Online Judge（OJ）搭建——2数据库构建

Posted 2020-09-19 Piers

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数据库EER图

数据库表、字段、约束解释

users 用户：

id 标识符，email 邮箱，password 密码，name 姓名，sex 性别，enabled 启用，role 角色

id primary key

advisors 指导：

id 标识符，uid_user 被指导者，uid_admin 指导者

id primary key

uid_user -> user(id)，uid_admin(id)

categories 类别：

id 标识符，name 名称，description 描述

id primary key

questions 问题：

id 标识符，qid 题号，description 描述，cid 类别标识符，name 名称

id primary key

cid -> categories(id)

test_data 测试数据：

id 标识符，qid 题号，input 一组输入，output 一组输出

id primary key

qid -> questions(id)

tests 测试信息：

id 标识符，uid 用户标识符，qid 题号，submit_time 提交时间，code 代码，correct_rate 正确率

id primary key

uid -> users(id)，qid -> questions(id)

scores 成绩：

id 标识符，uid 用户标识符，cid 类别标识符，usual_score 平时成绩

复杂SQL语句

selectSumScoreAndRank：

功能：根据用户 ID，查询用户 ID、用户所有题目总成绩、总成绩的排名。

实现：主要是利用了聚集函数，mysql 自带 @rowNum 属性。

代码：

       SELECT *
        FROM
            (SELECT
                 uid,
                 sum_correct_rate,
                 (@rowNum := @rowNum + 1) AS rank
             FROM (SELECT
                       uid,
                       sum(max_corrcet_rate) AS sum_correct_rate
                   FROM
                       (SELECT
                            uid,
                            qid,
                            max(correct_rate) AS max_corrcet_rate
                        FROM tests
                        GROUP BY uid, qid) AS max_tests
                   GROUP BY uid
                   ORDER BY sum_correct_rate DESC) AS rank_tests, (SELECT (@rowNum := 0)) AS rank) AS all_tests
        WHERE uid = #{uid}

selectPracticeAndUsualScoreFromAdmin：

功能：根据管理员 ID 和类别 ID，查询特定管理员的指导关系下用户的实践成绩和平时成绩。

实现：首先根据指导关系下的用户 ID 和特定题库 ID 选出特定题目，再根据特定题目选出特定测试，最后利用聚集函数进行成绩整合（例如，聚集函数 sum 计算特定用户特定题库下的总分）。

代码：

        SELECT
            scores.id,
            scores.uid,
            users.email,
            users.name                       AS userName,
            sum_tests.avg_correct_rate * 100 AS practice_score,
            scores.usual_score
        FROM
            (SELECT
                 uid,
                 avg(max_corrcet_rate) AS avg_correct_rate
             FROM
                 (SELECT
                      uid,
                      qid,
                      max(correct_rate) AS max_corrcet_rate
                  FROM (SELECT
                            uid,
                            qid,
                            correct_rate
                        FROM tests
                        WHERE
                            uid IN (SELECT uid_user
                                    FROM advisors
                                    WHERE uid_admin = #{uidAdmin}) AND qid IN (SELECT questions.id
                                                                               FROM categories
                                                                                   LEFT JOIN questions
                                                                                       ON categories.id = questions.cid
                                                                               WHERE cid = #{cid})
                       ) AS filter_tests
                  GROUP BY uid, qid) AS max_tests
             GROUP BY uid) AS sum_tests LEFT JOIN scores ON sum_tests.uid = scores.uid AND scores.cid = #{cid} LEFT JOIN users ON sum_tests.uid = users.id