我们有一个员工满意度调查系统,数据库中有一张表:
survey_scores
表:存储员工对公司的满意度打分。表结构如下:
-- 调查打分表
CREATE TABLE survey_scores (
score_id INT AUTO_INCREMENT PRIMARY KEY,
department VARCHAR(50), -- 部门
employee_id INT, -- 员工ID
score DECIMAL(5, 2) -- 满意度打分(0-10分)
);
插入一些测试数据:
-- 插入调查打分数据
INSERT INTO survey_scores (department, employee_id, score) VALUES
('HR', 1, 8.5),
('HR', 2, 9.0),
('HR', 3, 7.0),
('HR', 4, 8.0),
('HR', 5, 9.5),
('Engineering', 6, 7.5),
('Engineering', 7, 8.0),
('Engineering', 8, 6.5),
('Engineering', 9, 9.0),
('Engineering', 10, 8.5),
('Sales', 11, 6.0),
('Sales', 12, 7.0),
('Sales', 13, 6.5),
('Sales', 14, 8.0),
('Sales', 15, 7.5);
编写一个 SQL 查询,实现以下需求:
department
)分组,去除每个部门的最高分和最低分。department
(部门)avg_score
(去除最高分和最低分后的平均分)上述数据,查询结果类似于:
department | avg_score |
---|---|
HR | 8.17 |
Engineering | 8.00 |
Sales | 7.00 |
HR 部门:
[7.0, 8.0, 8.5, 9.0, 9.5]
[8.0, 8.5, 9.0]
(8.0 + 8.5 + 9.0) / 3 = 8.17
标记每个部门的最高分和最低分:
ROW_NUMBER()
对每个部门的打分按升序和降序排名,标记最高分和最低分。过滤掉每个部门的最高分和最低分:
计算每个部门的平均分:
-- 第一步:标记每个部门的最高分和最低分
WITH ranked_scores AS (
SELECT
department,
score,
ROW_NUMBER() OVER (PARTITION BY department ORDER BY score ASC) AS asc_rank,
ROW_NUMBER() OVER (PARTITION BY department ORDER BY score DESC) AS desc_rank
FROM
survey_scores
)
-- 第二步:过滤掉每个部门的最高分和最低分,计算平均分
SELECT
department,
AVG(score) AS avg_score
FROM
ranked_scores
WHERE
asc_rank > 1 AND desc_rank > 1 -- 去除最高分和最低分
GROUP BY
department;
上述数据,查询结果:
department | avg_score |
---|---|
HR | 8.17 |
Engineering | 8.00 |
Sales | 7.00 |
窗口函数:
ROW_NUMBER()
对每个部门的打分进行排名。子查询和 CTE(Common Table Expressions):
WITH
子句将复杂查询分解为多个步骤,提高可读性。不支持窗口函数和CTS
找到每个部门的最高分和最低分:
GROUP BY
和 MAX()
、MIN()
聚合函数找到每个部门的最高分和最低分。过滤掉每个部门的最高分和最低分:
计算每个部门的平均分:
-- 第一步:找到每个部门的最高分和最低分
SELECT
s.department,
AVG(s.score) AS avg_score
FROM
survey_scores s
JOIN (
SELECT
department,
MAX(score) AS max_score, -- 最高分
MIN(score) AS min_score -- 最低分
FROM
survey_scores
GROUP BY
department
) min_max_scores ON s.department = min_max_scores.department
WHERE
s.score != min_max_scores.max_score AND s.score != min_max_scores.min_score -- 去除最高分和最低分
GROUP BY
s.department;