MySQL(4)必备SQL

1.1条件where

select * from info where age > 30;
select * from info where id > 1;
select * from info where id = 1;
select * from info where id >= 1;
select * from info where id != 1;
select * from info where id between 2 and 4;   -- id大于等于2、且小于等于4
select * from article where id between 2 and 4;

select * from info where name = '武沛齐' and age = 19;
select * from info where name = 'alex' or age = 49;
select * from info where (name = '李杰' or email="pyyu@live.com")  and age=49;

select * from info where id in (1,4,6);
select * from info where id not in (1,4,6);
select * from info where id in (select id from depart);
# select * from info where id in (1,2,3);

# exists select * from depart where id=5，去查数据是否存在，如果存在，如果不存在。
select * from info where exists (select * from depart where id=5);
select * from info where not exists (select * from depart where id=5);

select * from (select * from info where id>2) as T where age > 10;

select * from info where info.id > 10;
select * from info where id > 10;

1.2通配符

一般用于模糊搜索

select * from info where name like "%沛%";
select * from info where name like "%沛";
select * from info where email like "%@live.com";
select * from info where name like "武%齐";
select * from info where name like "k%y";
select * from info where email like "wupeiqi%";

select * from info where email like "_@live.com";
select * from info where email like "_upeiqi@live.com";
select * from info where email like "__peiqi@live.com";
select * from info where email like "__peiqi_live.co_";

1.3 映射

想要获取的列。

select * from info;

select id, name				from info;
select id, name as NM 		from info;
select id, name as NM, 123  from info;
注意：少些select * ,自己需求。

select 
	id,
	name,
	666 as num,
	( select max(id) from depart ) as mid, -- max/min/sum
	( select min(id) from depart) as nid, -- max/min/sum
	age
from info;

select 
	id,
	name,
	( select title from depart where depart.id=info.depart_id) as x1
from info;

# 注意：效率很低

select 
	id,
	name,
	( select title from depart where depart.id=info.depart_id) as x1,
	( select title from depart where depart.id=info.id) as x2
from info;

select 
	id,
	name,
	case depart_id when 1 then "第1部门" end v1
from info;

select 
	id,
	name,
	case depart_id when 1 then "第1部门" else "其他" end v2
from info;
```sql

select 
	id,
	name,
	case depart_id when 1 then "第1部门" end v1,
	case depart_id when 1 then "第1部门" else "其他" end v2,
	case depart_id when 1 then "第1部门" when 2 then "第2部门" else "其他" end v3,
	case when age<18 then "少年" end v4,
	case when age<18 then "少年" else "油腻男" end v5,
	case when age<18 then "少年" when age<30 then "青年" else "油腻男" end v6
from info;

1.4 排序

select * from info order by age desc; -- 倒序
select * from info order by age asc;  -- 顺序

select * from info order by id desc;
select * from info order by id asc;
select * from info order by age asc,id desc; -- 优先按照age从小到大；如果age相同则按照id从大到小。

select * from info where id>10 order by age asc,id desc;
select * from info where id>6 or name like "%y" order by age asc,id desc;

1.5 取部分

select * from info limit 5;   										-- 获取前5条数据
select * from info order by id desc limit 3;						-- 先排序，再获取前3条数据
select * from info where id > 4 order by id desc limit 3;			-- 先排序，再获取前3条数据

select * from info limit 3 offset 2;	-- 从位置2开始，向后获取前3数据

1.6 分组

select age,max(id),min(id),count(id),sum(id),avg(id) from info group by age;

select age,count(1) from info group by age;

select depart_id,count(id) from info group by depart_id;

select depart_id,count(id) from info group by depart_id having count(id) > 2;

select age,count(id) from info where id > 2 group by age having count(id) > 1 order by age desc limit 1;
- 要查询的表info
- 条件 id>2
- 根据age分组
- 对分组后的数据再根据聚合条件过滤 count(id)>1
- 根据age从大到小排序
- 获取第1条

到目前为止SQL执行顺序：
    where 
    group by
    having 
    order by
    limit 

1.7 左右连表

主表 left outer join 从表 on 主表.x = 从表.id 

select * from info left outer join depart on info.depart_id = depart.id;

select info.id,info.name,info.email,depart.title from info left outer join depart on info.depart_id = depart.id;

从表 right outer join 主表 on 主表.x = 从表.id

select info.id,info.name,info.email,depart.title from info right outer join depart on info.depart_id = depart.id;

1.8 联合

select id,title from depart 
union
select id,name from info;


select id,title from depart 
union
select email,name from info;
-- 列数需相同

select id from depart 
union
select id from info;

-- 自动去重

select id from depart 
union all
select id from info;

-- 保留所有