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Essential MtSQL Selected Practice Questions with Answers

Noms de table et champs (MySQL)

  1. Table des étudiants
    Étudiant(s_id, s_name, s_birth, s_sex)
    Carte d'étudiant, nom de l'étudiant, date de naissance, sexe de l'étudiant

  2. Tableau des cours

    Cours(c_id, c_name, t_id)
    ID du cours, nom du cours, ID de l'enseignant

  3. Table des professeurs

    Professeur(t_id, t_name)
    ID de l'enseignant, nom de l'enseignant

  4. Tableau des scores

    Score(s_id, c_id, s_score)
    ID étudiant, ID de cours, score

Test Data - Creating Tables

  1. Student Table
CREATE TABLE  `Student`(  
`s_id`  VARCHAR(20),  
`s_name`  VARCHAR(20) NOT NULL DEFAULT '',  
`s_birth`  VARCHAR(20) NOT NULL DEFAULT '',  
`s_sex`  VARCHAR(10) NOT NULL DEFAULT '',  
PRIMARY KEY(`s_id`)  
);
  1. Course Table
CREATE TABLE  `Course`(  
`c_id`  VARCHAR(20),  
`c_name`  VARCHAR(20) NOT NULL DEFAULT '',  
`t_id`  VARCHAR(20) NOT NULL,  
PRIMARY KEY(`c_id`)  
);
  1. Teacher Table
CREATE TABLE  `Teacher`(  
`t_id`  VARCHAR(20),  
`t_name`  VARCHAR(20) NOT NULL DEFAULT '',  
PRIMARY KEY(`t_id`)  
);
  1. Score Table
CREATE TABLE  `Score`(  
`s_id`  VARCHAR(20),  
`c_id`  VARCHAR(20),  
`s_score`  INT(3),  
PRIMARY KEY(`s_id`,`c_id`)  
);
  1. Inserting Test Data into Student Table
INSERT INTO Student VALUES('01', 'John Doe', '1990-01-01', 'Male');  
INSERT INTO Student VALUES('02', 'Jane Smith', '1990-12-21', 'Male');  
INSERT INTO Student VALUES('03', 'Michael Brown', '1990-05-20', 'Male');  
INSERT INTO Student VALUES('04', 'Emily Davis', '1990-08-06', 'Male');  
INSERT INTO Student VALUES('05', 'Lucy Johnson', '1991-12-01', 'Female');  
INSERT INTO Student VALUES('06', 'Sophia Williams', '1992-03-01', 'Female');  
INSERT INTO Student VALUES('07', 'Olivia Taylor', '1989-07-01', 'Female');  
INSERT INTO Student VALUES('08', 'Victoria King', '1990-01-20', 'Female');
  1. Inserting Test Data into Course Table
INSERT INTO Course VALUES('01', 'Literature', '02');  
INSERT INTO Course VALUES('02', 'Mathematics', '01');  
INSERT INTO Course VALUES('03', 'English', '03');
  1. Inserting Test Data into Teacher Table
INSERT INTO Teacher VALUES('01', 'Andrew');  
INSERT INTO Teacher VALUES('02', 'Bethany');  
INSERT INTO Teacher VALUES('03', 'Charlie');
  1. Transcript Test Data
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

Exercise questions and SQL statements

  1. Retrieve the information and course scores of students who have a higher score in course '01' than in course '02'
SELECT a.*, b.s_score AS '01_score', c.s_score AS '02_score'  
FROM student a  
JOIN score b ON a.s_id = b.s_id AND b.c_id = '01'  
LEFT JOIN score c ON a.s_id = c.s_id AND c.c_id = '02'  
WHERE b.s_score > COALESCE(c.s_score, 0); -- Using COALESCE instead of OR c.c_id = NULL  

-- Alternatively  
SELECT a.*, b.s_score AS '01_score', c.s_score AS '02_score'  
FROM student a, score b, score c  
WHERE a.s_id = b.s_id  
AND a.s_id = c.s_id  
AND b.c_id = '01'  
AND c.c_id = '02'  
AND b.s_score > c.s_score;
  1. Retrieve the information and course scores of students who have a lower score in course '01' than in course '02'
SELECT a.*, b.s_score AS '01_score', c.s_score AS '02_score'  
FROM student a  
LEFT JOIN score b ON a.s_id = b.s_id AND b.c_id = '01'  
JOIN score c ON a.s_id = c.s_id AND c.c_id = '02'  
WHERE COALESCE(b.s_score, 0) 



<ol>
<li>Retrieve student IDs, names, and average scores for students with an average score of 60 or above
</li>
</ol>

<pre class="brush:php;toolbar:false">SELECT b.s_id, b.s_name, ROUND(AVG(a.s_score), 2) AS avg_score  
FROM student b  
JOIN score a ON b.s_id = a.s_id  
GROUP BY b.s_id, b.s_name  
HAVING AVG(a.s_score) >= 60;
  1. Retrieve student IDs, names, and average scores for students with an average score below 60 (including those with no scores)
SELECT b.s_id, b.s_name, ROUND(AVG(a.s_score), 2) AS avg_score  
FROM student b  
LEFT JOIN score a ON b.s_id = a.s_id  
GROUP BY b.s_id, b.s_name  
HAVING AVG(a.s_score) 



<ol>
<li>Retrieve student IDs, names, total courses selected, and total scores across all courses
</li>
</ol>

<pre class="brush:php;toolbar:false">SELECT a.s_id, a.s_name, COUNT(b.c_id) AS sum_course, SUM(b.s_score) AS sum_score  
FROM student a  
LEFT JOIN score b ON a.s_id = b.s_id  
GROUP BY a.s_id, a.s_name;
  1. Query the number of teachers with the surname "Smith"
SELECT  COUNT(t_id) FROM teacher WHERE t_name LIKE  'Smith%';
  1. Query the information of students who have taken classes taught by Teacher "John Doe"
SELECT a.*  
FROM student a  
JOIN score b ON a.s_id = b.s_id  
WHERE b.c_id IN (  
    SELECT c_id FROM course  
    WHERE t_id = (  
        SELECT t_id FROM teacher  
        WHERE t_name = 'John Doe'  
    )  
);
  1. Query the information of students who have not taken classes taught by Teacher "John Doe"
SELECT *  
FROM student c  
WHERE c.s_id NOT IN (  
    SELECT a.s_id  
    FROM student a  
    JOIN score b ON a.s_id = b.s_id  
    WHERE b.c_id IN (  
        SELECT a.c_id  
        FROM course a  
        JOIN teacher b ON a.t_id = b.t_id  
        WHERE t_name = 'John Doe'  
    )  
);
  1. Query the information of students who have taken both courses with IDs "Math101" and "Science101"
SELECT a.*  
FROM student a, score b, score c  
WHERE a.s_id = b.s_id  
AND a.s_id = c.s_id  
AND b.c_id = 'Math101'  
AND c.c_id = 'Science101';
  1. Query the information of students who have taken the course with ID "Math101" but have not taken the course with ID "Science101"
SELECT a.*
FROM student a
WHERE a.s_id IN (SELECT s_id FROM score WHERE c_id = 'Math101')
AND a.s_id NOT IN (SELECT s_id FROM score WHERE c_id = 'Science101');
  1. Query information of students who have not taken all courses
-- @wendiepei's approach
SELECT s.*
FROM student s
LEFT JOIN Score s1 ON s1.s_id = s.s_id
GROUP BY s.s_id
HAVING COUNT(s1.c_id) 



<ol>
<li>Query information of students who have taken at least one course in common with student ID '01'
</li>
</ol>

<pre class="brush:php;toolbar:false">SELECT *   
FROM student   
WHERE s_id IN (  
    SELECT DISTINCT a.s_id   
    FROM score a   
    WHERE a.c_id IN (  
        SELECT c_id   
        FROM score   
        WHERE s_id = '01'  
    )  
);
  1. Query information of students who have taken exactly the same courses as student ID '01'
SELECT
 t3.*
FROM
 (
  SELECT
   s_id,
   group_concat(c_id ORDER BY c_id) group1
  FROM
   score
  WHERE
   s_id <> '01'
  GROUP BY
   s_id
 ) t1
INNER JOIN (
 SELECT
  group_concat(c_id ORDER BY c_id) group2
 FROM
  score
 WHERE
  s_id = '01'
 GROUP BY
  s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN student t3 ON t1.s_id = t3.s_id
  1. Query the names of students who have not taken any course taught by Teacher "Tom"
select a.s_name from student a where a.s_id not in (
    select s_id from score where c_id = 
                (select c_id from course where t_id =(
                    select t_id from teacher where t_name = 'Tom')));
  1. Query student IDs, names, and average scores of students who have failed two or more courses
SELECT a.s_id, a.s_name, ROUND(AVG(b.s_score), 2) AS average_score  
FROM student a  
LEFT JOIN score b ON a.s_id = b.s_id  
WHERE a.s_id IN (  
    SELECT s_id  
    FROM score  
    WHERE s_score = 2  
)  
GROUP BY a.s_id, a.s_name;
  1. Retrieve student information for students who scored less than 60 on course "01", ordered by score in descending order.
SELECT a.*, b.c_id, b.s_score  
FROM student a  
JOIN score b ON a.s_id = b.s_id  
WHERE b.c_id = '01' AND b.s_score 



<ol>
<li>Display the scores of all courses and the average score for each student, ordered by their average score from highest to lowest.
</li>
</ol>

<pre class="brush:php;toolbar:false">SELECT   
    a.s_id,  
    MAX(CASE WHEN c_id = '01' THEN s_score END) AS Chinese,  
    MAX(CASE WHEN c_id = '02' THEN s_score END) AS Math,  
    MAX(CASE WHEN c_id = '03' THEN s_score END) AS English,  
    ROUND(AVG(s_score), 2) AS average_score  
FROM score a  
GROUP BY a.s_id  
ORDER BY average_score DESC;
  1. Query the highest score, lowest score, average score, pass rate, medium rate, good rate, and excellent rate for each course. Display in the following format: Course ID, Course Name, Highest Score, Lowest Score, Average Score, Pass Rate, Medium Rate, Good Rate, Excellent Rate. -- Pass is >=60, Medium is 70-80, Good is 80-90, Excellent is >=90
SELECT   
    a.c_id,  
    b.c_name,  
    MAX(s_score) AS HighestScore,  
    MIN(s_score) AS LowestScore,  
    ROUND(AVG(s_score), 2) AS AverageScore,  
    ROUND(100 * (SUM(CASE WHEN s_score >= 60 THEN 1 ELSE 0 END) / COUNT(s_score)), 2) AS PassRate,  
    ROUND(100 * (SUM(CASE WHEN s_score BETWEEN 70 AND 80 THEN 1 ELSE 0 END) / COUNT(s_score)), 2) AS MediumRate,  
    ROUND(100 * (SUM(CASE WHEN s_score BETWEEN 80 AND 90 THEN 1 ELSE 0 END) / COUNT(s_score)), 2) AS GoodRate,  
    ROUND(100 * (SUM(CASE WHEN s_score >= 90 THEN 1 ELSE 0 END) / COUNT(s_score)), 2) AS ExcellentRate  
FROM   
    score a   
LEFT JOIN   
    course b ON a.c_id = b.c_id   
GROUP BY   
    a.c_id, b.c_name;
  1. Sort scores by course and display rankings. MySQL does not have a built-in RANK() function, so we'll use variables to simulate it.
SELECT   
    a.s_id,  
    a.c_id,  
    @rank := IF(@prev_score = a.s_score, @rank, @rank + 1) AS rank_without_ties,  
    @prev_score := a.s_score AS score  
FROM   
    (SELECT s_id, c_id, s_score FROM score ORDER BY c_id, s_score DESC) a,  
    (SELECT @rank := 0, @prev_score := NULL) r  
ORDER BY   
    a.c_id, a.rank_without_ties;
  1. Query the total score of each student and rank them
SELECT   
    a.s_id,  
    @rank := IF(@prev_score = a.sum_score, @rank, @rank + 1) AS rank,  
    @prev_score := a.sum_score AS total_score  
FROM   
    (SELECT s_id, SUM(s_score) AS sum_score FROM score GROUP BY s_id ORDER BY sum_score DESC) a,  
    (SELECT @rank := 0, @prev_score := NULL) r  
ORDER BY   
    total_score DESC;
  1. Query the average score of different courses taught by different teachers, sorted from highest to lowest
SELECT   
    a.t_id,  
    c.t_name,  
    a.c_id,  
    ROUND(AVG(s_score), 2) AS avg_score   
FROM   
    course a  
LEFT JOIN   
    score b ON a.c_id = b.c_id   
LEFT JOIN   
    teacher c ON a.t_id = c.t_id  
GROUP BY   
    a.c_id, a.t_id, c.t_name   
ORDER BY   
    avg_score DESC;
  1. Query the information of students who rank second and third in all courses along with their scores
(SELECT   
    d.*,  
    c.ranking,  
    c.s_score,  
    c.c_id  
FROM   
    (SELECT   
        s_id,   
        s_score,   
        c_id,   
        @rank := IF(@prev_cid = c_id, @rank + 1, 1) AS ranking,  
        @prev_cid := c_id  
    FROM   
        score,   
        (SELECT @rank := 0, @prev_cid := NULL) AS var_init  
    WHERE   
        c_id = '01'  
    ORDER BY   
        c_id, s_score DESC  
    ) c  
LEFT JOIN   
    student d ON c.s_id = d.s_id  
WHERE   
    c.ranking BETWEEN 2 AND 3  
)  
UNION  
(SELECT   
    d.*,  
    c.ranking,  
    c.s_score,  
    c.c_id  
FROM   
    (SELECT similar structure as above but with c_id = '02' in the WHERE clause) c  
LEFT JOIN   
    student d ON c.s_id = d.s_id  
WHERE   
    c.ranking BETWEEN 2 AND 3  
)  
UNION  
(SELECT similar structure as above but with c_id = '03' in the WHERE clause);
  1. Count the number of students in each score range for each subject:
select distinct f.c_name, a.c_id,
       b.`85-100`, b.Percentage as `[85-100] Percentage`,
       c.`70-85`, c.Percentage as `[70-85] Percentage`,
       d.`60-70`, d.Percentage as `[60-70] Percentage`,
       e.`0-60`, e.Percentage as `[0-60] Percentage`
from score a
    left join (
        select c_id,
               SUM(case when s_score > 85 and s_score  85 and s_score  70 and s_score  70 and s_score  60 and s_score  60 and s_score = 0 and s_score = 0 and s_score 



<ol>
<li>Query average scores and their ranks for students:
</li>
</ol>

<pre class="brush:php;toolbar:false">select a.s_id,
       @i:=@i+1 as 'No Gaps in Ranking',
       @k:=(case when @avg_score=a.avg_s then @k else @i end) as 'With Gaps in Ranking',
       @avg_score:=avg_s as 'Average Score'
from (select s_id, ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC) a,
     (select @avg_score:=0, @i:=0, @k:=0) b;
  1. Query records of the top three students in each subject:
select a.s_id, a.c_id, a.s_score from score a 
    left join score b on a.c_id = b.c_id and a.s_score 



<ol>
<li>Query the number of students enrolled in each course:
</li>
</ol>

<pre class="brush:php;toolbar:false">select c_id, count(s_id) from score group by c_id;
  1. Query the student ID and name of students who have taken exactly two courses:
select s_id, s_name from student 
    where s_id in (select s_id from score group by s_id having count(c_id) = 2);
  1. Query the number of male and female students:
select s_sex, count(s_sex) as Count from student group by s_sex;
  1. Query student information whose name contains the character "Tom":
select * from student where s_name like '%Tom%';
  1. Query list of students with the same name and gender, and count of such names:
select a.s_name, a.s_sex, count(*) as Count from student a  
    join student b on a.s_id != b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
    group by a.s_name, a.s_sex;
  1. Query list of students born in 1990:
select s_name from student where s_birth like '1990%';
  1. Query average scores for each course, ordered by average score descending, and course ID ascending if average scores are the same:
select c_id, round(avg(s_score), 2) as avg_score from score group by c_id order by avg_score desc, c_id asc;
  1. Query student ID, name, and average score of students with average score >= 85:
select a.s_id, b.s_name, round(avg(a.s_score), 2) as avg_score from score a
    left join student b on a.s_id = b.s_id group by s_id having avg_score >= 85;
  1. Query names and scores of students who scored less than 60 in the course "mathematics":
select a.s_name, b.s_score from student a 
    join score b on a.s_id = b.s_id 
    where b.c_id = (select c_id from course where c_name = 'mathematics') 
    and b.s_score 



<ol>
<li>Query course-wise scores and total scores of all students:
</li>
</ol>

<pre class="brush:php;toolbar:false">select a.s_id, a.s_name,
    sum(case c.c_name when 'history' then b.s_score else 0 end) as 'history',
    sum(case c.c_name when 'mathematics' then b.s_score else 0 end) as 'mathematics',
    sum(case c.c_name when 'Politics' then b.s_score else 0 end) as 'Politics',
    sum(b.s_score) as 'Total score'
from student a 
left join score b on a.s_id = b.s_id 
left join course c on b.c_id = c.c_id 
group by a.s_id, a.s_name;
  1. Query names, course names, and scores of students scoring above 70 in any course:
select a.s_name, b.c_name, c.s_score from student a 
    left join score c on a.s_id = c.s_id 
    left join course b on c.c_id = b.c_id 
    where c.s_score >= 70;
  1. Query courses where students failed:
select a.s_id, a.c_id, b.c_name, a.s_score from score a 
    left join course b on a.c_id = b.c_id 
    where a.s_score 



<ol>
<li>Query student ID and name of students who scored above 80 in course '01':
</li>
</ol>

<pre class="brush:php;toolbar:false">select a.s_id, b.s_name from score a 
    left join student b on a.s_id = b.s_id 
    where a.c_id = '01' and a.s_score > 80;
  1. Count number of students in each course:
select count(*) from score group by c_id;
  1. Query information of the highest scoring student in courses taught by teacher "Tom": -- Get teacher ID
select c_id from course c, teacher d where c.t_id = d.t_id and d.t_name = 'Tom';

-- Get maximum score (could have ties)

select max(s_score) from score where c_id = '02';

-- Get information

select a.*, b.s_score, b.c_id, c.c_name from student a 
    left join score b on a.s_id = b.s_id 
    left join course c on b.c_id = c.c_id 
    where b.c_id = (select c_id from course c, teacher d where c.t_id = d.t_id and d.t_name = 'Tom')
    and b.s_score in (select max(s_score) from score where c_id = '02');
  1. Query student ID, course ID, and score where different courses have the same score:
select distinct b.s_id, b.c_id, b.s_score from score a, score b 
    where a.c_id != b.c_id and a.s_score = b.s_score;
  1. Query top two scores for each course:
select a.s_id, a.c_id, a.s_score from score a 
    where (select count(1) from score b where b.c_id = a.c_id and b.s_score >= a.s_score) 



<ol>
<li>Count number of students enrolled in each course (courses with more than 5 students):
</li>
</ol>

<pre class="brush:php;toolbar:false">select c_id, count(*) as total from score group by c_id having total > 5 order by total, c_id asc;
  1. Query student IDs who have enrolled in at least two courses:
select s_id, count(*) as sel from score group by s_id having sel >= 2;
  1. Query information of students who have enrolled in all courses:
select * from student where s_id in (select s_id from score group by s_id having count(*) = (select count(*) from course));
  1. Query age of each student: -- Calculate age based on birthdate; subtract one if current month/day is before birthdate's month/day
select s_birth, (date_format(now(), '%Y') - date_format(s_birth, '%Y') - 
    (case when date_format(now(), '%m%d') > date_format(s_birth, '%m%d') then 0 else 1 end)) as age
    from student;
  1. Query students whose birthday is this week:
select * from student where week(date_format(now(), '%Y%m%d')) = week(s_birth);
  1. Query students whose birthday is next week:
select * from student where week(date_format(now(), '%Y%m%d')) + 1 = week(s_birth);
  1. Query students whose birthday is this month:
select * from student where month(date_format(now(), '%Y%m%d')) = month(s_birth);
  1. Query students whose birthday is next month:
select * from student where month(date_format(now(), '%Y%m%d')) + 1 = month(s_birth);

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Quels sont les différents moteurs de stockage disponibles dans MySQL?Quels sont les différents moteurs de stockage disponibles dans MySQL?Apr 26, 2025 am 12:27 AM

MySQLoffersVariousStorageEngines, chacune, chacun desCasses: 1) InnodbisidealForApplications newedingacidcompenciance and highcurrency, soutenant les transactions et manqueurs

Quelles sont les vulnérabilités de sécurité communes dans MySQL?Quelles sont les vulnérabilités de sécurité communes dans MySQL?Apr 26, 2025 am 12:27 AM

Les vulnérabilités de sécurité courantes dans MySQL incluent l'injection SQL, les mots de passe faibles, la configuration d'autorisation incorrecte et les logiciels unpudés. 1. L'injection SQL peut être évitée en utilisant des instructions de prétraitement. 2. Les mots de passe faibles peuvent être évités en utilisant de force des stratégies de mot de passe solides. 3. Une mauvaise configuration d'autorisation peut être résolue par examen régulier et ajustement des autorisations utilisateur. 4. Les logiciels unpus peuvent être corrigés en vérifiant et mise à jour régulièrement la version MySQL.

Comment pouvez-vous identifier les requêtes lentes dans MySQL?Comment pouvez-vous identifier les requêtes lentes dans MySQL?Apr 26, 2025 am 12:15 AM

L'identification des requêtes lentes dans MySQL peut être réalisée en activant les journaux de requête lents et en définissant des seuils. 1. Activer les journaux de requête lents et définir les seuils. 2. Afficher et analyser les fichiers journaux de requête lente et utiliser des outils tels que MySqlDumpSlow ou Pt-Query-digest pour une analyse approfondie. 3. Optimisation des requêtes lentes peut être réalisée grâce à l'optimisation de l'index, à la réécriture de la requête et à l'évitement de l'utilisation de Select *.

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