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+# SQL窗口函数
+
+### 一. 什么是窗口函数
+
+#### 基本含义
+
+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。
+
+#### 基本语法
+
+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)
+
+```sql
+<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>)
+```
+
+
+
+**注意Mysql8才支持窗口函数**
+
+
+
+#### 演示表格
+
+| cid（班级id) | sname（学生姓名） | score（分数） |
+| ------------ | ----------------- | ------------- |
+| 001          | 张三              | 78            |
+| 001          | 李四              | 82            |
+| 002          | 小明              | 90            |
+| 001          | 王五              | 67            |
+| 002          | 小红              | 85            |
+| 002          | 小刚              | 62            |
+
+#### 演示脚本
+
+```sql
+CREATE TABLE SQL_5 (
+       cid varchar(4),
+       sname varchar(4),
+       score int
+);
+
+insert into SQL_5 (cid, sname, score) values ('001', '张三', 78);
+insert into SQL_5 (cid, sname, score) values ('001', '李四', 82);
+insert into SQL_5 (cid, sname, score) values ('002', '小明', 90);
+insert into SQL_5 (cid, sname, score) values ('001', '王五', 67);
+insert into SQL_5 (cid, sname, score) values ('002', '小红', 85);
+insert into SQL_5 (cid, sname, score) values ('002', '小刚', 62);
+```
+
+
+
+### 二. 窗口的确定
+
+例子：
+
+```sql
+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;
+```
+
+#### 分组子句(partition by)
+
+不分组可以写成partition by null或者直接不写
+
+后面可以跟多个列， 如 partition by cid, sname
+
+**注意 partition by与group by的区别**
+
+1）前者不会压缩行数但是后者会
+
+2）后者只能选取分组的列和聚合的列
+
+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
+
+#### 排序子句(order by)
+
+不排序可以写成order by null 或者直接不写
+
+asc或不写表示升序，desc表示降序
+
+后面可以跟多个列， 如 order by cid, sname
+
+#### 窗口子句(rows)
+
+窗口子句的描述
+
+1) 起始行:   N preceding/unbounded preceding
+
+2) 当前行:   current row
+
+3) 终止行:   N following/unbounded following
+
+举例：
+
+rows between unbounded  preceding and current row  从之前所有的行到当前行
+
+rows between 2 preceding and current row  从前面两行到当前行
+
+rows between current row and unbounded following  从当前行到之后所有的行
+
+rows between current row and 1following  从当前行到后面一行
+
+**注意：**
+
+**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**
+
+**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**
+
+#### 总体流程
+
+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）
+
+2） 通过row 子句针对每一行数据确定小窗口（滑动窗口）
+
+3） 对每行的小窗口内的数据执行函数并生成新的列
+
+
+
+### 三. 函数分类
+
+#### 排序类
+
+rank, dense_rank, row_number
+
+```sql
+-- 【排序类】
+-- 按班级分组后打上序号 不考虑并列
+select *, row_number() over (partition by cid order by score desc) as '不可并列排名' from SQL_5;
+-- 按班级分组后作跳跃排名 考虑并列
+select *, rank() over (partition by cid order by score desc) as '跳跃可并列排名' from SQL_5;
+-- 按班级分组后作连续排名 考虑并列
+select *, dense_rank() over (partition by cid order by score desc) as '连续可并列排名' from SQL_5;
+-- 合并起来对比
+select *, row_number() over (partition by cid order by score desc) as '不可并列排名' ,
+          rank() over (partition by cid order by score desc) as '跳跃可并列排名',
+          dense_rank() over (partition by cid order by score desc) as '连续可并列排名'
+from SQL_5;
+```
+
+#### 聚合类
+
+sum. avg, count, max, min
+
+```sql
+-- 【聚合类】
+-- 让同一班级每个学生都知道班级总分是多少
+select *, sum(score) over (partition by cid) as '班级总分' from SQL_5;
+-- 或者可以写成
+select *, sum(score) over (partition by cid rows between unbounded preceding and unbounded following) as '班级总分' from SQL_5;
+
+-- 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
+select *, sum(score) over (partition by cid order by score) '累加分数' from SQL_5;
+-- 或者可以写成 其中rows between ... and 是规定窗口大小
+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;
+```
+
+#### 跨行类
+
+lag, lead
+
+```sql
+-- 【跨行类】
+-- 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
+-- 同一班级内，成绩比自己低一名的分数是多少
+select *, lag(score, 1) over (partition by cid order by score) as '低一名的分数' from SQL_5;
+-- 或者写成
+select *, lag(score, 1, 0) over (partition by cid order by score) as '低一名的分数' from SQL_5;
+
+-- 同一班级内，成绩比自己高2名的分数是多少
+select *, lead(score, 2) over (partition by cid order by score) as '高两名的分数' from SQL_5;
+```
+
+
+
+### 四. 相关题目
+
+#### 表格
+
+| cid  | sname | course | score |
+| ---- | ----- | ------ | ----- |
+| 001  | 张三  | 语文   | 78    |
+| 002  | 小刚  | 语文   | 71    |
+| 001  | 李四  | 数学   | 56    |
+| 002  | 小明  | 数学   | 54    |
+| ...  | ...   | ...    | ...   |
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_6 (
+       cid varchar(4),
+       sname varchar(4),
+       course varchar(10),
+       score int
+);
+
+insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '语文', 78);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '语文', 71);
+insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '数学', 56);
+insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '数学', 97);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '数学', 54);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '数学', 67);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '数学', 82);
+insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '语文', 80);
+insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '数学', 77);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '语文', 58);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '语文', 87);
+insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '语文', 60);
+insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '英语', 66);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '英语', 50);
+insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '地理', 59);
+insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '地理', 88);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '地理', 45);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小刚', '地理', 66);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '地理', 82);
+insert into SQL_6 (cid, sname, course, score) values ('001', '王五', '英语', 81);
+insert into SQL_6 (cid, sname, course, score) values ('001', '张三', '地理', 77);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小明', '英语', 55);
+insert into SQL_6 (cid, sname, course, score) values ('002', '小红', '英语', 87);
+insert into SQL_6 (cid, sname, course, score) values ('001', '李四', '英语', 61);
+```
+
+##### 分组内topN
+
+问题1：求出每个学生成绩最高的三条记录
+
+```sql
+select * from 
+(
+    select *, row_number() over (partition by sname order by score desc) as rn from SQL_6   
+) temp
+where rn <= 3
+```
+
+###### 公式:
+
+```sql
+select * from
+(
+	select *, row_number() over (partition by 分组列 order by 比较列) as rn from table
+) as tmp
+where rn <= N;
+```
+
+
+
+##### 汇总分析
+
+问题2：找出每门课程都高于班级课程平均分的学生
+
+可以拆解成以下几个问题：
+
+1）求出每个班级，每门课程的平均分
+
+```sql
+with
+-- 1) 求出每个班级，每门课程的平均分
+t1 as
+(
+select *,
+       avg(score) over (partition by cid, course) as 'avg'
+from SQL_6
+),
+```
+
+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
+
+```sql
+t2 as (
+    select *,
+           score - avg  as 'del'
+    from t1
+)
+```
+
+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
+
+```sql
+select sname from t2
+group by sname
+having min(del) > 0;
+```
+
+合并后的SQL语句
+
+```sql
+with
+t1 as
+(
+select *,
+       avg(score) over (partition by cid, course) as 'avg'
+from SQL_6
+),
+t2 as (
+    select *,
+           score - avg  as 'del'
+    from t1
+)
+select sname from t2
+group by sname
+having min(del) > 0;
+
+-- 或者
+select sname from (
+                      select *,
+                             score - avg  as 'del'
+                      from (
+                               select *,
+                                      avg(score) over (partition by cid, course) as 'avg'
+                               from SQL_6
+                           ) t1
+                  ) t2
+group by sname
+having min(del) > 0;
+```
+
+
+
+#### 表格
+
+| empno | ename | hire_date  | salary | dept_no |
+| ----- | ----- | ---------- | ------ | ------- |
+| 001   | Adam  | 2018-03-01 | 1000   | A       |
+| 002   | Bill  | 2021-03-01 | 1200   | A       |
+| 003   | Cindy | 2016-03-01 | 1500   | A       |
+| 004   | Danny | 2020-03-01 | 5000   | A       |
+| 005   | Eason | 2020-03-01 | 4000   | B       |
+| 006   | Fred  | 2018-03-01 | 3500   | B       |
+| 007   | Gary  | 2017-03-01 | 1800   | B       |
+| 008   | Hugo  | 2020-03-01 | 2500   | B       |
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_7 (
+       empno varchar(4),
+       ename varchar(10),
+       hire_date varchar(10),
+       salary int,
+       dept_no varchar(2)
+);
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('001', 'Adam', '2018-03-01', 1000, 'A');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('002', 'Bill', '2021-03-01', 1200, 'A');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('003', 'Cindy', '2016-03-01', 1500, 'A');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('004', 'Danny', '2020-03-01', 5000, 'A');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('005', 'Eason', '2020-03-01', 4000, 'B');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('006', 'Fred', '2018-03-01', 3500, 'B');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('007', 'Gary', '2017-03-01', 1800, 'B');
+insert into SQL_7 (empno, ename, hire_date, salary, dept_no) values ('008', 'Hugo', '2020-03-01', 4500, 'B');
+
+select * from SQL_7;
+```
+
+##### 分组内topN
+
+问题一：求出每个部门工资最高的前三名员工
+
+```sql
+select * from
+    (
+        select *, row_number() over (partition by dept_no order by salary desc) as rn from SQL_7
+    ) as tmp
+where rn <= 3;
+```
+
+
+
+##### 汇总分析
+
+问题二：计算这些员工的工资占所属部门总工资的百分比
+
+```sql
+with
+t1 as (
+    select * , sum(salary) over (partition by dept_no) as 'sum_sal' from SQL_7
+),
+t2 as (
+    select *, round(salary*100/sum_sal,2) as 'percentage' from t1
+)
+select * from t2;
+```
+
+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
+
+```label
+with
+    t1 as (
+        select * , row_number() over (partition by dept_no order by salary) as cnt,
+               count(empno) over (partition by dept_no) as 'sum' from SQL_7
+    ),
+    t2 as (
+        select *, round(cnt/sum,2) as 'percentage' from t1
+    ),
+    t3 as (
+        select *, case when percentage <= 0.3 then '低层'
+                       when percentage <= 0.8 then '中层'
+                       when percentage <= 1 then '高层' end as 'label'
+        from t2
+    )
+select empno, ename, hire_date, salary, dept_no, label from t3;
+```
+
+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）
+
+```sql
+select year(hire_date) as hire_year, count(empno) as cnt
+    from SQL_7
+    group by year(hire_date) order by hire_year;
+```
+
+```sql
+with t1 as (
+    select year(hire_date) as hire_year, count(empno) as cnt from SQL_7 group by year(hire_date) order by hire_year
+)
+
+select *, sum(cnt) over(partition by null rows between unbounded preceding and current row) as sum from t1;
+```
+
+
+
+### 五. 技巧
+
+1）分组内topN公式
+
+```sql
+select * from
+(
+	select *, row_number() over (partition by 分组列 order by 比较列) as rn from table
+) as tmp
+where rn <= N;
+```
+
+2） 窗口函数 -> 生成辅助列（相当于高级语言的临时变量)
+
+3)   with 语句 -> 生成临时表（相当于高级语言的局部方法)
+
+​       把复杂的问题拆分成多个子问题并用临时表去表达
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/SQL-行转列与列转行.md b/SQL-行转列与列转行.md
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+# SQL 讲解 —— 行转列 与 列转行
+
+## 行转列
+
+### 题目1
+
+#### 描述
+
+```
+name  subject  score
+张三    语文     78
+张三    数学     88
+张三    英语 	 98
+李四    语文     89
+李四    数学     76
+李四    英语     90
+王五    语文	 99
+王五	  数学	 66
+王五	  英语	 91
+
+name  语文  数学  英语
+张三   78   88    98
+李四   89   76    90
+王五   99   66    91
+```
+
+
+
+#### 脚本
+
+```sql
+create table SQL_1
+(
+    name varchar(20),
+    subject varchar(20),
+    score float
+);
+insert into SQL_1 (name, subject, score) values ('张三', '语文', 78);
+insert into SQL_1 (name, subject, score) values ('张三', '数学', 88);
+insert into SQL_1 (name, subject, score) values ('张三', '英语', 98);
+insert into SQL_1 (name, subject, score) values ('李四', '语文', 89);
+insert into SQL_1 (name, subject, score) values ('李四', '数学', 76);
+insert into SQL_1 (name, subject, score) values ('李四', '英语', 90);
+insert into SQL_1 (name, subject, score) values ('王五', '语文', 99);
+insert into SQL_1 (name, subject, score) values ('王五', '数学', 66);
+insert into SQL_1 (name, subject, score) values ('王五', '英语', 91);
+select * from SQL_1;
+```
+
+
+
+#### 解题步骤
+
+1） 确定分组列，转换列，数据列
+
+2） 生成伪列
+
+3） 做分组查询
+
+4） 选择合适的聚合函数
+
+
+
+#### 两步法
+
+##### 公式：
+
+```sql
+select 分组列,
+       聚合函数(m1) as 列名1,
+       聚合函数(m2) as 列名2,
+       聚合函数(m3) as 列名3
+from (select *,
+             case 转换列 when 转换列值1 then 数据列 else ... end as m1,
+             case 转换列 when 转换列值2 then 数据列 else ... end as m2,
+             case 转换列 when 转换列值3 then 数据列 else ... end as m3
+      from 表名) 临时表名
+group by 分组列;
+```
+
+##### 解题SQL
+
+```sql
+select name,
+       sum(m1) as 语文,
+       sum(m2) as 数学,
+       sum(m3) as 英语
+from (select *,
+             case subject when '语文' then score else 0 end as m1,
+             case subject when '数学' then score else 0 end as m2,
+             case subject when '英语' then score else 0 end as m3
+      from sql_1) tmp
+group by name;
+```
+
+
+
+#### 一步法
+
+##### 公式：
+
+```sql
+select 分组列，
+       聚合函数(case 转换列 when 转换列值1 then 数据列 else ... end) as 列名1,
+	   聚合函数(case 转换列 when 转换列值2 then 数据列 else ... end) as 列名2,
+	   聚合函数(case 转换列 when 转换列值3 then 数据列 else ... end) as 列名3
+	   ...
+from 表名
+group by 分组列;
+
+select 分组列，
+       聚合函数(case when 转换列=转换列值1 then 数据列 else ... end) as 列名1,
+	   聚合函数(case when 转换列=转换列值2 then 数据列 else ... end) as 列名2,
+	   聚合函数(case when 转换列=转换列值3 then 数据列 else ... end) as 列名3
+	   ...
+from 表名
+group by 分组列;
+```
+
+##### 解题SQL
+
+```sql
+select name,
+       sum(case subject when '语文' then score else 0 end) as 语文,
+       sum(case subject when '数学' then score else 0 end) as 数学,
+       sum(case subject when '英语' then score else 0 end) as 英语
+from sql_1
+group by name;
+
+select name,
+       sum(case when subject = '语文' then score else 0 end) as 语文,
+       sum(case when subject = '数学' then score else 0 end) as 数学,
+       sum(case when subject = '英语' then score else 0 end) as 英语
+from sql_1
+group by name;
+```
+
+
+
+### 题目2
+
+#### 描述
+
+```txt
+# 日期        结果
+# 2022-01-01  胜
+# 2022-01-01  胜
+# 2022-01-02  负
+# 2022-01-02  负
+# 2022-01-01  负
+# 2022-01-02  负
+# 2022-01-02  胜
+
+# 日期         胜 负
+# 2022-01-01  2  1
+# 2022-01-02  1  3
+```
+
+#### 脚本
+
+```sql
+create table SQL_2(
+   ddate varchar(10), result varchar(2)
+);
+
+insert into SQL_2 (ddate, result) values('2022-01-01','胜');
+insert into SQL_2 (ddate, result) values('2022-01-01','胜');
+insert into SQL_2 (ddate, result) values('2022-01-02','负');
+insert into SQL_2 (ddate, result) values('2022-01-02','负');
+insert into SQL_2 (ddate, result) values('2022-01-01','负');
+insert into SQL_2 (ddate, result) values('2022-01-02','负');
+insert into SQL_2 (ddate, result) values('2022-01-02','胜');
+select * from SQL_2;
+```
+
+#### 解题SQL
+
+```sql
+select ddate,
+       sum(case when result = '胜' then 1 else 0 end) as 胜,
+       sum(case when result = '负' then 1 else 0 end) as 负
+from sql_2
+group by ddate;
+```
+
+
+
+## 列转行
+
+### 题目3
+
+#### 描述
+
+```txt
+name  语文  数学  英语 
+张三   78   88    98
+李四   89   76    90
+王五   99   66    91 
+
+name  subject  score
+张三    语文     78
+张三    数学     88
+张三    英语 	 98
+李四    语文     89
+李四    数学     76
+李四    英语     90
+王五    语文	 99
+王五	  数学	 66
+王五	  英语	 91
+```
+
+
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_3 (
+	name varchar(20),
+    语文 float,
+    数学 float,
+    英语 float
+);
+
+insert into SQL_3 (name, '语文', '数学', '英语') values ('张三', 78, 88, 98);
+insert into SQL_3 (name, '语文', '数学', '英语') values ('李四', 89, 76, 90);
+insert into SQL_3 (name, '语文', '数学', '英语') values ('王五', 99, 66, 91);
+```
+
+
+
+#### 解题步骤
+
+1） 确定转换列，非转换列
+
+2） 生成新列
+
+3） 使用union或union all 进行合并
+
+4） 根据需要进行order by
+
+
+
+#### 公式
+
+```sql
+SELECT 非转换列, '转换列1' AS 新转换列名, 转换列1 AS 新数据列名 FROM 表名
+UNION ALL
+SELECT 非转换列, '转换列2' AS 新转换列名, 转换列2 AS 新数据列名 FROM 表名
+UNION ALL
+SELECT 非转换列, '转换列3' AS 新转换列名, 转换列3 AS 新数据列名 FROM 表名
+ORDER BY ...;
+```
+
+
+
+#### 解题SQL
+
+```sql
+SELECT name,'语文' AS subject,语文 AS score FROM SQL_3
+UNION ALL
+SELECT name,'数学' AS subject,数学 AS score FROM SQL_3
+UNION ALL
+SELECT name,'英语' AS subject,英语 AS score FROM SQL_3
+ORDER BY name ASC, subject DESC;
+```
+
+
+
+### 题目4
+
+#### 描述
+
+```txt
+Q1     Q2     Q3     Q4
+1000   2000   3000   4000
+
+季度   业绩
+Q1    1000
+Q2    2000
+Q3    3000
+Q4    4000
+```
+
+
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_4 (
+	Q1 int, Q2 int, Q3 int, Q4 int
+);
+
+insert into SQL_4 values (1000, 2000, 3000, 4000);
+```
+
+
+
+#### 解题SQL
+
+```sql
+SELECT  'Q1' AS 季度, Q1 AS 业绩 FROM SQL_4
+UNION ALL
+SELECT 'Q2' AS 季度, Q2 AS 业绩 FROM SQL_4
+UNION ALL
+SELECT 'Q3' AS 季度, Q3 AS 业绩 FROM SQL_4
+UNION ALL
+SELECT 'Q4' AS 季度, Q4 AS 业绩 FROM SQL_4
+ORDER BY 季度;
+```
+
+
+
+### 技巧：
+
+扩展列：select ... as 新列名
+
+减少列：直接不写
+
+扩展行：union/ union all
+
+减少行:  聚合函数
+
+
+
diff --git a/SQL窗口函数（二）.md b/SQL窗口函数（二）.md
new file mode 100644
index 0000000..aabfe46
--- /dev/null
+++ b/SQL窗口函数（二）.md
@@ -0,0 +1,231 @@
+# SQL窗口函数（二）—— 连续问题
+
+### 题目一
+
+#### 表格
+
+| user_id | login_date |
+| ------- | ---------- |
+| A       | 2022-09-02 |
+| A       | 2022-09-03 |
+| A       | 2022-09-04 |
+| B       | 2021-11-25 |
+| B       | 2021-12-31 |
+| C       | 2022-01-01 |
+| C       | 2022-04-04 |
+| C       | 2022-09-03 |
+| C       | 2022-09-04 |
+| C       | 2022-09-05 |
+| A       | 2022-09-03 |
+| D       | 2022-10-20 |
+| D       | 2022-10-21 |
+| A       | 2022-10-03 |
+| D       | 2022-10-22 |
+| D       | 2022-10-23 |
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_8
+(
+     user_id 	varchar(2),
+     login_date date
+);
+INSERT INTO SQL_8 (user_id,login_date)
+VALUES ('A', '2022-09-02'), ('A', '2022-09-03'), ('A', '2022-09-04'), ('B', '2021-11-25'),
+       ('B', '2021-12-31'), ('C', '2022-01-01'), ('C', '2022-04-04'), ('C', '2022-09-03'),
+       ('C', '2022-09-05'), ('C', '2022-09-04'), ('A', '2022-09-03'), ('D', '2022-10-20'),
+       ('D', '2022-10-21'), ('A', '2022-10-03'), ('D', '2022-10-22'), ('D', '2022-10-23');
+```
+
+#### 问题
+
+找出这张表中所有的连续3天登录用户
+
+#### 分析
+
+连续N天登录用户，要求数据行满足以下条件：
+
+1) userid 要相同，表示同一用户
+
+2) 同一用户每行记录以登录时间从小到大排序
+
+3) 后一行记录比前一行记录的登录时间多一天
+
+4) 数据行数大于等于N
+
+#### 解答
+
+```sql
+-- 方法一
+with t1 as (
+    select distinct user_id, login_date from SQL_8
+),
+t2 as (
+    select *, row_number() over (partition by user_id order by login_date) as rn from t1
+),
+t3 as (
+    select *, DATE_SUB(login_date, interval rn day) as sub from t2
+)
+select distinct user_id from t3 group by user_id, sub having count(user_id) >= 3;
+
+-- 方法二
+with t1 as (
+    select distinct user_id, login_date from SQL_8
+),
+t2 as (
+    select *, DATEDIFF(login_date, lag(login_date, 1) over (partition by user_id order by login_date)) as diff from t1
+)
+select user_id from t2 where diff = 1 group by user_id having count(user_id) >= 2;
+
+```
+
+
+
+### 题目二
+
+#### 表格
+
+| player_id | score | score_time          |
+| --------- | ----- | ------------------- |
+| B3        | 1     | 2022-09-20 19:00:14 |
+| A2        | 1     | 2022-09-20 19:01:04 |
+| A2        | 3     | 2022-09-20 19:01:16 |
+| A2        | 3     | 2022-09-20 19:02:05 |
+| A2        | 2     | 2022-09-20 19:02:25 |
+| B5        | 2     | 2022-09-20 19:02:54 |
+| A4        | 3     | 2022-09-20 19:03:10 |
+| B1        | 2     | 2022-09-20 19:03:34 |
+| B1        | 2     | 2022-09-20 19:03:58 |
+| B1        | 3     | 2022-09-20 19:04:07 |
+| A2        | 1     | 2022-09-20 19:04:19 |
+| A3        | 2     | 2022-09-20 19:04:31 |
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_9
+(
+     player_id 	varchar(2),
+     score      int,
+     score_time datetime
+);
+INSERT INTO SQL_9 (player_id, score, score_time)
+VALUES ('B3', 1, '2022-09-20 19:00:14'), ('A2', 1, '2022-09-20 19:01:04'),
+       ('A2', 3, '2022-09-20 19:01:16'), ('A2', 3, '2022-09-20 19:02:05'),
+       ('A2', 2, '2022-09-20 19:02:25'), ('B3', 2, '2022-09-20 19:02:54'),
+       ('A4', 3, '2022-09-20 19:03:10'), ('B1', 2, '2022-09-20 19:03:34'),
+       ('B1', 2, '2022-09-20 19:03:58'), ('B1', 3, '2022-09-20 19:04:07'),
+       ('A2', 1, '2022-09-20 19:04:19'), ('B3', 2, '2022-09-20 19:04:31');
+```
+
+#### 问题
+
+统计出连续三次（及以上）为球队得分的球员名单
+
+#### 分析
+
+连续N次以上为球队得分， 要求数据行满足以下条件：
+
+1) player_id 要相同表示同一球员
+
+2) 每行记录以得分时间从小到大排序
+
+3) 数据行数大于等于N
+
+#### 解答
+
+```sql
+-- 方法一
+with t1 as (
+    select *, lag(player_id, 1) over (order by score_time) as last_play_id from SQL_9
+)
+select distinct player_id from t1 where player_id = last_play_id group by player_id having count(player_id) >= 2;
+
+-- 方法二
+with t1 as (
+    select *, row_number() over (order by score_time) as rn from SQL_9
+),
+ t2 as (
+     select *, row_number() over (order by score_time) + 1 as rn from SQL_9
+ ),
+ t3 as (
+     select t1.player_id  as player_id from t1 join t2 on t1.rn = t2.rn and t1.player_id = t2.player_id
+ )
+select distinct player_id from t3 group by player_id having count(player_id) >= 2;
+```
+
+
+
+### 题目三
+
+#### 表格
+
+| log_id |
+| :----: |
+|   1    |
+|   2    |
+|   3    |
+|   7    |
+|   8    |
+|   10   |
+
+#### 脚本
+
+```sql
+CREATE TABLE SQL_10
+(
+     log_id  int
+);
+INSERT INTO SQL_10 (log_id) VALUES (1), (2), (3), (7), (8), (10);
+```
+
+#### 问题
+
+编写SQL 查询得到 Logs 表中的连续区间的开始数字和结束数字。按照 start_id 排序。查询结果格式如下:
+
+| start_id | end_id |
+| -------- | ------ |
+| 1        | 3      |
+| 7        | 8      |
+| 10       | 10     |
+
+#### 解答
+
+```sql
+-- 方法一
+with t1 as (
+    select *, log_id - row_number() over (order by log_id) as gr from SQL_10
+)
+
+select min(log_id) as start_id, max(log_id) as end_id from t1 group by gr
+
+-- 方法二
+with t1 as (
+    select *, log_id - row_number() over (order by log_id) as gr, log_id - lag(log_id,1) over () as diff from SQL_10
+),
+t2 as (
+    select log_id, gr from t1 where ifnull(diff,-1) <> 1
+),
+t3 as (
+    select *, log_id - row_number() over (order by log_id) as gr, log_id - lead(log_id,1) over () as diff from SQL_10
+),
+t4 as (
+ select log_id, gr from t3 where ifnull(diff, 1) <> -1
+)
+select t2.log_id as start_id, t4.log_id as end_id from t2, t4 where t2.gr = t4.gr;
+```
+
+### 技巧
+
+如何求连续区间？
+
+1）行号过滤法
+
+​      通过row_number() 生成连续行号，与区间列进行差值运算，得到的临时结果如果相同表示为同一连续区间
+
+2） 错位比较法
+
+​      通过row_number() / row_number()  + 1 分别生成原生的和错位的连续行号列，进行连表操作
+
+​      也可以通过lag/lead函数直接生成错位列
\ No newline at end of file
diff --git a/centos  安装 docker.md b/centos  安装 docker.md
index 6910a39..e4de180 100644
--- a/centos  安装 docker.md	
+++ b/centos  安装 docker.md	
@@ -1,6 +1,6 @@
  
 
-# Centos 安装 Docker
+# Linux 安装 Docker
 
 
 
@@ -9,7 +9,7 @@
 一键安装脚本！Linux系统都支持！
 
 ```bash
-   curl  -sSL https://get.docker.com/ | sh
+    curl  -sSL https://get.docker.com/ | sh
     #国内阿里云镜像
     curl  -fsSL https://get.docker.com | bash -s docker --mirror Aliyun  
     #Azure源(中国区azure)
@@ -18,6 +18,39 @@
 
 
 
+## oracle linxu 
+
+1. 执行以下命令，以安装Docker依赖项：
+
+```
+sudo yum -y install yum-utils device-mapper-persistent-data lvm2
+```
+
+2. 执行以下命令，以添加Docker官方GPG key：
+
+```
+sudo yum-config-manager --add-repo https://download.docker.com/linux/centos/docker-ce.repo
+```
+
+3. 执行以下命令，以安装Docker CE：
+
+```
+sudo yum install docker-ce
+```
+
+4. 启动并设置Docker服务为开机自启：
+
+```
+sudo systemctl start docker
+sudo systemctl enable docker
+```
+
+通过`docker --version`命令，您可以检查Docker是否已经安装。如果命令输出了版本信息，说明已经成功安装Docker CE。
+
+希望这次能给您提供一个更简洁的方法。
+
+
+
 ## 启动服务
 
 ```bash
@@ -64,14 +97,17 @@ Docker中国：https://registry.docker-cn.com
 ### 写入配置文件 重启服务
 
 ```
-   sudo mkdir -p /etc/docker
-   
-    sudo tee /etc/docker/daemon.json <<-'EOF'
-    {
-      "registry-mirrors": ["https://mirror.ccs.tencentyun.com"]
-    }
-    EOF
+sudo mkdir -p /etc/docker
 
+#注意EOF前不能有空格    
+sudo tee /etc/docker/daemon.json <<-'EOF'
+{
+"registry-mirrors": ["https://mirror.ccs.tencentyun.com"]
+}
+EOF
+```
+
+```
     sudo systemctl daemon-reload
     sudo systemctl restart docker
 ```
diff --git a/linux/linux使用sed命令.md b/linux/linux使用sed命令.md
index b978516..155fdca 100644
--- a/linux/linux使用sed命令.md
+++ b/linux/linux使用sed命令.md
@@ -22,7 +22,6 @@ s ：取代，可以直接进行取代的工作哩！通常这个 s 的动作可
 可以使用正则  特殊字符\转义 
 
 
-sed ”s/要被取代的字串/新的字串/g“ fileName
+sed "s/要被取代的字串/新的字串/g" fileName
 ```
 
-
diff --git a/postgresql_and_edb/postgresql导入导出.md b/postgresql_and_edb/postgresql导入导出.md
index e1ba8ae..df4882e 100644
--- a/postgresql_and_edb/postgresql导入导出.md
+++ b/postgresql_and_edb/postgresql导入导出.md
@@ -637,7 +637,15 @@ psql -Uicctestedb -dicctestedb -h192.168.53.123 -f ~/
 # -O 不设置表归属,
 # -F c 自定义压缩
 # -v 显示详情
-./pg_dump -Uenterprisedb -diccedb -h192.168.53.118 -O -v -F c -f ~/diccedb_202207_29.data.sql
+./pg_dump -Uenterprisedb -diccedb -h192.168.53.118 -O -v -F c -f ~/alg.da
+
+```
+
+
+
+#### 数据导入
+
+```shell
 
 #导入数据
 #-c 指定恢复过程中清空目标数据库中的现有表
@@ -655,10 +663,6 @@ psql -Uicctestedb -dicctestedb -h192.168.53.123 -f ~/
 
 
 
-
-
-
-
 #### COPY 导出导出部分数据
 
 ```sh
diff --git a/redis/redis-3.0.3,哨兵集群.md b/redis/redis-3.0.3,哨兵集群.md
new file mode 100644
index 0000000..5817184
--- /dev/null
+++ b/redis/redis-3.0.3,哨兵集群.md
@@ -0,0 +1,214 @@
+## redis-v3.0.3哨兵集群搭建文档
+
+
+
+### 准备工作
+
+### 主机
+
+准备三台全新 centos linux 服务器 
+
+固定IP 
+
+```
+192.168.1.11 
+192.168.1.12
+192.168.1.13
+```
+
+**以下命令如无特殊说明 在三台主机中均要执行**
+
+准备环境 
+
+```bash
+yum install make gcc wget
+```
+
+创建用户 
+
+```bash
+useradd  redis
+```
+
+创建相关目录
+
+```bash
+mkdir -p /opt/{app/redis,applog}
+```
+
+授权
+
+```bash
+chown -R redis:redis /opt/app/redis/
+```
+
+```bash
+chown -R redis:redis /opt/applog/redis/
+```
+
+
+
+#### 编译源码
+
+切换用户
+
+```bash
+su - redis
+```
+
+下载源码
+
+```bash
+cd /opt/app/redis/
+wget https://download.redis.io/releases/redis-3.0.3.tar.gz
+```
+
+
+
+解压源码
+
+```bash
+tar -zxvf redis-3.0.3.tar.gz
+```
+
+
+
+编译源码 这里为了和生产一致 使用libc 内存分配器
+
+```bash
+cd redis-3.0.3
+make MALLOC=libc
+```
+
+安装
+
+```bash
+make PREFIX=/opt/app/redis/ install
+```
+
+
+
+配置 path
+
+```bash
+echo 'export PATH=$PATH:/opt/app/redis/bin' >> ~/.bashrc
+
+source ~/.bashrc
+```
+
+
+
+### 配置redis
+
+复制配置文件
+
+```bash
+cp /opt/app/redis/redis-3.0.3/src/redis.conf /opt/app/redis/
+
+cp /opt/app/redis/redis-3.0.3/src/sentinel.conf /opt/app/redis/
+```
+
+编辑配置文件 redis.conf 
+
+```bash
+sed -i 's/daemonize no/daemonize yes/g' redis.conf
+
+sed -i 's|^logfile ""$|logfile "/opt/applog/redis/redis.log"|g' redis.conf
+
+sed -i 's|^dir "./"$|dir "/opt/app/redis/"|g' redis.conf
+
+sed -i 's/appendonly no/appendonly yes/g' redis.conf
+
+sed -i 's/^# cluster-node-timeout 15000$/cluster-node-timeout 5000/g' redis.conf
+#设置主从复制密码
+sed -i 's/^# masterauth <master-password>$/masterauth dUw~7a)6/g' redis.conf
+#设置 节点密码
+sed -i 's/^# requirepass foobared$/requirepass dUw~7a)6/g' redis.conf
+```
+
+
+
+配置主从节点
+
+在 192.168.1.12，192.168.1.13 两台机子中执行
+
+```bash
+echo "slaveof 10.23.101.3 6379" >> redis.conf
+```
+
+
+
+编辑配置文件 sentinel.conf 
+
+```bash
+sed -i 's|^\(sentinel monitor mymaster\) 127.0.0.1|\1 181.168.1.11|' sentinel.conf
+
+#设置哨兵密码
+sed -i 's/^#sentinel auth-pass <master-name> <password>/sentinel auth-pass mymaster HmpYZ2KB/g' sentinel.conf
+
+echo "daemonize yes" >> sentinel.conf
+
+```
+
+
+
+
+
+编写动脚本
+
+```bash
+echo "bin/redis-server redis.config">/opt/app/redis/start.sh
+chmod 775 start.sh
+
+echo "bin/redis-sentinel sentinel.config">/opt/app/redis/start-sentinel.sh
+chmod 775 start.sh
+```
+
+指定用户启动
+
+```bash
+#!/bin/bash
+su -s /bin/bash -c "$(pwd)/bin/redis-sentinel $(pwd)/sentinel.conf " redis
+```
+
+
+
+### 启动
+
+先启动三个redis节点
+
+```bash
+bash start.sh
+```
+
+再启动 sentinel j节点
+
+```bash
+bash start-sentinel.sh
+```
+
+
+
+### 验证
+
+登录三个redis节点 分别写入key
+
+192.168.1.11  成功写入  并同步至 192.168.1.12，192.168.1.13
+192.168.1.12	写入失败
+192.168.1.13	写入失败
+
+
+
+登录哨兵节点 查看哨兵信息
+
+```bash
+redis-cli  -p 26379
+
+sentinel sentinels mymaster
+```
+
+
+
+杀死 主节点
+
+分别查看 另外两台redis 的info 信息 是否发生切换
\ No newline at end of file