Question

我有以下代码用于正在进行的聊天机器人工作。但是，它不能按预期工作。当我运行调试器时，它显示“再见”，由于一些奇怪的原因，不断被分配到“关键字”。从逻辑上讲，一切看起来应该按预期工作。但显然不是。任何帮助表示赞赏。我是Java编程的新手。

sum = 0, list = grid.getListofDots()
1. Find nearest dot from starting position (or previous dot that was removed) using manhattan distance
2. add to sum
3. Remove dot from list of possible dots
4. repeat steps 1 - 3 until list is empty
5. return the sum

Answer 1

1，请阅读substring的javadoc。

返回一个新字符串，该字符串是此字符串的子字符串。子串从指定的beginIndex开始并延伸到at处的字符 index endIndex - 1

根据您的逻辑，您必须修改

 String sub = statement.substring( position, position + keyword.length() + 1 );

到

String sub = statement.substring( position, position + keyword.length() + 2 );

2，因为您已经创建了Chatbot的实例，所以无需在Chatbot中使用静态方法。

3，当你不再使用它时，请记得关闭Scanner。

请参阅以下更新的代码：

public class Chatbot {


    public Chatbot( )
    {
    }

    /*
     * Generates a variety of responses, based on what the user has stated
     */
    public void respond( String statement )
    {        
        // use the findKeyword method to check for various cases of user statements
        if( statement.length() == 0 )
        {
            System.out.println( "Please say something :)" );
        }

        else if( findKeyword( statement, "hi" ) > 0 ||
            findKeyword( statement, "hello" ) > 0 ||
            findKeyword( statement, "hey" ) > 0 ||
            findKeyword( statement, "hiya" ) > 0 ||
            findKeyword( statement, "heya" ) > 0 )
            {
                System.out.println( "Hello to you too!" );
            }

        else if( findKeyword( statement, "how are you" ) > 0 ||
                 findKeyword( statement, "hows it going" ) > 0 ||
                 findKeyword( statement, "howre you" ) > 0 ||
                 findKeyword( statement, "how ya doing" ) > 0 ||
                 findKeyword( statement, "yo wassup" ) > 0 ||
                 findKeyword( statement, "hey whats up" ) > 0 || 
                 findKeyword( statement, "whats up" ) > 0 )
            {
                System.out.println( "I'm good, how are you?" );
            }
    }



    /*
     * findKeyword method, returns either a 0 or a 1
     * @ 0 -- keyword not found
     * @ 1 -- keyword found
     */
    public int findKeyword( String statement, String keyword )
    {

                // This is in case the keyword is not in the statement at all
        if( !statement.contains( keyword ) )
        {
            return 0;
        }


        int position = statement.toLowerCase().indexOf( keyword.toLowerCase() );        // position of the keyword in the statement
        statement = " " + statement.toLowerCase().replaceAll( "\'\",.?", "") + " ";                   // the purpose of this statement is to allow for us to search for specific phrases w/ spaces before and after the keyword

        String sub = statement.substring( position, position + keyword.length() + 2 );  // isolates the keyword with 1 character before and after

        String charBeforeKeyword = sub.substring( 0, 1 );                               // the character before the keyword
        String charAfterKeyword = sub.substring( sub.length() - 1, sub.length());      // the character after the keyword



        /*
         * Now, we check to see if the characters we isolated before are letters; if they are        * 
         * @ If they are letters...then our keyword is part of a bigger word (e.g. if we searched for "success" and it brought us "successful"
         * @ If they are not letters, then we have found our keyword with punctuation and/or spaces before/after it
         */
        if( (charBeforeKeyword.compareTo( "a" ) < 0 || charBeforeKeyword.compareTo( "z" ) > 0 )
                && (charAfterKeyword.compareTo( "a" ) < 0 || charAfterKeyword.compareTo( "z" ) > 0 ))
        {
            return 1;
        }

        return 0;  

    }


}

还有Execute类。

import java.util.Scanner;

public class Execute 
{
    public static void main( String [] args )
    {   
        // Variables and Objects
        Chatbot bot = new Chatbot();
        Scanner input = new Scanner( System.in );
        String statement = "";

        // Prompt and get the user's first input
        System.out.println( "Type text to start chatting!" );
        statement = input.nextLine();

        // While the user doesn't say goodbye or some other form of it, respond to user and then get their next response
        while( bot.findKeyword( statement, "bye" ) != 1 &&
               bot.findKeyword( statement, "cya" ) != 1 &&
               bot.findKeyword( statement, "goodbye" ) != 1 &&
               bot.findKeyword( statement, "gtg" ) != 1 )
        {
            bot.respond( statement );
            statement = input.nextLine();
        }
        input.close();
        System.out.println( "Goodbye!" );
    }
}

Answer 2

您可以尝试将“单词”初始化为字符串单词

示例

public class Chatbot {


    public Chatbot( )
    {
      String hi= "hi"
      String hello= "Hello"
      String hey= "Hey"
      String hiya= "Hiya"
      String heya= "Heya"
      String n;



    }

    /*
     * Generates a variety of responses, based on what the user has stated
     */
    public static void respond( String statement )
    {        
        // use the findKeyword method to check for various cases of user statements
        if( n = sc1.next()== "")
        {
            System.out.println( "Please say something :)" );
        }

        else if( findKeyword( statement, hi ) ||
            findKeyword( statement, hello )  ||
            findKeyword( statement,hey ) ||
            findKeyword( statement,  hiya)  ||
            findKeyword( statement, heya) )
            {
                System.out.println( "Hello to you too!" );
            }

这是我的猜测......

Java chatbot - 代码无法正常工作

2 个答案: