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JavaWeb 项目 --- 在线 OJ 平台 (二)

转载 作者:知者 更新时间:2024-03-12 23:50:31 28 4
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1. 使用数据库实现题目管理

1.1 设计数据库

题目表一般有的属性:

  • 题号
  • 标题
  • 难度
  • 描述
  • 测试用例
  • 自带的代码
create database if not exists Oj_Online;

use Oj_Online;

drop table if exists oj_problem;
-- 题目表
create table oj_problem(
    -- 题目的序号
    id int primary key auto_increment,
    -- 题目的标题
    title varchar(50),
    -- 题目的难度
    level varchar(50),
    -- 题目的描述
    description varchar(4096),
    -- 题目给定的代码
    templateCode varchar(4096),
    -- 题目的测试用例
    testCode varchar(4096)
);

1.2 封装 DBUtil

在 common 中创建 DBUtil

package common;

import com.mysql.jdbc.jdbc2.optional.MysqlDataSource;

import javax.sql.DataSource;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;

/**
 * 数据库操作
 */
public class DBUtil {
    public static final String URL = "jdbc:mysql://127.0.0.1:3306/Oj_Online?characterEncoding=utf8&useSSL=false";
    public static final String USERNAME = "root";
    public static final String PASSWORD = "";

    private static volatile DataSource dataSource = null;

    private static DataSource getDataSource() {
        if(dataSource == null){
            synchronized (DBUtil.class){
                if(dataSource == null){
                    dataSource = new MysqlDataSource();
                    ((MysqlDataSource) dataSource).setURL(URL);
                    ((MysqlDataSource) dataSource).setUser(USERNAME);
                    ((MysqlDataSource) dataSource).setPassword(PASSWORD);
                }
            }
        }
        return dataSource;
    }

    public static Connection getConnection() throws SQLException {
        return getDataSource().getConnection();
    }

    public static void close(Connection connection, PreparedStatement statement, ResultSet resultSet){
        if(resultSet != null){
            try {
                resultSet.close();
            } catch (SQLException e) {
                e.printStackTrace();
            }
        }
        if(statement != null){
            try {
                statement.close();
            } catch (SQLException e) {
                e.printStackTrace();
            }
        }
        if(connection != null) {
            try {
                connection.close();
            } catch (SQLException e) {
                e.printStackTrace();
            }
        }
    }
}

1.3 创建 Problem类

这个类表示一个题目

package dao;

public class Problem {
    private int id;
    private String title;
    private String level;
    private String description;
    private String templateCode;
    private String testCode;
	
	// ...一堆getter和setter
}

1.4 创建 ProblemDao类

这个类封装了对 Problem类的增删改查操作.

1.4.1 实现新增题目的功能

/**
     * 新增题目
     * @param problem 题目
     */
    public void insert(Problem problem){
        Connection connection = null;
        PreparedStatement statement =null;
        try {
            // 1. 建立连接
            connection = DBUtil.getConnection();
            // 2. 拼装 SQL 语句
            String sql = "insert into oj_problem values(null,?,?,?,?,?)";
            statement = connection.prepareStatement(sql);
            statement.setString(1,problem.getTitle());
            statement.setString(2,problem.getLevel());
            statement.setString(3,problem.getDescription());
            statement.setString(4,problem.getTemplateCode());
            statement.setString(5,problem.getTestCode());
            // 3. 执行 SQL 语句
            int ret = statement.executeUpdate();
            if(ret == 1){
                System.out.println("插入成功!");
            }else {
                System.out.println("插入失败!");
            }
        } catch (SQLException e) {
            e.printStackTrace();
        } finally {
            // 4. 关闭释放资源
            DBUtil.close(connection,statement,null);
        }
    }

1.4.2 实现删除题目的功能

/**
     * 删除对应id的题目
     * @param id 题目的序号
     */
    public void delete(int id){
        Connection connection = null;
        PreparedStatement statement = null;
        try {
            // 1. 建立连接
            connection = DBUtil.getConnection();
            // 2. 拼装 SQL 语句
            String sql = "delete from oj_problem where id = ?";
            statement = connection.prepareStatement(sql);
            statement.setInt(1,id);
            // 3. 执行 SQL 语句
            int ret = statement.executeUpdate();
            if(ret == 1){
                System.out.println("删除成功!");
            }else {
                System.out.println("删除失败!");
            }
        } catch (SQLException e) {
            e.printStackTrace();
        } finally {
            // 4. 关闭释放资源
            DBUtil.close(connection,statement,null);
        }
    }

1.4.3 实现查询题目列表的功能

/**
     * 查询题目列表
     * @return 题目列表
     */
    public List<Problem> selectAll(){
        Connection connection = null;
        PreparedStatement statement = null;
        ResultSet resultSet = null;
        List<Problem> list = new ArrayList<>();
        try {
            // 1. 建立连接
            connection = DBUtil.getConnection();
            // 2. 拼装 SQL 语句
            String sql = "select id,title,level from oj_problem";
            statement = connection.prepareStatement(sql);
            // 3. 执行 SQL 语句
            resultSet = statement.executeQuery();
            // 4. 遍历结果集
            while (resultSet.next()){
                Problem problem = new Problem();
                problem.setId(resultSet.getInt("id"));
                problem.setTitle(resultSet.getString("title"));
                problem.setLevel(resultSet.getString("level"));
                list.add(problem);
            }
            return list;
        } catch (SQLException e) {
            e.printStackTrace();
        }finally {
            // 5. 关闭释放资源
            DBUtil.close(connection,statement,resultSet);
        }
        return null;
    }

1.4.4 实现查找对应id题目的功能

/**
     * 查找对应id的题目
     * @param id 题目序号
     * @return 返回一个题目
     */
    public Problem selectOne(int id){
        Connection connection = null;
        PreparedStatement statement = null;
        ResultSet resultSet = null;
        try {
            // 1. 建立连接
            connection = DBUtil.getConnection();
            // 2. 拼装 SQL 语句
            String sql = "select * from oj_problem where id = ?";
            statement = connection.prepareStatement(sql);
            statement.setInt(1,id);
            // 3. 执行 SQL 语句
            resultSet =  statement.executeQuery();
            // 4. 遍历结果集
            if (resultSet.next()){
                Problem problem = new Problem();
                problem.setId(resultSet.getInt("id"));
                problem.setTitle(resultSet.getString("title"));
                problem.setLevel(resultSet.getString("level"));
                problem.setDescription(resultSet.getString("description"));
                problem.setTemplateCode(resultSet.getString("templateCode"));
                problem.setTestCode(resultSet.getString("testCode"));
                return problem;
            }
        } catch (SQLException e) {
            e.printStackTrace();
        } finally {
            // 5. 关闭释放资源
            DBUtil.close(connection,statement,resultSet);
        }
        return null;
    }

1.5 测试这几个功能

1.5.1 设计测试用例测试新增功能

public static void main(String[] args) {
        ProblemDao problemDao = new ProblemDao();
        Problem problem = new Problem();
        problem.setTitle("两数之和");
        problem.setLevel("简单");
        problem.setDescription("给定一个整数数组 nums和一个整数目标值 target,请你在该数组中找出 和为目标值 target 的那两个整数,并返回它们的数组下标。\n" +
                "你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。\n" +
                "你可以按任意顺序返回答案。\n" +
                "示例 1:\n" +
                "输入:nums = [2,7,11,15], target = 9\n" +
                "输出:[0,1]\n" +
                "解释:因为 nums[0] + nums[1] == 9 ,返回 [0, 1] 。\n" +
                "示例 2:\n" +
                "输入:nums = [3,2,4], target = 6\n" +
                "输出:[1,2]\n" +
                "示例 3:\n" +
                "输入:nums = [3,3], target = 6\n" +
                "输出:[0,1]\n" +
                "\n" +
                "提示:\n" +
                "2 <= nums.length <= 104\n" +
                "-109 <= nums[i] <= 109\n" +
                "-109 <= target <= 109\n" +
                "只会存在一个有效答案\n" +
                "进阶:你可以想出一个时间复杂度小于 O(n2) 的算法吗?\n" +
                "来源:力扣(LeetCode)\n" +
                "链接:https://leetcode.cn/problems/two-sum\n" +
                "著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。");
        problem.setTemplateCode("class Solution {\n" +
                "    public int[] twoSum(int[] nums, int target) {\n" +
                "\n" +
                "    }\n" +
                "}");
        problem.setTestCode("    public static void main(String[] args) {\n" +
                "        Solution solution = new Solution();\n" +
                "        int[] test1 = {2,7,11,15};\n" +
                "        int target1 = 9;\n" +
                "        int[] res1 = solution.twoSum(test1, target1);\n" +
                "        if (res1.length == 2 && res1[0] == 0 && res1[1] == 1){\n" +
                "            System.out.println(\"testcase1 OK!\");\n" +
                "        }else {\n" +
                "            System.out.println(\"testcase1 FAILED!\");\n" +
                "        }\n" +
                "        int[] test2 = {3,2,4};\n" +
                "        int target2 = 6;\n" +
                "        int[] res2 = solution.twoSum(test2, target2);\n" +
                "        if (res2.length == 2 && res2[0] == 1 && res2[1] == 2){\n" +
                "            System.out.println(\"testcase2 OK!\");\n" +
                "        }else {\n" +
                "            System.out.println(\"testcase2 FAILED!\");\n" +
                "        }\n" +
                "    }");
        problemDao.insert(problem);
    }

查看数据库中的内容

1.5.2 测试删除功能

再次新增一个内容

执行以下代码

ProblemDao problemDao = new ProblemDao();
     problemDao.delete(2);

再次查看数据库

1.5.3 测试查询列表功能

ProblemDao problemDao = new ProblemDao();
    List<Problem> list = problemDao.selectAll();
    System.out.println(list);

1.5.4 测试查询对应id的功能

ProblemDao problemDao = new ProblemDao();
    Problem problem = problemDao.selectOne(1);
    System.out.println(problem);

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