SQL 질문 기록

다른

MySQL의 @ 및 @@ 기호

Mysql @ @ @@ 부호
Mysql @和@@

조인, 왼쪽 조인, 오른쪽 조인, 전체 외부 조인

MySQL 내부 조인, 왼쪽 및 오른쪽 외부 조인 그림

창 기능 연습

mysql 창 기능 연습

SQL 579 직원의 누적 급여 쿼리 어려움

SELECT 
    id, 
    Month,
    SUM(Salary) OVER(PARTITION BY id ORDER BY Month RANGE 2 PRECEDING) AS Salary
FROM
    Employee
WHERE 
    (id, Month) NOT IN (
        SELECT
            id,
            MAX(Month) AS Month
        FROM
            Employee
        GROUP BY 
            id
    )
ORDER BY
    id, Month DESC

SQL162 2021년 11월 일인당 기사 검색 시간

MySQL 날짜 관련 함수

mysql의 DATE 관련 함수 세부정보
MySQL SUBDATE() 사용법 및 코드 예
MySQL SUBDATE() 함수
MySQL DATE_SUB 및 SUBDATE 함수: 날짜 빼기 연산

MySQL 타임스탬프diff() 함수

시차 함수—TIMESTAMPDIFF()
[Mysql] TIMESTAMPDIFF 함수

MySQL date_format() 함수

MySQL DATE_FORMAT() 함수
date_format: 형식 날짜 날짜 함수
date_format의 공통 매개변수 설명:
%Y-%m-%d %H:%i:%s 2023-4-24 02:41:20

%a 缩写星期名 
%b 缩写月名 
%c 月，数值 
%D 带有英文前缀的月中的天 
%d 月的天，数值(00-31) 
%e 月的天，数值(0-31) 
%f 微秒 
%H 小时 (00-23) 
%h 小时 (01-12) 
%I 小时 (01-12) 
%i 分钟，数值(00-59) 
%j 年的天 (001-366) 
%k 小时 (0-23) 
%l 小时 (1-12) 
%M 月名 
%m 月，数值(00-12) 
%p AM 或 PM 
%r 时间，12-小时（hh:mm:ss AM 或 PM） 
%S 秒(00-59) 
%s 秒(00-59) 
%T 时间, 24-小时 (hh:mm:ss) 
%U 周 (00-53) 星期日是一周的第一天 
%u 周 (00-53) 星期一是一周的第一天 
%V 周 (01-53) 星期日是一周的第一天，与 %X 使用 
%v 周 (01-53) 星期一是一周的第一天，与 %x 使用 
%W 星期名 
%w 周的天 （0=星期日, 6=星期六） 
%X 年，其中的星期日是周的第一天，4 位，与 %V 使用 
%x 年，其中的星期一是周的第一天，4 位，与 %v 使用 
%Y 年，4 位 
%y 年，2 位

SELECT
    DATE (in_time) dt,
    ROUND(
        SUM(timestampdiff (second, in_time, out_time)) / COUNT(distinct uid),
        1
    ) avg_lensec
from
    tb_user_log
where
    date_format (in_time, "%Y-%m") = "2021-11"
    and artical_id != 0
group by
    dt
order by
    avg_lensec

각 비디오의 SQL156 평균 완료율

SQL 날짜 함수 day(), Month(), year()

SQL 날짜 함수 day(), Month(), year() 다양한 사용 방법

MySQL IF 문의 세 가지 용도

MySQL if 문의 세 가지 용도

MySQL의 count(1)과 count(*)의 차이점

MySQL sum(1) count(1) 사이의 차이점과 연결에 대해 간략히 설명합니다.
mysql에서 count(1)와 count(*) 사이의 차이점

MySQL SUM() 함수

mysql sum() 함수

MySQL 기능 사용법 FORMAT

MySQL 기능 사용법 FORMAT

SELECT
    a.video_id,
    ROUND(
        SUM(IF (end_time - start_time >= duration, 1, 0)) / COUNT(start_time),
        3
    ) AS avg_comp_play_rate
FROM
    tb_user_video_log a
    LEFT JOIN tb_video_info b ON a.video_id = b.video_id
WHERE
    YEAR (start_time) = 2021
GROUP BY
    a.video_id
ORDER BY
    avg_comp_play_rate DESC;

SQL157 평균 재생 진행률이 60%를 초과하는 비디오 카테고리

SELECT
    tag,
    CONCAT (avg_play_progress, "%") AS avg_play_progress
FROM
    (
        SELECT
            tag,
            ROUND(
                AVG(
                    IF (  
                        TIMESTAMPDIFF (SECOND, start_time, end_time) > duration,
                        1,
                        TIMESTAMPDIFF (SECOND, start_time, end_time) / duration
                    )
                ) * 100,
                2
            ) AS avg_play_progress
        FROM 
            tb_user_video_log
            JOIN tb_video_info USING (video_id)
        GROUP BY
            tag
        HAVING
            avg_play_progress > 60
        ORDER BY
            avg_play_progress DESC
    ) as t_progress

SQL180 특정 매장의 SPU 수

SELECT
    style_id,
    COUNT(DISTINCT item_id) AS SPU_num
FROM 
    product_tb
GROUP BY 
    style_id
ORDER BY 
    SPU_num DESC

문의

SQL 1873 단순 특별 보너스 계산

LIKE +% 나머지

SELECT 
    employee_id,
    IF(
        employee_id % 2 = 1 AND name NOT LIKE "M%",
        salary,
        0
    ) AS bonus
FROM
    employees
ORDER BY
    employee_id;

Mod 및 LEFT 기능

SELECT 
    employee_id,
    IF(
        MOD(employee_id, 2) = 1 AND LEFT(name, 1) != 'M',
        salary,
        0
    ) AS bonus
FROM
    employees
ORDER BY
    employee_id;

가지 구조의 CASE

SELECT 
    employee_id,
    CASE
        WHEN MOD(employee_id, 2) != 0 AND LEFT(name, 1) != 'M' THEN salary
        WHEN MOD(employee_id, 2) = 0 OR LEFT(name, 1) = 'M' THEN 0
        ELSE 0
    END AS bonus
FROM
    employees
ORDER BY
    employee_id;

REGEXP 키워드는 정규식을 지정합니다.

SELECT 
    employee_id,
    CASE
        WHEN MOD(employee_id, 2) = 1 AND name NOT REGEXP '^M' THEN salary 
        ELSE 0 
    END AS bonus 
FROM
    employees
ORDER BY
    employee_id;

SQL 1112 학생당 최고 성적 보통

윈도우 기능 - 정렬 기능

SELECT
    student_id,
    course_id,
    grade
FROM 
    (
        SELECT 
            student_id,
            course_id,
            grade,
            RANK() OVER (PARTITION BY student_id ORDER BY grade DESC, course_id ASC) AS ranking
        FROM
            enrollments
    ) t1
WHERE
    ranking = 1
ORDER BY
    student_id

SQL1398 A상품과 B상품을 구매했으나 C상품은 구매하지 않은 고객

MySQL SUM(1)과 SUM(0)의 차이점

MySQL SUM(1)과 SUM(0)의 차이점

SELECT
    o.customer_id,
    c.customer_name
FROM 
    orders o
    JOIN customers c ON o.customer_id = c.customer_id
GROUP BY
    customer_id
HAVING
    SUM(IF(product_name = 'A', 1, 0)) > 0 AND
    SUM(IF(product_name = 'B', 1, 0)) > 0 AND
    SUM(IF(product_name = 'C', 1, 0)) = 0

연결하다

SQL 607 영업사원 단순

SELECT
    name
FROM 
    salesperson s
WHERE 
    sales_id NOT IN (
        SELECT
            sales_id
        FROM 
            company c RIGHT JOIN orders o
            ON  c.com_id = o.com_id
        WHERE c.name = "RED"
    )

Simple을 판매하지 않은 SQL 1607 판매자

존재하지 않음/ IN이 아님/ LEFT JOIN 키워드

MySQL IN 및 NOT IN 사용법에 대한 자세한 설명
MySql에서 NOT IN 사용 시 주의사항

방법 1: NOT EXISTS 사용법이 익숙하지 않은 경우 내부 테이블과 외부 테이블에 대한 연결 조건을 사용해야 합니다. 외부 레이어에서는 내부 레이어에서 해당 레코드를 찾을 수 있습니다.

-- 使用 NOT EXISTS 来查找，所有销售日期中年份不等于 2020 的记录，注意 NOT EXISTS 需要使用连接字段
-- 将外层内层查询连接起来

SELECT
    seller_name
FROM
    seller s
WHERE NOT EXISTS(
    SELECT
        sale_date
    FROM 
        orders o
    WHERE
        DATE_FORMAT(sale_date, "%Y") = "2020"
        AND s.seller_id = o.seller_id
)
ORDER BY 
    seller_name

방법 2: 2020년에 판매한 직원을 필터링한 다음 외부 쿼리를 중첩하여 내부 목록에 없는 Seller_name을 필터링합니다.

-- 筛选出在 2020 年销售过的员工，然后嵌套外层查询筛选出不在内层名单的 seller_name

SELECT
    seller_name
FROM
    seller
WHERE
    seller_id NOT IN (
        SELECT
            DISTINCT o.seller_id
        FROM 
            orders o
        WHERE
            DATE_FORMAT(o.sale_date, "%Y") = "2020"
    )
ORDER BY
    seller_name

SQL 1407 최고의 여행자 단순

MySQL IFNULL() 함수

MySQL IFNULL() 함수

SELECT
    name,
    IFNULL(SUM(distance), 0) AS travelled_distance
FROM 
    users u LEFT JOIN
    rides r ON u.id = r.user_id
GROUP BY
    user_id
ORDER BY 
    travelled_distance DESC,
    name

SQL 1440 부울 표현식 매체 평가

왼쪽 조인 두 번

SELECT 
    e.left_operand,
    e.operator,
    e.right_operand,
    CASE WHEN (
        (e.operator = '=' AND v1.value = v2.value) OR
        (e.operator = '>' AND v1.value > v2.value) OR
        (e.operator = '<' AND v1.value < v2.value)
    ) THEN "true" ELSE "false"
    END AS value
FROM
    expressions e
    LEFT JOIN variables v1
    ON e.left_operand = v1.name
    LEFT JOIN variables v2
    ON e.right_operand = v2.name

SQL 1212 쿼리 팀 포인트 중간

UNION ALL + JOIN + 하위 쿼리

SELECT
    t.team_id, t.team_name, IFNULL(score, 0) num_points
FROM(
    SELECT team_id, SUM(score) score
    FROM(
        SELECT
            host_team team_id,
            SUM(CASE 
            WHEN host_goals > guest_goals THEN 3
            WHEN host_goals = guest_goals THEN 1
            ELSE 0
            END) score
        FROM 
            matches
        GROUP BY 
            host_team
        UNION ALL
        SELECT
            guest_team team_id,
            SUM(CASE 
            WHEN host_goals > guest_goals THEN 0
            WHEN host_goals = guest_goals THEN 1
            ELSE 3
            END) score
        FROM 
            matches
        GROUP BY 
            guest_team
    ) b
    GROUP BY
        team_id 
) a
RIGHT JOIN teams t ON t.team_id = a.team_id
ORDER BY num_points DESC, t.team_id

중합

SQL 1571 창고 관리자 단순

조인 + 집계 함수 SUM()

SELECT 
    w.name warehouse_name,
    SUM(w.units * p.width * p.length * p.height) VOLUME
FROM 
    warehouse w LEFT JOIN products p
    ON w.product_id = p.product_id
GROUP BY
    w.name

SQL 586 가장 많은 주문을 받은 고객 (중요!) Simple

LIMIT 및 COUNT() 함수 사용

SELECT
    customer_number
FROM
    orders
GROUP BY
    customer_number
ORDER BY
    COUNT(*) DESC
LIMIT 0, 1

DENSE_RANK() 창 함수 및 PARTITION BY 세부 정보 사용

586. 최다 주문 고객 (창순위 기능 내 PARTITION BY 상세 내용)

SELECT 
    customer_number
FROM
    (SELECT
        *,
        DENSE_RANK() OVER (ORDER BY COUNT(order_number) DESC) AS ranking
    FROM 
        orders
    GROUP BY 
        customer_number
    ) tmp
WHERE ranking = 1

SQL 1076 프로젝트 직원 II

창 전체 순위+개수

SELECT 
    project_id
FROM
    (SELECT
        project_id,
        DENSE_RANK() OVER (ORDER BY COUNT(*) DESC) AS ranking
    FROM 
        project
    GROUP BY 
        project_id
    ) tmp
WHERE ranking = 1

SQL 1082 판매 분석 I 단순

창 전체 순위 + 합계

SELECT 
    seller_id
FROM
    (
    SELECT
        seller_id,
        DENSE_RANK() OVER (ORDER BY SUM(price) DESC) AS ranking
    FROM 
        sales
    GROUP BY
        seller_id
    ) tmp
WHERE ranking = 1

SQL 176 두 번째로 높은 급여 보통

창 전체 순위+IFNULL

SELECT
    IFNULL(
        (SELECT 
            DISTINCT salary
        FROM
            (SELECT
                salary,
                DENSE_RANK() OVER (ORDER BY salary DESC) AS ranking
            FROM
                employee
            ) tmp
        WHERE ranking = 2),
        NULL
    ) AS SecondHighestSalary

가장 높은 응답률을 보이는 SQL 578 쿼리 질문 중간

윈도우 전체 순위 + 합계(if) 사용량 + 지표 계산

SELECT
    question_id survey_log
FROM 
    (SELECT
        question_id,
        DENSE_RANK() OVER (ORDER BY SUM(IF(action = "answer", 1, 0)) 
        / SUM(IF(action = "show", 1, 0 )) DESC) AS ranking
    FROM 
        surveylog
    GROUP BY
        question_id
    ) tmp
WHERE ranking = 1
ORDER BY question_id
LIMIT 0, 1

SQL 602 친구 응용 프로그램 II: 친구가 가장 많은 사람은 누구입니까?

창 종합 순위 + UNION ALL

SELECT 
    requester_id AS id,
    num
FROM
    (SELECT 
        requester_id,
        DENSE_RANK() OVER (ORDER BY COUNT(*) DESC) AS ranking,
        COUNT(*) AS num
    FROM
        (
        SELECT * FROM requestaccepted r1
        UNION
        SELECT
            accepter_id AS requester_id,
            requester_id AS accepter_id,
            accepter_id
        FROM requestaccepted r2
        ) t1
    GROUP BY
        requester_id
    ) t2
WHERE ranking = 1

SQL 1741 각 직원이 보낸 총 시간 찾기

GROUP BY를 두 번

SELECT 
    event_day AS day,
    emp_id,
    SUM(out_time - in_time) AS total_time
FROM 
    employees e
GROUP BY
    event_day, emp_id

SQL 1173 즉석음식배달 I

ROUND()는 소수점 이하 자릿수 기능을 유지합니다.

SELECT 
    ROUND(
        (SELECT COUNT(*) FROM delivery WHERE order_date = customer_pref_delivery_date) /
        (SELECT COUNT(*) FROM delivery) * 100,
        2
    ) AS immediate_percentage

정렬 및 그룹화

SQL 1587 은행 계좌 요약 II

SELECT 
    name,
    SUM(amount) AS balance
FROM 
    users u RIGHT JOIN transactions t
    ON u.account = t.account
GROUP BY
    t.account
HAVING
    SUM(amount) > 10000

SQL 1050 3회 이상 함께 작업한 배우 및 감독

SELECT 
    actor_id,
    director_id
FROM 
    actordirector ad
GROUP BY
    actor_id,
    director_id
HAVING 
    COUNT(*) >= 3

SQL 1511 소비자 주문 빈도 단순

MySQL LEFT() 함수

MySQL LEFT() 함수

HAVING, SUM IF

SELECT 
    customer_id,
    name
FROM 
    customers
WHERE
    customer_id IN(
        SELECT 
            customer_id
        FROM
            product p RIGHT JOIN orders o
            ON p.product_id = o.product_id
        GROUP BY 
            customer_id
        HAVING 
            SUM(IF(LEFT(order_date, 7) = '2020-06',  price * quantity, 0)) >= 100 AND
            SUM(IF(LEFT(order_date, 7) = '2020-07',  price * quantity, 0)) >= 100
    )

SQL 1693 일상적인 리더 및 파트너

개수(개별 xxx)

SELECT
    date_id,
    make_name,
    COUNT(DISTINCT lead_id) AS unique_leads,
    COUNT(DISTINCT partner_id) AS unique_partners
FROM
    dailysales d
GROUP BY 
    date_id,
    make_name

SQL 1501 안심하고 투자할 수 있는 국가 (중요) Medium

SUBSTR() 차단 문자열 함수

MySQL 문자열 작업

UNION+JOIN 작성방법

SELECT
    name as country
FROM 
    (SELECT caller_id AS call_id, duration FROM calls
    UNION ALL
    SELECT callee_id AS call_id, duration FROM calls
    ) a
    LEFT JOIN
    (SELECT id, c.name
    FROM person p LEFT JOIN country c
    ON LEFT(p.phone_number, 3) = c.country_code
    ) b
    ON call_id = id
GROUP BY 
    name
HAVING AVG(duration) > (SELECT AVG(duration) FROM calls)

고급 쿼리 및 조인

SQL 603 연속 공석

셀프 조인 JOIN 및 ABS() 사용

SELECT
    DISTINCT a.seat_id
FROM 
    cinema a JOIN cinema b
    ON abs(a.seat_id - b.seat_id) = 1 AND a.free = true AND b.free = true
ORDER BY
    a.seat_id;

SQL 1795 다양한 매장의 각 제품 가격

열에서 행으로

SELECT product_id, 'store1' store, store1 price FROM products WHERE store1 IS NOT NULL
UNION
SELECT product_id, 'store2' store, store2 price FROM products WHERE store2 IS NOT NULL
UNION
SELECT product_id, 'store3' store, store3 price FROM products WHERE store3 IS NOT NULL;

행에서 열로

SELECT 
  product_id,
  SUM(IF(store = 'store1', price, NULL)) 'store1',
  SUM(IF(store = 'store2', price, NULL)) 'store2',
  SUM(IF(store = 'store3', price, NULL)) 'store3' 
FROM
  Products1 
GROUP BY product_id ;

SQL 613 직선에서 가장 가까운 거리

ABS() 및 MIN() 함수 사용

SELECT
    MIN(ABS(p1.x - p2.x)) AS shortest
FROM
    point p1 JOIN point p2
    ON p1.x != p2.x
;

정보가 누락된 SQL 1965 직원

UNION 및 JOIN 사용

SELECT
    employee_id
FROM(
    SELECT employee_id FROM employees e
    UNION
    SELECT employee_id FROM salaries s
) t
WHERE
    employee_id NOT IN (
        SELECT 
            employee_id
        FROM 
            employees e JOIN salaries s
            USING(employee_id)
    )
ORDER BY
    employee_id

SQL 1264 페이지 권장 사항 매체

SELECT
    DISTINCT page_id recommended_page
FROM 
    likes
WHERE 
    user_id IN (
        SELECT
            user2_id
        FROM (
            SELECT user1_id, user2_id FROM friendship
            UNION ALL
            SELECT user2_id, user1_id FROM friendship
        ) t
        WHERE
            user1_id = 1
    ) AND page_id NOT IN (SELECT page_id FROM likes WHERE user_id = 1)

SQL 608 트리 노드 매체

SELECT
    id,
    CASE 
        WHEN COUNT(*) = 1 AND SUM(IF(p_id IS NULL, 1, 0)) = 0 THEN 'Leaf'
        WHEN COUNT(*) > 1 AND SUM(IF(p_id IS NULL, 1, 0)) = 0 THEN 'Inner'
        WHEN COUNT(*) >= 1 AND SUM(IF(p_id IS NULL, 1, 0)) > 0 THEN 'Root'
    END AS type
FROM(
    SELECT id, p_id FROM tree
    UNION ALL
    SELECT p_id, id FROM tree
) t
WHERE 
    id IS NOT NULL
GROUP BY
    id

SQL 534 게임플레이 분석 III 중간

창 집계 함수의 누적 합계

SELECT 
    player_id,
    event_date,
    SUM(games_played) OVER (PARTITION BY player_id ORDER BY event_date ASC) games_played_so_far
FROM 
    activity;

SQL 1783 그랜드 슬램 수 매체

전통적인 데카르트 곱

SELECT
    player_id,
    player_name,
    SUM(p.player_id = c.Wimbledon) + SUM(p.player_id = c.Fr_open) 
    + SUM(p.player_id = c.US_open) + SUM(p.player_id = c.Au_open) AS grand_slams_count
FROM 
    championships c JOIN players p
GROUP BY
    p.player_id
HAVING
    grand_slams_count > 0

SQL 1747 Leetflex 계정은 금지되어야 합니다

데카르트 곱, BETWEEN 키워드

MySQL 간의 사용법

SELECT
    DISTINCT a.account_id
FROM
    LogInfo a JOIN LogInfo b
WHERE
    a.account_id = b.account_id 
    AND a.ip_address != b.ip_address
    AND (a.logout BETWEEN b.login AND b.logout)

하위 쿼리

SQL 1303 팀원 수 찾기 (중요)

윈도우 함수 COUNT()

group by 문으로 인해 행 수가 단일 행으로 줄어드는 문제를 방지하려면 윈도우 기능을 사용하는 것이 좋습니다.

SELECT
    employee_id,
    COUNT(employee_id) OVER(PARTITION BY team_id) AS team_size
FROM
    Employee

고급 주제: 창 함수 및 CTE(공용 테이블 표현식)

MySQL 공통 테이블 표현식(CTE)

MySQL 공통 테이블 표현식(CTE)
MySQL CTE(공통 테이블 표현식)

SQL 1077. 프로젝트 직원 III 중간

윈도우 순위 기능, CTE

WITH cte AS(
    SELECT
        p.project_id,
        e.employee_id,
        DENSE_RANK() OVER (PARTITION BY p.project_id ORDER BY e.experience_years DESC) AS ranking
    FROM 
        Project p JOIN Employee e
        USING(employee_id)
)

SELECT
    project_id,
    employee_id
FROM 
    cte
WHERE
    ranking = 1

SQL 1285 연속 범위의 시작 및 끝 번호 찾기

연속 간격 및 창 함수 ROW_NUMBER()

SELECT
    MIN(log_id) start_id,
    MAX(log_id) end_id
FROM
    (SELECT
        log_id,
        log_id - ROW_NUMBER() OVER() AS diff
    FROM 
        Logs
    ) t
GROUP BY
    diff

SQL 1596 고객당 가장 자주 주문되는 품목(중요)

PARTITION BY + ORDER BY + 외부 GROUP BY + COUNT()

SELECT 
    customer_id,
    product_id,
    product_name
FROM
    (SELECT
        customer_id,
        product_id,
        product_name,
        DENSE_RANK() OVER (PARTITION BY customer_id ORDER BY COUNT(*) DESC) AS ranking
    FROM
        Orders o JOIN Products p
        USING(product_id)
    GROUP BY
        customer_id, product_id
    ) t
WHERE
    ranking = 1

SQL 1709 액세스 날짜 사이의 최대 간격 기간

LEAD() 함수 및 LAG() 함수

MySQL의 LAG() 함수, LEAD() 함수 사용법
MYSQL lag(), Lead() 함수 사용법

SELECT
    user_id,
    MAX(TIMESTAMPDIFF(DAY, visit_date, next_day)) AS biggest_window
FROM
    (SELECT
        user_id,
        visit_date,
        LEAD(visit_date, 1, "2021-1-1") OVER (PARTITION BY user_id ORDER BY visit_date ASC) AS next_day
    FROM 
        UserVisits
    ) t
GROUP BY
    user_id
ORDER BY 
    user_id

SQL 1270 회사의 CEO에게 보고하는 모든 사람

방법 1: UNION ALL, 하위 쿼리

SELECT DISTINCT employee_id FROM (
    SELECT employee_id
    FROM Employees WHERE manager_id = 1
    UNION ALL
    SELECT employee_id
    FROM Employees WHERE manager_id IN (
        SELECT employee_id FROM Employees WHERE manager_id = 1
    )
    UNION ALL
    SELECT employee_id
    FROM Employees WHERE manager_id IN (
        SELECT employee_id FROM Employees WHERE manager_id IN (
            SELECT employee_id FROM Employees WHERE manager_id = 1
        )
    )
) T WHERE employee_id != 1

방법 2: 가입

SELECT 
	e1.employee_id
FROM 
	Employees e1 JOIN Employees e2 
	ON e1.manager_id = e2.employee_id
	JOIN Employees e3 
	ON e2.manager_id = e3.employee_id
WHERE 
	e1.employee_id != 1 AND e3.manager_id = 1

SQL 1412 난이도가 중간 수준인 학생 찾기

밀도_순위()

Windowing 기능은 결과를 위쪽으로 정렬한 다음 뒤쪽으로 정렬합니다. 두 열 모두 1이 아닌 한 중앙에 있습니다.

WITH cte AS(
    SELECT 
        student_id,
        IF(DENSE_RANK() OVER (PARTITION BY exam_id ORDER BY score DESC) = 1, 1, 0) AS d_rank,
        IF(DENSE_RANK() OVER (PARTITION BY exam_id ORDER BY score) = 1, 1, 0) AS a_rank
    FROM 
        Exam
)

SELECT
    student_id,
    student_name
FROM
    Student s JOIN cte
    USING(student_id)
GROUP BY
    student_id
HAVING
    SUM(d_rank) = 0 AND SUM(a_rank) = 0
ORDER BY
    student_id

SQL 1767 실행되지 않은 작업 쌍을 찾는 것이 어렵습니다.

CTE 재귀

WITH RECURSIVE cte(task_id, subtask_id) AS (
    SELECT task_id, subtasks_count FROM Tasks
    UNION
    SELECT task_id, subtask_id - 1 FROM cte WHERE subtask_id > 1
)

SELECT 
    task_id,
    subtask_id
FROM
    cte LEFT JOIN Executed e
    USING (task_id, subtask_id)
WHERE 
    e.subtask_id IS NULL
ORDER BY 
    task_id, subtask_id

SQL 1225 연속 날짜의 시스템 상태 보고 어려움

차이 diff, SUBDATE() 날짜 함수를 사용한 연속 간격

SELECT
    tktype AS period_state,
    MIN(date_time) AS start_date,
    MAX(date_time) AS end_date
FROM
(
    SELECT
        tktype,
        date_time,
        SUBDATE(date_time, INTERVAL ROW_NUMBER() OVER (PARTITION BY tktype ORDER BY date_time) DAY) AS diff
    FROM
    (
        SELECT
            'failed' tktype,
            fail_date date_time
        FROM
            Failed
        UNION ALL
        SELECT
            'succeeded' tktype,
            success_date date_time
        FROM 
            Succeeded
    ) t1
) t2
WHERE
    date_time BETWEEN "2019-01-01" AND "2019-12-31"
GROUP BY
    tktype, diff
ORDER BY
    start_date

SQL136 시험 유형별 상위 3개 점수 중간

ROW_NUMBER(), PARTITION BY, GROUP BY

SELECT 
    tag,
    uid,
    ranking
FROM
    (SELECT
        b.tag,
        a.uid,
        ROW_NUMBER() OVER (PARTITION BY tag ORDER BY MAX(a.score) DESC, MIN(a.score) DESC, a.uid DESC) AS ranking
    FROM
        exam_record a LEFT JOIN examination_info b 
        ON a.exam_id = b.exam_id
    GROUP BY
        b.tag, a.uid
    ) t
WHERE 
    ranking <= 3

SQL137 난이도 두 번째로 빠른/느린 시간 차이가 시험지 길이의 절반보다 깁니다.

TIMESTAMPDIFF()、ROW_NUMBER()、CASE WHEN

SELECT 
    exam_id,
    duration,
    release_time
FROM
    (
        SELECT
            exam_id,
            duration,
            release_time,
            SUM(
                CASE
                    WHEN rank1 = 2 THEN costtime
                    WHEN rank2 = 2 THEN -costtime
                    ELSE 0
                END
            ) AS sub
        FROM
            (
                SELECT
                    e_i.exam_id,
                    duration, 
                    release_time,
                    TIMESTAMPDIFF(MINUTE, start_time, submit_time) AS costtime,
                    ROW_NUMBER() OVER (
                        PARTITION BY e_r.exam_id 
                        ORDER BY TIMESTAMPDIFF(MINUTE, start_time, submit_time) DESC
                    ) AS rank1,
                    ROW_NUMBER() OVER (
                        PARTITION BY e_r.exam_id 
                        ORDER BY TIMESTAMPDIFF(MINUTE, start_time, submit_time) ASC
                    ) AS rank2
                FROM 
                    exam_record e_r LEFT JOIN examination_info e_i
                    ON e_r.exam_id = e_i.exam_id
                WHERE 
                    submit_time IS NOT NULL
            ) t1
        GROUP BY
            exam_id
    ) t2
WHERE 
    sub * 2 >= duration
ORDER BY    
    exam_id DESC

SQL138 난이도 시험지를 연속으로 두 번 답할 수 있는 최대 시간 창

WITH t2 AS(
    SELECT 
        uid,
        COUNT(start_time) AS total, -- 用户2021年作答的次数
        DATEDIFF(MAX(start_time), MIN(start_time)) + 1 AS diff_time, -- 头尾作答时间窗
        MAX(DATEDIFF(next_time, start_time)) + 1 AS days_window -- 最大间隔天数
    FROM
        (
            SELECT
                uid,
                start_time,
                LEAD(start_time, 1) OVER (PARTITION BY uid ORDER BY start_time) AS next_time -- 第二次作答时间
            FROM 
                exam_record
            WHERE 
                YEAR(start_time) = 2021
        ) t1
    GROUP BY
        uid
)

SELECT
    uid,
    days_window,
    ROUND(total * days_window / diff_time, 2) AS avg_exam_cnt
FROM 
    t2
WHERE 
    diff_time > 1
ORDER BY 
    days_window DESC, avg_exam_cnt DESC