php - 根据考试、无薪、年度等休假类型从 mysql 数据库中获取数据

Question

This is my Database table

:

This is my front end where i want display data:

我需要的是基于休假类型和半天全天的员工休假记录。从左到右员工姓名然后年度计数基于 half day = 1 in db 和 leave type = annual 依此类推，最后显示 total 列和 half_day annual，exam 和 unpaid 中所有列的一些列乘以 0.5。

我尝试过的:

SELECT users.name, leave_type, count( leaves.id )
  FROM leaves
       INNER JOIN users ON users.id = leaves.employee
 GROUP BY leaves.leave_type, users.name

---

SELECT leaves.*, count( users.id ) 
  FROM leaves 
       LEFT JOIN users ON users.id=leaves.employee 
 GROUP BY users.name

在此先感谢，请帮忙。

Answer 1

好的，那么，让我们试试这个，看看它是否有效。

这是我掌握的架构。它非常简单，但支持您的用例。

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)
);

一些测试数据...

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);

不要过分关注列名称和定义，我根本没有完善架构，因为我对您的应用了解不够，无法做到这一点。

一旦创建了架构并在其中存储了数据，这个非常简单的查询就可以完成您想要的操作。

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

看看你的想法。这可能是一个高性能或重负载的应用程序吗？

编辑

此查询涉及更多，可能会有一些性能缺陷，但可能更准确。

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;