这是我的数据库表
这是我需要显示数据的前端:
我需要的是员工离职记录基本的假期类型和半天全天。从左到右的员工姓名然后根据半天的年度计数= 1在db和假类型=年度等等和最后一个显示总列中的所有列和半日年度,检查和未付的乘以0.5。< / p>
我尝试了什么:
SELECT leaves.*, count( users.id )
FROM leaves
LEFT JOIN users ON users.id=leaves.employee
GROUP BY users.name
{{1}}
提前致谢,请帮忙。
答案 0 :(得分:1)
好的,那么,让我们试一下,看看它是否有效。
这是我有一些模式。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);
不要过多关注列名和定义,我根本没有完善架构,因为我对你的应用程序知之甚少。
一旦构建了架构并在那里创建了数据,这个非常简单的查询就可以做你想要的了。
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
看一看,看看你的想法。这可能是高性能或高负载的应用程序吗?
修改强>
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;