fetch data from mysql database on the basis of leave types such as Exam, Unpaid, Annual

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轻奢々
轻奢々 2021-01-23 12:41

This is my Database table

:

This is my front end where i want display data:

What I ne

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  • 2021-01-23 13:09

    Ok, so, lets try this and see if it works.

    This is the schema I have some up with. Its very simple but supports your use case.

    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)
    );
    

    Some test data ...

    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);
    

    Dont focus too much on column names and definitions, I did not polish the schema at all as I don't know enough about your app to do this.

    Once the schema is made and data in there, this pretty simple query should do what you want.

    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
    

    Have a look and see what you think. Is this likely to be a high performance or heavily loaded application?

    EDIT

    This query is more involved and would probably have some performance drawbacks but may be more accurate.

    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;
    
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