提交 785557c9 编写于 作者: M Mars Liu

join

上级 f411bfa8
...@@ -2,9 +2,14 @@ ...@@ -2,9 +2,14 @@
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# 自连接
现有 node 表如下:
```mysql
create table node(
id int primary key auto_increment,
pid int,
content varchar(256)
)
```
现在Joe 想要给出 content 以 `fork-` 开头的所有节点,和它们的子节点,输出 `parent_id, parent_content, child_id, child_content`
他应该怎么做?
## 答案
```mysql
select l.id as parent_id,
l.content as parent_content,
r.id as child_id,
r.content as child_content
from node as l
join node as r on l.id = r.pid
where l.content like 'fork-%';
```
## 选项
### A
```mysql
select l.id as parent_id,
l.content as parent_content,
r.id as child_id,
r.content as child_content
from node as l
right join node as r on l.id = r.pid
where l.content like 'fork-%';
```
### B
```mysql
select l.id as parent_id,
l.content as parent_content,
r.id as child_id,
r.content as child_content
from node as l, node as r
where l.id =(+) r.pid l.content like 'fork-%';
```
### C
```mysql
select l.id as parent_id,
l.content as parent_content,
r.id as child_id,
r.content as child_content
from node as l
cross join node as r on l.id = r.pid
where l.content like 'fork-%';
```
### D
```mysql
select l.id as parent_id,
l.content as parent_content,
r.id as child_id,
r.content as child_content
from node as l
join node as r on l.pid = r.id
where l.content like 'fork-%';
```
\ No newline at end of file
...@@ -2,9 +2,14 @@ ...@@ -2,9 +2,14 @@
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"left_join.json"
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......
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\ No newline at end of file
# 左连接
现有部门表
```mysql
create table department(
id int primary key auto_increment,
name varchar(256)
)
```
和员工表
```mysql
create table employee(
id int primary key auto_increment,
dept int,
name varchar(256),
post varchar(16)
)
```
Joe 想要列出所有的部门,如果这个部门有部门助理(post 为 `assistant`),则将 stuff 的名字也列出来,那么这个查询应该是:
## 答案
```mysql
select d.id, d.name, e.name as assistant
from department as d
left join employee as e on e.dept = d.id
where e.post = 'assistant'
```
## 选项
### A
```mysql
select d.id, d.name, e.name as assistant
from department as d
cross join employee as e on e.dept = d.id
where e.post = 'assistant'
```
### B
```mysql
select d.id, d.name, e.name as assistant
from department as d
join employee as e on e.dept = d.id
where e.post = 'assistant'
```
### C
```mysql
select d.id, d.name, e.name as assistant
from department as d
cross join employee as e on e.dept = d.id
where e.post = 'assistant'
```
### D
```mysql
select d.id, d.name, e.name as assistant
from employee as e
left join department as d on e.dept = d.id
where e.post = 'assistant'
```
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......
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\ No newline at end of file
# 右连接
现有部门表
```mysql
create table department(
id int primary key auto_increment,
name varchar(256)
)
```
和员工表
```mysql
create table employee(
id int primary key auto_increment,
dept int,
name varchar(256),
post varchar(16)
)
```
公司经过了一轮调整后,员工信息有些混乱,现在 Joe 要写一个查询,找出部门信息写
错的员工,这些员工所在的部门在 department 表中没有对应记录。
## 答案
```mysql
select e.id, e.name, e.dept
from department as d
right join employee as e on d.id = e.dept
where d.id is null;
```
## 选项
### A
```mysql
select e.id, e.name, e.dept
from employee as e
right join department as d on d.id = e.dept
where e.id is null;
```
### B
```mysql
select e.id, e.name, e.dept
from employee as e
right join department as d on d.id = e.dept
where d.id is null;
```
### C
```mysql
select e.id, e.name, e.dept
from department as d
join employee as e on d.id = e.dept
where d.id is null;
```
### D
```mysql
select e.id, e.name, e.dept
from department as d
right join employee as e on d.id = e.dept
where d.id is null;
```
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...@@ -2,9 +2,14 @@ ...@@ -2,9 +2,14 @@
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"keywords": [], "keywords": [],
"children": [], "children": [],
"export": [], "export": [
"cross_join.json"
],
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["mysql", "cross join"] [
"mysql",
"cross join"
]
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......
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\ No newline at end of file
# Cross Join
Joe 需要生成 goods 表
```mysql
create table goods(
id int primary key auto_increment,
category varchar(64),
name varchar(256),
price decimal(12, 4),
stock int,
upper_time timestamp
)
```
中所有T恤(category为`T-Shirt`)的所有尺寸,尺寸信息在 size 表
```mysql
create table size(
id int primary key auto_increment,
name varchar(16)
)
```
中,那么这个查询应该是:
## 答案
```mysql
select g.id, g.name, s.name as size
from goods as g
cross join size as s
where g.category = 'T-Shirt';
```
## 选项
### A
```mysql
select g.id, g.name, s.name as size
from goods as g
full join size as s
where g.category = 'T-Shirt';
```
### B
```mysql
select g.id, g.name, s.name as size
from goods as g
left join size as s
where g.category = 'T-Shirt';
```
### C
```mysql
select g.id, g.name, s.name as size
from goods as g
left join size as s
where g.category = 'T-Shirt';
```
### D
```mysql
select g.id, g.name, s.name as size
from goods as g
right join size as s
where g.category = 'T-Shirt';
```
### E
```mysql
select g.id, g.name, s.name as size
from goods as g
outter join size as s
where g.category = 'T-Shirt';
```
# 延迟复制 # 延迟复制
`主从复制`一节里,Joe实现了一个点对点的主从复制架构,其中standby 是 trade的从库,现在, `主从复制`一节里,Joe实现了一个点对点的主从复制架构,其中 standby 是 trade 的从库,现在,
Joe 要添加一个名为 backup 的新的复制节点,这个节点的同步进度要比 trade 晚半小时。他应该怎么做? Joe 要添加一个名为 backup 的新的复制节点,这个节点的同步进度要比 trade 晚半小时。他应该怎么做?
## 答案 ## 答案
......
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...@@ -2448,50 +2448,6 @@ ...@@ -2448,50 +2448,6 @@
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" 嵌套查询": {
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"GROUP BY优化": {
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"performance",
"优化",
"group by"
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{ {
" SHOW STATUS": { " SHOW STATUS": {
"node_id": "mysql-3574b2e5c9ca475789d9d582d7726906", "node_id": "mysql-3574b2e5c9ca475789d9d582d7726906",
......
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