io.PrintStream;
import java.util.Scanner;

public class NewClass1
{
  public static void main(String[] args)
  {
      int num;
      double num2;
      String name;
      char c;
          Scanner sc = new Scanner(System.in);
          PrintStream ps = new PrintStream(System.out);

      //for integer
      System.out.println("Enter a number: ");
      num = sc.nextInt();
      ps.printf("%d\n", num);

      //for float
      System.out.println("Enter a float value: ");
      num2 = sc.nextDouble();
      ps.printf("%.2f\n", num2);

      //for name w/o white space
      System.out.print("Enter your first name: ");
      name = sc.next();
      ps.printf("Hello %s, welcome to Scanner\n", name);


      //for character
      System.out.print("Enter a character: ");
      c = sc.findWithinHorizon(".", 0).charAt(0);
      System.out.print(“%c”, c);

      //for name w/ white space
      System.out.print("Enter your full name: ");
      name = sc.nextLine();
      System.out.print(“%s”, name);
  }
}

package aw;

import java.io.PrintStream;
import java.util.Scanner;

public class NewClass1
{
  public static void main(String[] args)
  {
      int num;
      double num2;
      String name;
      char c;
          Scanner sc = new Scanner(System.in);
          PrintStream ps = new PrintStream(System.out);

      //for integer
      System.out.println("Enter a number: ");
      num = sc.nextInt();
      ps.printf("%d\n", num);

      //for float
      System.out.println("Enter a float value: ");
      num2 = sc.nextDouble();
      ps.printf("%.2f\n", num2);

      //for name w/o white space
      System.out.print("Enter your first name: ");
      name = sc.next();
      ps.printf("Hello %s, welcome to Scanner\n", name);


      //for character
      System.out.print("Enter a character: ");
      c = sc.findWithinHorizon(".", 0).charAt(0);
      System.out.print(“%c”, c);

      //for name w/ white space
      System.out.print("Enter your full name: ");
      name = sc.nextLine();
      System.out.print(“%s”, name);
  }
}

I hope you can help me. Thanks!

我希望你能帮助我。谢谢！

采纳答案by Ravi Thapliyal

First, there's no need to wrap System.outin a PrintStreambecause outalready supports formatting with format()or printf()methods.

首先，不需要包含System.out在 a 中，PrintStream因为out已经支持使用format()或printf()方法进行格式化。

Next, you need to understand that when you input a line of data you also terminate it with a new line\n. The next<Type>()methods only consume the <Type>and nothing else. So, if a next<Type>()call may match \n, you need to skip over any extra new lines\nwith another nextLine()before.

接下来，您需要了解，当您输入一行数据时，您也会以新行终止它\n。这些next<Type>()方法只消耗<Type>，没有别的。所以，如果一个next<Type>()电话可以匹配\n，你需要跳过任何额外的新线\n与另一nextLine()前。

Here's your code with fixes:

这是您的修复代码：

  int num;
  double num2;
  String name;
  char c;

  Scanner sc = new Scanner(System.in);

  //for integer
  System.out.print("Enter a number: ");
  num = sc.nextInt();
  System.out.printf("%d\n", num);

  //for float
  System.out.print("Enter a float value: ");
  num2 = sc.nextDouble();
  System.out.printf("%.2f\n", num2);

  //for name w/o white space
  System.out.print("Enter your first name: ");
  name = sc.next();
  System.out.printf("Hello %s, welcome to Scanner\n", name);

  //for character
  System.out.print("Enter a character: ");
  c = sc.findWithinHorizon(".", 0).charAt(0);
  System.out.printf("%c\n", c);

  sc.nextLine(); // skip

  //for name w/ white space
  System.out.print("Enter your full name: ");
  name = sc.nextLine();
  System.out.printf("%s", name);

回答by slanecek

Use this:

用这个：

  //for a single char
  char Character = sc.findWithinHorizon(".", 0).charAt(0);

  //for a name with white space
  System.out.print("Enter your full name: ");
      String name2 = sc.next();
      String surname = sc.next();
      System.out.println(name2 + " " + surname);

回答by Luca Basso Ricci

Use Scanner.next(Pattern)and pass Pattern.compile("[A-Za-z0-9]")to let scanner accept only 1 character defined.
You can pass any regex as argument and check for next()
Scanner.next();for next line with spaces

使用Scanner.next(Pattern)传递Pattern.compile("[A-Za-z0-9]")让扫描器只接受定义的 1 个字符。
您可以将任何正则表达式作为参数传递并检查next()
Scanner.next();是否有空格