提交 d2fc511a 编写于 作者: M Mars Liu

cte

上级 4bf40e72
...@@ -7,7 +7,8 @@ ...@@ -7,7 +7,8 @@
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\ No newline at end of file
# 树结构溯根
现有一个表 node
```mysql
create table node
(
id int primary key auto_increment,
pid integer,
val integer
);
```
其 pid 列引用 id 列,形成一个树结构,根节点的 pid 为 0。
现在我们希望写一个查询,找到某一个给定id的记录,其父节点、父节点的父节点,直至根节点的路径。那么这个查询应该是:
## 答案
```mysql
with recursive t(id, pid, val) as (
select id, pid, val
from node
where id = $1
union all
select node.id, node.pid, node.val
from node
join t on node.id = t.pid)
select node.id, node.pid, node.val
from node
join t on node.id = t.id;
```
## 选项
### 没有递归定义
```mysql
with t as (
select id, pid, val
from node
where id = $1
union all
select node.id, node.pid, node.level
from node
join t on node.id = t.pid)
select node.id, node.pid, node.val
from node
join t on node.id = t.id;
```
### 平凡的连接查询无法处理递归问题
```mysql
select node.id, node.pid, node.val
from node
join node as p on node.pid = p.id
where id = $1;
```
### 子查询无法处理递归问题
```mysql
select node.id, node.pid, node val
from node as t
where t.pid = (select id from t where id = t.pid)
```
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...@@ -20,7 +20,7 @@ mysql> desc payment; ...@@ -20,7 +20,7 @@ mysql> desc payment;
| customer_id | int | YES | | NULL | | | customer_id | int | YES | | NULL | |
| staff_id | int | YES | | NULL | | | staff_id | int | YES | | NULL | |
| rental_id | int | YES | | NULL | | | rental_id | int | YES | | NULL | |
| amount | decimal(12,8) | YES | | NULL | | | amount | decimal(12,4) | YES | | NULL | |
| pathment_date | timestamp | YES | | NULL | | | pathment_date | timestamp | YES | | NULL | |
+---------------+---------------+------+-----+---------+----------------+ +---------------+---------------+------+-----+---------+----------------+
6 rows in set (0.12 sec) 6 rows in set (0.12 sec)
......
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\ No newline at end of file
# 每日报表优化
# 每日报表
分析过去一段时间的查询数据,Joe 发现 payment 表
```mysql
create table payment(
payment_id int primary key auto_increment,
customer_id int,
staff_id int,
rental_id int,
amount decimal(12, 4),
payment_date timestamp
)
```
每天的订单量很大,下面这个查询的统计过程占用了大多数数据库资源,查询不会
早于当天,总是在历史上某一个日期段内查询
```mysql
select date(payment_date) as day, sum(amount)
from payment
where date(payment_date) between $1 and $2
group by date(payment_date);
```
怎样优化最有效?
## 答案
建立中间表并建立索引。
```postgresql
create table daily_payment(day date primary key , amount decimal(12, 4));
insert into daily_payment(day, amount)
select payment_date::date as day, sum(amount) as amount from payment group by day;
```
使用
```mysql
select day, amount from view_daily_payment where day between $1 and $2;
```
进行查询。并且每天定时执行一次刷新命令
```mysql
insert into daily_payment(day, amount)
select date(payment_date) as day, sum(amount) as amount
from payment
where date(payment_date) between DATE_SUB(CURDATE(), INTERVAL 2 DAY) and DATE_SUB(CURDATE(), INTERVAL 1 DAY)
group by day;
```
## 选项
### 不会优化 sum 和 group by
在 payment_date 列上建立索引
```mysql
create index idx_payment_date on payment(payment_date);
```
### 不会优化 sum
建立计算列
```mysql
alter table payment add day date generated always as ( payment_date::date ) stored
```
然后使用它改写查询
```mysql
select day as day, sum(amount)
from payment
where day between $1 and DATE_SUB(CURDATE(), INTERVAL 1 DAY)
group by day;
```
### 优化了日期查询,但是不会优化统计过程
建立表达式索引
```mysql
create index idx_payment_day on payment((date(payment_date)));
```
### 不做物化,对查询速度不会有显著改善
建立视图
```mysql
create view view_daily_payment as select date(payment_date) as day, amount from payment;
```
然后在视图 view_daily_payment 上执行
```mysql
select day, sum(amount)
from view_daily_payment
where day between $1 and date('yesterday')
group by day
```
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