url 连接中的 java.lang.nullpointerException

Question

我是java编程的初学者。我正在尝试从 url 读取响应并插入到数据库中。我运行了下面发布的程序，但它返回了此错误:

Exception in thread "main" java.lang.NullPointerException
    at javaapp.JavaApplication1.parseResponseString(JavaApplication1.java:41)
    at javaapp.JavaApplication1.main(JavaApplication1.java:71)

我一直试图通过研究自己解决这个问题，但我无法解决。我想知道是否有人可以帮助我。如果您确实需要有关我的计划的任何其他信息，请直接询问。提前致谢!

这是我编写的代码

在此处输入代码 package javaapp;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.SQLException;
import java.util.HashMap;
import java.util.Map;

public class JavaApplication1 
{

Map<String,String> responseMap=new HashMap<String, String>();
    static String input;

    public void getResponseFromUrl() throws IOException
    {
        URL url = new URL("http://localhost:8084/home/Home");
        HttpURLConnection con = (HttpURLConnection) url.openConnection();

        BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream())) ;
       String inputLine;
        while ((inputLine = in.readLine()) != null) 
        {
           if (inputLine.contains("<h1>")) 
            {
                String input = inputLine;
                input = input.substring(input.indexOf("<h1>") + 4, input.indexOf("</h1>"));   
            }
        }

    }
    public void parseResponseString(String input)
    {
        String params[]=input.split(",");

        for(String param:params)
        {
            String key= param.substring(0,param.indexOf('='));
            String value= param.substring(param.indexOf('=')+1,param.length());
            responseMap.put(key, value);
            System.out.println(param);
        }
    }
    public void insertToDatabase() throws SQLException, ClassNotFoundException
    {
        Connection conn=null;
        Class.forName("oracle.jdbc.driver.OracleDriver");

        conn=DriverManager.getConnection("jdbc:oracle:thin:@localhost:1521:xe","jai","jaikiran");
        String insertQuery = " INSERT INTO value_1 (username1,username2,username3,username4,username5) "+
                             " values (?,?,?,?,?)";
        PreparedStatement pstmt = conn.prepareStatement(insertQuery);
        pstmt.setString(1,responseMap.get("username1"));
        pstmt.setString(2,responseMap.get("username2"));
        pstmt.setString(3,responseMap.get("username3"));
        pstmt.setString(4,responseMap.get("username4"));
        pstmt.setString(5,responseMap.get("username5"));
        pstmt.executeUpdate();
    }
    public static void main(String[] args) throws MalformedURLException, IOException, SQLException, ClassNotFoundException 
    {
        JavaApplication1 application =new JavaApplication1();
        application.getResponseFromUrl();
        application.parseResponseString(input);//shows exception here
        try {
            application.insertToDatabase();
        } catch (Exception e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }   
}





Thanks in advance,please helpme..

Answer 1

static String input; 默认情况下为 null，您将其传递给 parseResponseString() 方法。

您需要为input变量分配一个值。

我相信您希望在 getResponseFromUrl() 方法中完成分配，您已在该方法中创建了一个新的 input 变量。为了给 static String input; 分配一个值，您需要使用 this 关键字引用它。

if (inputLine.contains("<h1>")) 
{
    String input = inputLine;
    this.input = input.substring(input.indexOf("<h1>") + 4, input.indexOf("</h1>"));   
}