Exam、Unpaid、Annualなどの休暇タイプに基づいてmysqlデータベースからデータをフェッチします

では、これを試して、機能するかどうかを確認しましょう。

これは私が持っているスキーマです。非常にシンプルですが、ユースケースをサポートします。

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)
);

いくつかのテストデータ...

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);

列名と定義にあまり焦点を当てないでください。これを行うためのアプリについて十分な知識がないため、スキーマをまったく磨きませんでした。

スキーマが作成され、そこにデータが含まれると、この非常に単純なクエリで必要な処理が実行されます。

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

見て、あなたがどう思うか見てください。これは、高性能または高負荷のアプリケーションである可能性がありますか？

編集

このクエリはより複雑で、おそらくパフォーマンス上の欠点がいくつかありますが、より正確な場合があります。

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;