• 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'


     

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