


Noms de table et champs (MySQL)
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'étudiantTableau des cours
Cours(c_id, c_name, t_id)
ID du cours, nom du cours, ID de l'enseignantTable des professeurs
Professeur(t_id, t_name)
ID de l'enseignant, nom de l'enseignantTableau des scores
Score(s_id, c_id, s_score)
ID étudiant, ID de cours, score
Test Data - Creating Tables
- 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`) );
- 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`) );
- Teacher Table
CREATE TABLE `Teacher`( `t_id` VARCHAR(20), `t_name` VARCHAR(20) NOT NULL DEFAULT '', PRIMARY KEY(`t_id`) );
- Score Table
CREATE TABLE `Score`( `s_id` VARCHAR(20), `c_id` VARCHAR(20), `s_score` INT(3), PRIMARY KEY(`s_id`,`c_id`) );
- 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');
- 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');
- Inserting Test Data into Teacher Table
INSERT INTO Teacher VALUES('01', 'Andrew'); INSERT INTO Teacher VALUES('02', 'Bethany'); INSERT INTO Teacher VALUES('03', 'Charlie');
- 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
- 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;
- 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;
- 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;
- Query the number of teachers with the surname "Smith"
SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE 'Smith%';
- 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' ) );
- 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' ) );
- 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';
- 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');
- 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' ) );
- 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
- 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')));
- 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;
- 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;
- 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;
- 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;
- 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;
- 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;
- 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);
- 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;
- 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;
- 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);
- Query the number of male and female students:
select s_sex, count(s_sex) as Count from student group by s_sex;
- Query student information whose name contains the character "Tom":
select * from student where s_name like '%Tom%';
- 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;
- Query list of students born in 1990:
select s_name from student where s_birth like '1990%';
- 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;
- 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;
- 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;
- 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;
- 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;
- Count number of students in each course:
select count(*) from score group by c_id;
- 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');
- 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;
- 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;
- 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;
- 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));
- 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;
- Query students whose birthday is this week:
select * from student where week(date_format(now(), '%Y%m%d')) = week(s_birth);
- Query students whose birthday is next week:
select * from student where week(date_format(now(), '%Y%m%d')) + 1 = week(s_birth);
- Query students whose birthday is this month:
select * from student where month(date_format(now(), '%Y%m%d')) = month(s_birth);
- 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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MySQL est un système de gestion de la base de données relationnel open source, principalement utilisé pour stocker et récupérer les données rapidement et de manière fiable. Son principe de travail comprend les demandes des clients, la résolution de requête, l'exécution des requêtes et les résultats de retour. Des exemples d'utilisation comprennent la création de tables, l'insertion et la question des données et les fonctionnalités avancées telles que les opérations de jointure. Les erreurs communes impliquent la syntaxe SQL, les types de données et les autorisations, et les suggestions d'optimisation incluent l'utilisation d'index, les requêtes optimisées et la partition de tables.

MySQL est un système de gestion de base de données relationnel open source adapté au stockage, à la gestion, à la requête et à la sécurité des données. 1. Il prend en charge une variété de systèmes d'exploitation et est largement utilisé dans les applications Web et autres domaines. 2. Grâce à l'architecture client-serveur et à différents moteurs de stockage, MySQL traite efficacement les données. 3. L'utilisation de base comprend la création de bases de données et de tables, d'insertion, d'interrogation et de mise à jour des données. 4. L'utilisation avancée implique des requêtes complexes et des procédures stockées. 5. Les erreurs courantes peuvent être déboguées par le biais de la déclaration Explication. 6. L'optimisation des performances comprend l'utilisation rationnelle des indices et des instructions de requête optimisées.

MySQL est choisi pour ses performances, sa fiabilité, sa facilité d'utilisation et son soutien communautaire. 1.MySQL fournit des fonctions de stockage et de récupération de données efficaces, prenant en charge plusieurs types de données et opérations de requête avancées. 2. Adoptez l'architecture client-serveur et plusieurs moteurs de stockage pour prendre en charge l'optimisation des transactions et des requêtes. 3. Facile à utiliser, prend en charge une variété de systèmes d'exploitation et de langages de programmation. 4. Avoir un solide soutien communautaire et fournir des ressources et des solutions riches.

Les mécanismes de verrouillage d'InnoDB incluent des verrous partagés, des verrous exclusifs, des verrous d'intention, des verrous d'enregistrement, des serrures d'écart et des mèches suivantes. 1. Le verrouillage partagé permet aux transactions de lire des données sans empêcher d'autres transactions de lire. 2. Lock exclusif empêche les autres transactions de lire et de modifier les données. 3. Le verrouillage de l'intention optimise l'efficacité de verrouillage. 4. Enregistrement de l'indice de verrouillage d'enregistrement. 5. Écart d'enregistrement de l'indice des verrous de verrouillage de l'espace. 6. Le verrouillage de la touche suivante est une combinaison de verrouillage des enregistrements et de verrouillage de l'écart pour garantir la cohérence des données.

Les principales raisons des mauvaises performances de requête MySQL incluent le non-utilisation d'index, la mauvaise sélection du plan d'exécution par l'optimiseur de requête, la conception de table déraisonnable, le volume de données excessif et la concurrence de verrouillage. 1. Aucun indice ne provoque une requête lente et l'ajout d'index ne peut améliorer considérablement les performances. 2. Utilisez la commande Expliquez pour analyser le plan de requête et découvrez l'erreur Optimizer. 3. Reconstruire la structure de la table et l'optimisation des conditions de jointure peut améliorer les problèmes de conception de la table. 4. Lorsque le volume de données est important, les stratégies de partitionnement et de division de table sont adoptées. 5. Dans un environnement de concurrence élevé, l'optimisation des transactions et des stratégies de verrouillage peut réduire la concurrence des verrous.

Dans l'optimisation de la base de données, les stratégies d'indexation doivent être sélectionnées en fonction des exigences de requête: 1. Lorsque la requête implique plusieurs colonnes et que l'ordre des conditions est fixe, utilisez des index composites; 2. Lorsque la requête implique plusieurs colonnes mais que l'ordre des conditions n'est pas fixe, utilisez plusieurs index mono-colonnes. Les index composites conviennent à l'optimisation des requêtes multi-colonnes, tandis que les index mono-colonnes conviennent aux requêtes à colonne unique.

Pour optimiser la requête lente MySQL, SlowQueryLog et Performance_Schema doivent être utilisées: 1. Activer SlowQueryLog et définir des seuils pour enregistrer la requête lente; 2. Utilisez Performance_schema pour analyser les détails de l'exécution de la requête, découvrir les goulots d'étranglement des performances et optimiser.

MySQL et SQL sont des compétences essentielles pour les développeurs. 1.MySQL est un système de gestion de base de données relationnel open source, et SQL est le langage standard utilisé pour gérer et exploiter des bases de données. 2.MySQL prend en charge plusieurs moteurs de stockage via des fonctions de stockage et de récupération de données efficaces, et SQL termine des opérations de données complexes via des instructions simples. 3. Les exemples d'utilisation comprennent les requêtes de base et les requêtes avancées, telles que le filtrage et le tri par condition. 4. Les erreurs courantes incluent les erreurs de syntaxe et les problèmes de performances, qui peuvent être optimisées en vérifiant les instructions SQL et en utilisant des commandes Explication. 5. Les techniques d'optimisation des performances incluent l'utilisation d'index, d'éviter la numérisation complète de la table, d'optimiser les opérations de jointure et d'améliorer la lisibilité du code.


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