java - 扫描仪输入数据类型

Answer 1

First, there's no need to wrap System.out in a PrintStream because out already supports formatting with format() or printf() methods.

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 \n with another nextLine() before.

Here's your code with fixes:

  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);

Answer 2

Use this:

  //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);

Answer 3

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