如何在Java中使用错误解决扫描器类？

Question

我一直在使用扫描器类此码输入获得。当过我运行这段代码保存它仅前两个值。我不能进入第三个变量值。

 //Input Code
 import java.util.Scanner;
 public class Input
 {
  public static void main(String[] args)
 {
 try{
Scanner s = new Scanner(System.in);
 int i;
float f;
String str;
 System.out.println("Enter the Integer value"); //getting.   
 input from user
 i=s.nextInt();   //store the use entered value
 System.out.println("Enter the Float value");
f=s.nextFloat();
System.out.println("Enter the String value");
str=s.nextLine();
 System.out.println("\nInt: "+i+"\nFloat: "+f+"\nString 
 "+str);  //print the final result
 }
  catch(Exception e)
 {
 System.out.print(e);   
 }
 }
 }

- - - 结束 - - -

输出： 输入整数值 4 输入的float值 4.4 输入字符串值

诠释4 浮法4.10 串

Answer 1

请更改s.nextline（）来s.next（）。有关详细信息，你可以参考下面的完整代码

import java.util.Scanner;

public class Input {
  public static void main(String[] args) {
    try {
      Scanner s = new Scanner(System.in);
      int i;
      float f;
      String str;
      System.out.println("Enter the Integer value"); // getting.
      // input from user
      i = s.nextInt(); // store the use entered value
      System.out.println("Enter the Float value");
      f = s.nextFloat();
      System.out.println("Enter the String value");
      str = s.next();
      System.out.println("\nInt: " + i + "\nFloat: " + f + "\nString  " + str); // print the final result
    } catch (Exception e) {
      System.out.print(e);
    }
  }
}