mySQL

1创建一个表
CREATE TABLE person3(
firstName varchar(222),
lastName varchar(222),
age varchar(222)
)

2删除
drop table if exists `person3`

3 插入数据
insert into person3(firstName,lastName,age)
values('zhang','xiaolei','23');
insert into person3(firstName,lastName,age)
values('jia','ningning','23');
insert into person3(firstName,lastName,age)
values('zhang','lei','25');
insert into person3(firstName,lastName,age)
values('zhao','mengyao','21');
insert into person3(firstName,lastName,age)
values('du','ruihong','22');
insert into person3(firstName,lastName,age)
values('zheng','zongjiao','17');
insert into person3(firstName,lastName,age)
values('tian','zhen','18');
insert into person3(firstName,lastName,age)
values('wu','shuaipeng','19');
insert into person3(firstName,lastName,age)
values('li','cong','20')

4查询前两个数据
SELECT 'firstName','lastName' FROM `person3`

5查询id>2
SELECT * FROM `person3` WHERE age<18

6按年龄从小到大排序
SELECT * FROM `person3` order by age

7按年龄从大到小排序
SELECT * FROM `person3` order by age desc

8改变已设的年龄
UPDATE `person3` SET age='24'
WHERE firstName='zhang'and lastName='xiaolei'

9删除firstName为du的
DELETE FROM `person3` WHERE firstName='du'

10相同的合并distinct
SELECT distinct firstName FROM `person3` WHERE 1

11取18岁到23岁的between  and
SELECT * FROM `person3` WHERE age between 18 and 23

12查找不同年龄人的个数group  by
SELECT age as age,count(1) FROM `person3` group by age

13将firstName和lastName合并在一起concat
SELECT concat(firstName,lastName) as 名字,age as 年龄 FROM `person3` WHERE 1

老师的作业

1.查询person表所有成年人的信息。

SELECT * FROM `person3` WHERE age>18

2.查询person表中firstName以zh开头的所有人的 name(包括firstName+lastname)和年龄。

SELECT concat(firstName,lastName) as name,age as age FROM `person3` WHERE firstName like 'zh%'


3.查询person表所有人的年龄段和该年龄段的人数。

SELECT age as age FROM `person3` WHERE group by age


4.查询person 表中 年龄在16,17,18岁的人数的个数。

SELECT * FROM `person3` WHERE age in(16,17,18)

5.把姓zhang的人加1岁。

UPDATE `person3` SET age=age+1
 WHERE firstName='zhang'


 

原文地址:https://www.cnblogs.com/xiaoleidiv/p/3003939.html