sql

sql有些生疏，特

对于满足SQL92标准的SQL语句

select foo,count(foo) from pokes where foo>10 group by foo having count(*)>5 order bu foo

执行顺序

from --> where  --> group  by--> having  --> select  --> order by

from计算关系

where就要指定条件对记录进行筛选

group by将数据划分为多个分组

having 筛选分组

select 产生查询结果

order by 对结果集进行排序

挥手建一个表

create table sb
(
id VARCHAR(50),
user VARCHAR(50)
);

随意插入一些数据

INSERT into sb(id,user) VALUES('node1','a+sqld');
INSERT into sb(id,user) VALUES('node1','a');
INSERT into sb(id,user) VALUES('node2','b+SQL');
INSERT into sb(id,user) VALUES('node2','b-SQL');
INSERT into sb(id,user) VALUES('node3','c+SQLd');
INSERT into sb(id,user) VALUES('node3','c-sqld');
INSERT into sb(id,user) VALUES('node4','d+SQL');
INSERT into sb(id,user) VALUES('node4','d-SQL');
INSERT into sb(id,user) VALUES('node4','DSQL');

a、统计id出现次数大于2
思路：以id进行分组，having筛选分组,查询分组后次数大于2的

select id  from sb GROUP BY id HAVING COUNT(id)>2 ;

b、统计id出现次数大于2，并以user降序

思路：order by 进行排序 可取值desc asc

select id ,user from sb GROUP BY id HAVING COUNT(id)>2 ORDER BY user desc ;

c、统计id出现次数大于2，与对应的is
注意: 打印次数，必须给having后的字段起别名

select  COUNT(id) idNum ,id,user from sb GROUP BY id HAVING idNum>2 ;

d、模糊查询

注意：也不用关注查询数据的大小写

select * from sb WHERE user like '%sql%'

e、查看字段中含有sqld,并且sqld出现次数多于1次的id,

(1)、找出字段中含有sqld的id

select id,user from sb where user like '%sqld%'

(2)、在结果集中找sql出现次数多于1的id 

注意：中间结果集作为一张表进行查询要起别名，select * from table

第一步：统计所有id以及id出现次数

select id, count(*) idCount ,user from sb where user like '%sql%' group by id

第二步：查找第一步结果集中id大于1的。count(*)=count(id)

select id ,idCount,user from 
(select id, count(*) idCount ,user from sb where user like '%sql%' group by id) a 
where idCount>1

             

或者：

select id, user from (select id,user from sb where user like '%sqld%')  group by id HAVING count(id)>1

(3)、如果打印次数，要起别名,having 做统计用--(2)后者比较笨的办法了，不建议使用

select count(id) countId , id, user from (select id,user from sb where user like '%sqld%') as sqls group by id HAVING countId>1


如下图

select *     
from (
select user,month,pay_amount,
ROW_COUNT() over ( partition by user order by pay_amount) as pay_amount_rank
from(
select user,pay1 as pay_amount,1 as month from user_pay_log
union all 
select user,pay2 as pay_amount,2 as month from user_pay_log
union all
select user,pay3 as pay_amount,3 as month from user_pay_log
union all 
select user,pay4 as pay_amount,4 as month from user_pay_log
union all 
select user,pay5 as pay_amount,5 as month from user_pay_log
union all 
select user,pay6 as pay_amount,6 as month from user_pay_log
) t1
)t2
where pay_amount_rank=1;