récupérer les données de la base de données mysql sur la base des types de congés tels que examen, non rémunéré, annuel

Ok, alors, essayons ceci et voyons si cela fonctionne.

C'est le schéma que j'en ai. C'est très simple mais prend en charge votre cas d'utilisation.

CREATE TABLE employees (
  id int unsigned auto_increment,
  name varchar(255),
  PRIMARY KEY(id)
);

CREATE TABLE leave_type (
  id int unsigned auto_increment,
  name varchar(255),
  PRIMARY KEY(id)
);

CREATE TABLE leave_log (
  id int unsigned auto_increment,
  leave_type_id int unsigned,
  employee_id int unsigned,
  is_full_day int unsigned,
  is_half_day int unsigned,
  PRIMARY KEY(id)
);

Quelques données de test...

INSERT INTO employees VALUES (14, 'Lisa'), (15, 'Homer'), (13, 'Bart');
INSERT INTO leave_type VALUES (1, 'Annual'), (2, 'Unpaid'), (3, 'Exam');
INSERT INTO leave_log VALUES (NULL, 3, 14, 1, 0), (NULL, 1, 14, 1, 0), (NULL, 1, 14, 0, 1), (NULL, 1, 14, 0, 1);
INSERT INTO leave_log VALUES (NULL, 2, 15, 0, 1);
INSERT INTO leave_log VALUES (NULL, 3, 13, 1, 0), (NULL, 1, 13, 1, 0);

Ne vous concentrez pas trop sur les noms de colonnes et les définitions, je n'ai pas du tout peaufiné le schéma car je ne connais pas assez votre application pour le faire.

Une fois le schéma créé et les données dedans, cette requête assez simple devrait faire ce que vous voulez.

SELECT e.name, SUM(annual.is_half_day), SUM(unpaid.is_half_day), SUM(exam.is_half_day), 
       SUM(annual.is_full_day), SUM(unpaid.is_full_day), SUM(exam.is_full_day)
  FROM employees e
       LEFT JOIN leave_log annual ON annual.leave_type_id = 1 AND annual.employee_id = e.id
       LEFT JOIN leave_log unpaid ON unpaid.leave_type_id = 2 AND unpaid.employee_id = e.id
       LEFT JOIN leave_log exam ON exam.leave_type_id = 3 AND exam.employee_id = e.id
 GROUP BY e.id

Jetez un œil et voyez ce que vous en pensez. S'agit-il probablement d'une application très performante ou très chargée ?

MODIFIER

Cette requête est plus complexe et aurait probablement des inconvénients en termes de performances, mais peut être plus précise.

SELECT e.name, e.id,     
       IFNULL(annual_half.total, 0) annual_half,
       IFNULL(unpaid_half.total, 0) unpaid_half,
       IFNULL(exam_half.total, 0) exam_half,
       IFNULL(annual_full.total, 0) annual_full,
       IFNULL(unpaid_full.total, 0) unpaid_full,
       IFNULL(exam_full.total, 0) exam_full
  FROM employees e    
       LEFT JOIN ( SELECT SUM(is_full_day) as total, employee_id, leave_type_id FROM leave_log WHERE is_full_day = 1 AND leave_type_id = 1 GROUP BY 3, 2) annual_full ON annual_full.employee_id = e.id    
       LEFT JOIN ( SELECT SUM(is_full_day) as total, employee_id, leave_type_id FROM leave_log WHERE is_full_day = 1 AND leave_type_id = 2 GROUP BY 3, 2) unpaid_full ON unpaid_full.employee_id = e.id    
       LEFT JOIN ( SELECT SUM(is_full_day) as total, employee_id, leave_type_id FROM leave_log WHERE is_full_day = 1 AND leave_type_id = 3 GROUP BY 3, 2) exam_full ON exam_full.employee_id = e.id    
       LEFT JOIN ( SELECT SUM(is_half_day) as total, employee_id, leave_type_id FROM leave_log WHERE is_half_day = 1 AND leave_type_id = 1 GROUP BY 3, 2) annual_half ON annual_half.employee_id = e.id    
       LEFT JOIN ( SELECT SUM(is_half_day) as total, employee_id, leave_type_id FROM leave_log WHERE is_half_day = 1 AND leave_type_id = 2 GROUP BY 3, 2) unpaid_half ON unpaid_half.employee_id = e.id    
       LEFT JOIN ( SELECT SUM(is_half_day) as total, employee_id, leave_type_id FROM leave_log WHERE is_half_day = 1 AND leave_type_id = 3 GROUP BY 3, 2) exam_half ON exam_half.employee_id = e.id
 GROUP BY 1;