459 lines
13 KiB
Markdown
459 lines
13 KiB
Markdown
# SQL窗口函数
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### 一. 什么是窗口函数
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#### 基本含义
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窗口限定一个范围，它可以理解为满足某些条件的记录集合，窗口函数也就是在窗口范围内执行的函数。
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#### 基本语法
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窗口函数有over关键字，指定函数执行的范围，可分为三部分：分组子句(partition by)，排序子句(order by)，窗口子句(rows)
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```sql
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<函数名> over (partition by <分组的列> order by <排序的列> rows between <起始行> and <终止行>)
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```
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**注意Mysql8才支持窗口函数**
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#### 演示表格
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| cid（班级id) | sname（学生姓名） | score（分数） |
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| ------------ | ----------------- | ------------- |
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| 001 | 张三 | 78 |
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| 001 | 李四 | 82 |
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| 002 | 小明 | 90 |
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| 001 | 王五 | 67 |
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| 002 | 小红 | 85 |
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| 002 | 小刚 | 62 |
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#### 演示脚本
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```sql
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CREATE TABLE SQL_5 (
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cid varchar(4),
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sname varchar(4),
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score int
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);
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insert into SQL_5 (cid, sname, score) values ('001', '张三', 78);
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insert into SQL_5 (cid, sname, score) values ('001', '李四', 82);
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insert into SQL_5 (cid, sname, score) values ('002', '小明', 90);
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insert into SQL_5 (cid, sname, score) values ('001', '王五', 67);
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insert into SQL_5 (cid, sname, score) values ('002', '小红', 85);
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insert into SQL_5 (cid, sname, score) values ('002', '小刚', 62);
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```
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### 二. 窗口的确定
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例子：
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```sql
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select *, sum(score) over (partition by cid order by score rows between unbounded preceding and unbounded following) as '班级总分' from SQL_5;
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```
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#### 分组子句(partition by)
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不分组可以写成partition by null或者直接不写
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后面可以跟多个列， 如 partition by cid, sname
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**注意 partition by与group by的区别**
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1）前者不会压缩行数但是后者会
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2）后者只能选取分组的列和聚合的列
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也就是说group by 后生成的结果集与原表的行数和列数都不同
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#### 排序子句(order by)
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不排序可以写成order by null 或者直接不写
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asc或不写表示升序，desc表示降序
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后面可以跟多个列， 如 order by cid, sname
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#### 窗口子句(rows)
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窗口子句的描述
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1) 起始行: N preceding/unbounded preceding
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2) 当前行: current row
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3) 终止行: N following/unbounded following
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举例：
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rows between unbounded preceding and current row 从之前所有的行到当前行
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rows between 2 preceding and current row 从前面两行到当前行
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rows between current row and unbounded following 从当前行到之后所有的行
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rows between current row and 1following 从当前行到后面一行
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**注意：**
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**排序子句后面缺少窗口子句，窗口规范默认是 rows between unbounded preceding and current row**
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**排序子句和窗口子句都缺失，窗口规范默认是 rows between unbounded preceding and unbounded following**
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#### 总体流程
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1） 通过partition by 和 order by 子句确定大窗口（ 定义出上界unbounded preceding和下界unbounded following）
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2） 通过row 子句针对每一行数据确定小窗口（滑动窗口）
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3） 对每行的小窗口内的数据执行函数并生成新的列
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### 三. 函数分类
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#### 排序类
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rank, dense_rank, row_number
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```sql
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-- 【排序类】
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-- 按班级分组后打上序号 不考虑并列
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select *, row_number() over (partition by cid order by score desc) as '不可并列排名' from SQL_5;
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-- 按班级分组后作跳跃排名 考虑并列
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select *, rank() over (partition by cid order by score desc) as '跳跃可并列排名' from SQL_5;
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-- 按班级分组后作连续排名 考虑并列
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select *, dense_rank() over (partition by cid order by score desc) as '连续可并列排名' from SQL_5;
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-- 合并起来对比
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select *, row_number() over (partition by cid order by score desc) as '不可并列排名' ,
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rank() over (partition by cid order by score desc) as '跳跃可并列排名',
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dense_rank() over (partition by cid order by score desc) as '连续可并列排名'
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from SQL_5;
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```
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#### 聚合类
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sum. avg, count, max, min
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```sql
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-- 【聚合类】
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-- 让同一班级每个学生都知道班级总分是多少
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select *, sum(score) over (partition by cid) as '班级总分' from SQL_5;
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-- 或者可以写成
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select *, sum(score) over (partition by cid rows between unbounded preceding and unbounded following) as '班级总分' from SQL_5;
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-- 计算同一班级，每个同学和比他分数低的同学的累计总分是多少
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select *, sum(score) over (partition by cid order by score) '累加分数' from SQL_5;
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-- 或者可以写成 其中rows between ... and 是规定窗口大小
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select *, sum(score) over (partition by cid order by score rows between unbounded preceding and current row) as '累加分数' from SQL_5;
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```
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#### 跨行类
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lag, lead
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```sql
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-- 【跨行类】
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-- lag/lead 函数 参数1:比较的列 参数2: 偏移量 参数3:找不到的默认值
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-- 同一班级内，成绩比自己低一名的分数是多少
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select *, lag(score, 1) over (partition by cid order by score) as '低一名的分数' from SQL_5;
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-- 或者写成
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select *, lag(score, 1, 0) over (partition by cid order by score) as '低一名的分数' from SQL_5;
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-- 同一班级内，成绩比自己高2名的分数是多少
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select *, lead(score, 2) over (partition by cid order by score) as '高两名的分数' from SQL_5;
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```
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### 四. 相关题目
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#### 表格
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| cid | sname | course | score |
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| ---- | ----- | ------ | ----- |
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| 001 | 张三 | 语文 | 78 |
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| 002 | 小刚 | 语文 | 71 |
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| 001 | 李四 | 数学 | 56 |
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| 002 | 小明 | 数学 | 54 |
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| ... | ... | ... | ... |
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#### 脚本
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```sql
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CREATE TABLE SQL_6 (
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cid varchar(4),
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sname varchar(4),
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course varchar(10),
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score int
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);
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insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '语文', 78);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '语文', 71);
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insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '数学', 56);
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insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '数学', 97);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '数学', 54);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '数学', 67);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '数学', 82);
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insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '语文', 80);
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insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '数学', 77);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '语文', 58);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '语文', 87);
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insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '语文', 60);
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insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '英语', 66);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '英语', 50);
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insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '地理', 59);
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insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '地理', 88);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '地理', 45);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '地理', 66);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '地理', 82);
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insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '英语', 81);
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insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '地理', 77);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '英语', 55);
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insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '英语', 87);
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insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '英语', 61);
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```
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##### 分组内topN
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问题1：求出每个学生成绩最高的三条记录
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```sql
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select * from
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(
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select *, row_number() over (partition by sname order by score desc) as rn from SQL_6
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) temp
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where rn <= 3
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```
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###### 公式:
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```sql
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select * from
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(
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select *, row_number() over (partition by 分组列 order by 比较列) as rn from table
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) as tmp
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where rn <= N;
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```
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##### 汇总分析
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问题2：找出每门课程都高于班级课程平均分的学生
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可以拆解成以下几个问题：
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1）求出每个班级，每门课程的平均分
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```sql
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with
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-- 1) 求出每个班级，每门课程的平均分
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t1 as
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(
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select *,
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avg(score) over (partition by cid, course) as 'avg'
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from SQL_6
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),
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```
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2）将学生每门课程的成绩与所在班级的对应课程平均分相减，结果大于0就说明该学生的这门成绩高于课程平均分
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```sql
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t2 as (
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select *,
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score - avg as 'del'
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from t1
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)
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```
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3）“找出每门课程都高于班级课程平均分的学生”说明对于学生来说，最小的“相减结果”都是大于0的
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```sql
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select sname from t2
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group by sname
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having min(del) > 0;
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```
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合并后的SQL语句
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```sql
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with
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t1 as
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(
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select *,
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avg(score) over (partition by cid, course) as 'avg'
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from SQL_6
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),
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t2 as (
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select *,
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score - avg as 'del'
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from t1
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)
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select sname from t2
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group by sname
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having min(del) > 0;
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-- 或者
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select sname from (
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select *,
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score - avg as 'del'
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from (
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select *,
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avg(score) over (partition by cid, course) as 'avg'
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from SQL_6
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) t1
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) t2
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group by sname
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having min(del) > 0;
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```
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#### 表格
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| empno | ename | hire_date | salary | dept_no |
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| ----- | ----- | ---------- | ------ | ------- |
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| 001 | Adam | 2018-03-01 | 1000 | A |
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| 002 | Bill | 2021-03-01 | 1200 | A |
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| 003 | Cindy | 2016-03-01 | 1500 | A |
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| 004 | Danny | 2020-03-01 | 5000 | A |
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| 005 | Eason | 2020-03-01 | 4000 | B |
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| 006 | Fred | 2018-03-01 | 3500 | B |
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| 007 | Gary | 2017-03-01 | 1800 | B |
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| 008 | Hugo | 2020-03-01 | 2500 | B |
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#### 脚本
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```sql
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CREATE TABLE SQL_7 (
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empno varchar(4),
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ename varchar(10),
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hire_date varchar(10),
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salary int,
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dept_no varchar(2)
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);
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('001', 'Adam', '2018-03-01', 1000, 'A');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('002', 'Bill', '2021-03-01', 1200, 'A');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('003', 'Cindy', '2016-03-01', 1500, 'A');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('004', 'Danny', '2020-03-01', 5000, 'A');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('005', 'Eason', '2020-03-01', 4000, 'B');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('006', 'Fred', '2018-03-01', 3500, 'B');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('007', 'Gary', '2017-03-01', 1800, 'B');
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insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('008', 'Hugo', '2020-03-01', 4500, 'B');
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select * from SQL_7;
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```
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##### 分组内topN
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问题一：求出每个部门工资最高的前三名员工
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```sql
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select * from
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(
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select *, row_number() over (partition by dept_no order by salary desc) as rn from SQL_7
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) as tmp
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where rn <= 3;
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```
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##### 汇总分析
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问题二：计算这些员工的工资占所属部门总工资的百分比
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```sql
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with
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t1 as (
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select * , sum(salary) over (partition by dept_no) as 'sum_sal' from SQL_7
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),
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t2 as (
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select *, round(salary*100/sum_sal,2) as 'percentage' from t1
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)
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select * from t2;
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```
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问题三：对各部门员工的工资进行从小到大排序，排名前30%为低层，30%-80%为中层，高于80%为高层，并打上标签
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```label
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with
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t1 as (
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select * , row_number() over (partition by dept_no order by salary) as cnt,
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count(empno) over (partition by dept_no) as 'sum' from SQL_7
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),
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t2 as (
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select *, round(cnt/sum,2) as 'percentage' from t1
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),
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t3 as (
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select *, case when percentage <= 0.3 then '低层'
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when percentage <= 0.8 then '中层'
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when percentage <= 1 then '高层' end as 'label'
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from t2
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)
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select empno, ename, hire_date, salary, dept_no, label from t3;
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```
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问题四：统计每年入职总数以及截至本年累计入职总人数（本年总入职人数 + 本年之前所有年的总入职人数之和）
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```sql
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select year(hire_date) as hire_year, count(empno) as cnt
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from SQL_7
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group by year(hire_date) order by hire_year;
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```
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```sql
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with t1 as (
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select year(hire_date) as hire_year, count(empno) as cnt from SQL_7 group by year(hire_date) order by hire_year
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)
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select *, sum(cnt) over(partition by null rows between unbounded preceding and current row) as sum from t1;
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```
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### 五. 技巧
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1）分组内topN公式
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```sql
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select * from
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(
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select *, row_number() over (partition by 分组列 order by 比较列) as rn from table
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) as tmp
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where rn <= N;
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```
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2） 窗口函数 -> 生成辅助列（相当于高级语言的临时变量)
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3) with 语句 -> 生成临时表（相当于高级语言的局部方法)
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把复杂的问题拆分成多个子问题并用临时表去表达
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