從掃描儀輸入輸出多個字符串

Question

我的程序的目的是接受來自用戶的整數，雙精度和字符串，並且當通過輸入單詞「quit」終止程序時，程序平均整數和雙精度，並輸出提交的字符串。這是我到目前爲止：

import java.util.*; 

public class Lab09 { 
    public static void main(String [] args) { 
     Scanner console = new Scanner(System.in); 

     double sumI = 0; 
     double sumD = 0; 
     String words = ""; 
     int numInputInt = 0; 
     int numInputDoub = 0; 
     do { 
      System.out.print("Enter something: "); 
      if (console.hasNextInt()) { 
       int numI = console.nextInt(); 

      if (numI >= -100 && numI <= 100) { 
       sumI += numI; 
       numInputInt++; 
      } 
      else { 
       System.out.println("Integer out of range!(-100 .. 100)"); 
      } 

      } 
     else if (console.hasNextDouble()) { 
      double numD = console.nextDouble(); 
      if (numD >= -10.0 && numD <= 10.0) { 
       sumD += numD; 
       numInputDoub++; 
      } 
      else { 
       System.out.println("Double out of range!(-10.0 .. 10.0)"); 
      } 

     } 
     else { 
      words = console.next(); 
     } 

     } while (!words.equalsIgnoreCase("quit")); 
     System.out.println("Program terminated..."); 
     double avgInt = sumI/numInputInt; 
     double avgDoub = sumD/numInputDoub; 

     if (numInputInt > 0) { 
     System.out.println("\tAveragae of Integers: " + avgInt); 
     } 
     else { 
     System.out.println("\tNo intergers submitted"); 
     } 
     if (numInputDoub > 0) { 
     System.out.println("\tAverage of Doubles: " + avgDoub); 
     } 
     else { 
     System.out.println("\tNo doubles submitted"); 
     } 
     System.out.println(words); 


    } 

} 

整數和雙打得到處理好，但我卡在字符串。有關如何去做的任何想法？ 在此先感謝！

Answer 1

你可以使用

String input = ""; 
do { 
    //... 
    } 
    else { 
     input = console.next();    
     words += input + " "; 
    } 
    console.nextLine(); // Read the carriage return 
} while (!input.equalsIgnoreCase("quit")); 
//... 
System.out.println(words.trim()); 

來串聯文本，但是當一個循環中完成的，這是相當低效。

一個更好的解決辦法是使用StringBuilder ......

StringBuilder work = new StringBuilder(128); 
String input = ""; 
do { 
    //... 
    } 
    else { 
     input = console.next();    
     words.append(" ").append(input); 
    } 
    console.nextLine(); // Read the carriage return 
} while (!input.equalsIgnoreCase("quit")); 
//... 
System.out.println(words); 

Answer 2

雖然你可以建立單詞的List<String>看起來你只是想每一個新詞串聯到您的String words一樣，

if (words.length() > 0) words += " "; // <-- add a space. 
words += console.next(); // <-- += not = 

然後你的循環測試更改爲類似，

while (!words.trim().toLowerCase().endsWith("quit")); 

它應該像你期望的那樣工作。

Answer 3

我剛剛看了你的意見，你不能使用任何其它類，試試這個：

import java.util.*; 

public class Lab09 { 
    public static void main(String [] args) { 
     Scanner console = new Scanner(System.in); 

     double sumI = 0; 
     double sumD = 0; 
     String words = ""; 
     int numInputInt = 0; 
     int numInputDoub = 0; 
     do { 
      System.out.print("Enter something: "); 
      if (console.hasNextInt()) { 
       int numI = console.nextInt(); 

      if (numI >= -100 && numI <= 100) { 
       sumI += numI; 
       numInputInt++; 
      } 
      else { 
       System.out.println("Integer out of range!(-100 .. 100)"); 
      } 

      } 
     else if (console.hasNextDouble()) { 
      double numD = console.nextDouble(); 
      if (numD >= -10.0 && numD <= 10.0) { 
       sumD += numD; 
       numInputDoub++; 
      } 
      else { 
       System.out.println("Double out of range!(-10.0 .. 10.0)"); 
      } 

     } 
     else { 
      words = words.concat(" ").concat(console.next()); 

     } 

     } while (!words.contains("quit")); 
     System.out.println("Program terminated..."); 
     double avgInt = sumI/numInputInt; 
     double avgDoub = sumD/numInputDoub; 

     if (numInputInt > 0) { 
     System.out.println("\tAveragae of Integers: " + avgInt); 
     } 
     else { 
     System.out.println("\tNo intergers submitted"); 
     } 
     if (numInputDoub > 0) { 
     System.out.println("\tAverage of Doubles: " + avgDoub); 
     } 
     else { 
     System.out.println("\tNo doubles submitted"); 
     } 
     System.out.println(words); 


    } 

}