如何在java中尝试并捕获异常处理后恢复代码

Question

我实际上遇到了我的程序问题。其实我在学校学习（11年级 - 大专），所以请用非常简单的语言来解释我。我正在开发一个测验，非常基础的学校项目，所以程序就这样继续下去......我希望用户通过输入1-4的数字输入他/她的选择。我不希望用户输入字母，字母或特殊字符或任何其他编号。除了1-4之外。我尝试使用try和catch，但是我的程序在抛出异常后停止。即使在显示错误信息System.out.println（“输入无效，请在1-4之间输入一个数字”）之后，我想让程序代码运行。 示例程序

import java.io.*; 
    import java.lang; 
    public class 
    { 
    public static void main()throws IOException 
    { 
    InputStreamReader read =new InputStreamReader (System.in); 
    BufferedReader in =new BufferedReader (read); 
    System.out.println(" Select your category"); 
    System.out.println(" 1. Food"); 
    System.out.println(" 2. National"); 
    System.out.println(""); 
    System.out.println(" Enter your choice"); 
    int choice=Integer.parseInt(in.readLine()); 
    if(choice==1)//food category 
    { 
     int score = 0; 
     System.out.println("what are dynamites made?"); 
     System.out.println("1. peanuts");System.out.println("2. grapes"); 
     System.out.println("3. flaxseeds");System.out.println("4. fish"); 
     System.out.println(" Enter your choice"); 
     int food1= Integer.parseInt(in.readline()); 
     if(c1=1) 
     { System.out.println(" Correct answer"); 
     score=score+10 
     } 
     else 
     { 
     System.out.println(" Wronge answer"); 
     score=score+0 
     } 
     //then i have the second question with the same format 
    } 
     if(choice==2)//natioanl category with the same format as above 
     {//two question with if else statements in them } 
    } 
}// also help me where to add the try and catch statements 

Answer 1

这是对此问题的更优雅的解决方案。没有尝试涉及。

import java.io.*; 
    import java.lang; 
    public class 
    { 
    //Compares response to the string representations of all numbers between 1 and numOptions, inclusive. 
    //Returns the matching integer or -1 otherwise. 
    public static int parseResponse(String response, int numOptions) 
    { 
      for(int i=1;i<=numOptions;i++) { 
       if(response.equals(Integer.toString(i))) return i; 
      } 
      return -1; 
    } 

    public static void main()throws IOException 
    { 
     InputStreamReader read =new InputStreamReader (System.in); 
     BufferedReader in =new BufferedReader (read); 
     System.out.println(" Select your category"); 
     System.out.println(" 1. Food"); 
     System.out.println(" 2. National"); 
     System.out.println(""); 
     System.out.println(" Enter your choice"); 

     //New code 
     int choice=-1; 
     while(choice==-1) { 
      choice=parseResponse(in.readLine(),2); 
      if(choice==-1) 
      { 
       System.out.println(" Invalid input. Please enter a number between 1-2"); 
      } 
     } 


     if(choice==1)//food category 
     { 
      int score = 0; 
      System.out.println("what are dynamites made?"); 
      System.out.println("1. peanuts");System.out.println("2. grapes"); 
      System.out.println("3. flaxseeds");System.out.println("4. fish"); 
      System.out.println(" Enter your choice"); 

      //New code 
      int food1=-1; 
      while(food1==-1) { 
       food1=parseResponse(in.readLine(),4); 
       if(food1==-1) 
       { 
        System.out.println(" Invalid input. Please enter a number between 1-4"); 
       } 
      } 

      if(food1==1) 
      { System.out.println(" Correct answer"); 
       score=score+10 
      } 
      else 
      { 
       System.out.println(" Wronge answer"); 
       score=score+0 
      } 
      //then i have the second question with the same format 
     } 
     if(choice==2)//natioanl category with the same format as above 
     {//two question with if else statements in them 
     } 
    } 
}// also help me where to add the try and catch statements 

如果你不知道：

if(response.equals(Integer.toString(i))) return i; 

相同

if(response.equals(Integer.toString(i))) 
{ 
     return i; 
} 

Answer 2

嘿这里是你可以运行的代码。

import java.io.*; 

import java.lang.*; 

public class TestTryCatch 
{ 

public static void main(String args[]) 
{ 
    try 
    { 
     InputStreamReader read =new InputStreamReader (System.in); 
     BufferedReader in =new BufferedReader (read); 
     System.out.println(" Select your category"); 
     System.out.println(" 1. Food"); 
     System.out.println(" 2. National"); 
     System.out.println(""); 
     System.out.println(" Enter your choice"); 
     int choice=0; 
     try 
     { 
      choice = Integer.parseInt(in.readLine()); 
     } 
     catch(Exception e) 
     { 
      choice=0; 
     } 
     if(choice==0) 
     { 
      System.out.println("Invalid Input");   
     } 
     if(choice==1)//food category 
     { 
      int score = 0; 
      System.out.println("what are dynamites made?"); 
      System.out.println("1. peanuts");System.out.println("2. grapes"); 
      System.out.println("3. flaxseeds");System.out.println("4. fish"); 
      System.out.println(" Enter your choice"); 
      int food1= Integer.parseInt(in.readLine()); 
      if(food1==1) 
      { System.out.println(" Correct answer"); 
      score=score+10; 
      } 
      else 
      { 
       System.out.println(" Wronge answer"); 
       score=score+0; 
      } 

     } 
     if(choice==2) 
     { 
     } 
    } 
    catch(Exception e) 
    { 
     e.printStackTrace(); 
    } 
} 
} 

把你尝试转换成整型。如果你的代码给出了异常，它会进入异常块并处理任何你想要的值。在这里，我把它作为0，你也可以分配-1。

如果此代码有帮助还原，请回复。

Answer 3

程序发生崩溃的地方就在这里：

int food1= Integer.parseInt(in.readline()); 
if(c1=1) 
{ System.out.println(" Correct answer"); 
    score=score+10 
} 

“C1”不是一个定义的变量，所以使用它会导致程序，无论用户输入什么崩溃。您应该将“c1”替换为“food1”。此外，赋值运算符“=”应替换为比较运算符“==”。

退房整数的Javadoc（谷歌“整型的javadoc”），你会发现，

public static int parseInt(String s) 
       throws NumberFormatException 

那么NumberFormatException的是我们需要赶上。在运行程序时，您还可以通过查看堆栈跟踪来找出抛出的异常和位置。

每当引发异常时，错误之后和下一个'}'之前的所有内容都会被跳过。由于大部分代码应该在错误发生后执行很好，我们要保持我们的try-catch小，即

try { 
    int choice=Integer.parseInt(in.readLine()); 
} catch (NumberFormatException ex) { 
    //Not a number 
} 

我们需要的变量，“选择”是我们尝试的访问之外，让我们拭目以待在我们的尝试之外声明它。

int choice; 

try { 
    choice=Integer.parseInt(in.readLine()); 
} catch (NumberFormatException ex) { 
    //Not a number 
} 

但是，NumberFormatException不会告诉您用户是否输入了除1-4之外的数字。所以你必须添加你自己的验证。

int choice; 

try { 
    choice=Integer.parseInt(in.readLine()); 
    if(choice<1 || choice>4) ; //Not 1-4 
} catch (NumberFormatException ex) { 
    //Not a number 
} 

最后，我们需要跟踪我们的输入是否有效，并在需要时循环。

boolean valid=false; 

while(!valid) { 

    int choice; 

    try { 
     choice=Integer.parseInt(in.readLine()); 
     if(choice<1 || choice>4) valid=false; //Not 1-4 
     else valid=true; 
    } catch (NumberFormatException ex) { 
     //Not a number 
     valid=false; 
    } 

    if(valid) { 
     //Handle valid response here 
    } else { 
     System.out.println(" Invalid input. Please enter a number between 1-4"); 
    } 
} 

祝你好运！