找不到变量？请高手指点我啊~~

Question

今天小菜鸟我写了一段简单的代码：import java.util.*;public class ChangeByPunctuation
{
public static void main(String[] args){
  
  Scanner keyboard=new Scanner(System.in);
  do{
 System.out.println("Enter a sentence with punctuations:");
     String sentenceInput=keyboard.nextLine();
     int lengthOfSentence=sentenceInput.length();
 char lastPunctuation=sentenceInput.charAt(lengthOfSentence-1);
     if((lastPunctuation=='?')&&(lengthOfSentence%2==0)){
      System.out.println("Yes");
     }else if((lastPunctuation=='?')&&(lengthOfSentence%2==1)){
      System.out.println("No");
     }else if(lastPunctuation=='!'){
      System.out.println("Wow");
     }else{
      System.out.println("You always say \""+sentenceInput+" \"");
     }
 System.out.println("Again?Yes or No?");
     String answer=keyboard.nextLine();
  }while(answer.equalsIgnoreCase("yes")); 
}
}结果它说找不到变量answer？执行流不是先经过“String answer=keyboard.nextLine();”再进行循环的检测吗？怎么会编译不通过，而且说找不到呢？请高手指点啊~~

Answer 1

import java.util.*; public class ChangeByPunctuation 
{ 
public static void main(String[] args){ 
  
  Scanner keyboard=new Scanner(System.in); 
  //扩大你的answer变量范围
  String answer=null;
  do{ 
System.out.println("Enter a sentence with punctuations:"); 
    String sentenceInput=keyboard.nextLine(); 
    int lengthOfSentence=sentenceInput.length(); 
char lastPunctuation=sentenceInput.charAt(lengthOfSentence-1); 
    if((lastPunctuation=='?')&&(lengthOfSentence%2==0)){ 
      System.out.println("Yes"); 
    }else if((lastPunctuation=='?')&&(lengthOfSentence%2==1)){ 
      System.out.println("No"); 
    }else if(lastPunctuation=='!'){ 
      System.out.println("Wow"); 
    }else{ 
      System.out.println("You always say \""+sentenceInput+" \""); 
    } 
System.out.println("Again?Yes or No?"); 
    answer=keyboard.nextLine(); 
  }while(answer.equalsIgnoreCase("yes")); 
} 
} 

Answer 2

import java.util.*; public class ChangeByPunctuation 
{ 
public static void main(String[] args){ 
  
  Scanner keyboard=new Scanner(System.in); 
  String answer ="";
  do{ 
System.out.println("Enter a sentence with punctuations:"); 
    String sentenceInput=keyboard.nextLine(); 
    int lengthOfSentence=sentenceInput.length(); 
    char lastPunctuation=sentenceInput.charAt(lengthOfSentence-1); 
    if((lastPunctuation=='?')&&(lengthOfSentence%2==0)){ 
      System.out.println("Yes"); 
    }else if((lastPunctuation=='?')&&(lengthOfSentence%2==1)){ 
      System.out.println("No"); 
    }else if(lastPunctuation=='!'){ 
      System.out.println("Wow"); 
    }else{ 
      System.out.println("You always say \""+sentenceInput+" \""); 
    } 
    System.out.println("Again?Yes or No?"); 
    answer=keyboard.nextLine(); 
  }while(answer.equalsIgnoreCase("yes")); 
} 
} answer要定义在循环的外面

Answer 3

}while(answer.equalsIgnoreCase("yes")); 当while是真后,执行啥?

Answer 4

你的answer在do{}范围里,
while()里看不到

Answer 5

while为真后，再进行一次do-while循环啊~

Answer 6

先执行do{}里面的，然后再执行while()，
判断条件是否成立，如果成立继续do{}里面的，不成立则结束。

Answer 7

小弟冒充一下高手！！
你的answer在do循环中定义只能在循环中用
在while中调不到它的
在do循环上一行定义，运行看看是不是你想要的结果！

Answer 8

尽量不要用do...while
直接用while吧,do...while可读性差