SQL开发技巧

1. 同一属性的多值过滤

user1表

user_skill表

关联方式1

#获取同时拥有“变化”和“念经”技能的用户
select a.user_name, b.skill, c.skill from user1 a join user_skills b on a.id = b.user_id and b.skill = "念经"
join user_skills c on c.user_id = b.user_id and c.skill = "变化"
where b.skill_level>0 and c.skill_level>0

执行结果

关联方式2

#获取同时拥有4项技能的用户，如果没有显示null，过滤少于2项技能的用户
select a.user_name, b.skill, c.skill, d.skill, e.skill from user1 a 
left join user_skills b on a.id = b.user_id and b.skill = "念经" and b.skill_level>0
left join user_skills c on a.id = c.user_id and c.skill = "变化" and c.skill_level>0
left join user_skills d on a.id = d.user_id and d.skill = "腾云" and d.skill_level>0
left join user_skills e on a.id = e.user_id and e.skill = "浮水" and e.skill_level>0
where 
(case when b.skill is not null then 1 else 0 end) 
+(case when c.skill is not null then 1 else 0 end)
+(case when d.skill is not null then 1 else 0 end)
+(case when e.skill is not null then 1 else 0 end) >= 2

执行结果

 group by方式

#查询出至少大于等于两项技能的用户
select a.id, a.user_name
from user1 a 
join user_skills b on a.id = b.user_id 
where b.skill in ("念经","变化","腾云","浮水") and b.skill_level>0
group by  a.user_name having count(*) >= 2

执行结果

2. 如何在子查询中匹配两个值

user1表

user_kills

join方法

select a.user_name, b.timestr, b.kills from user1 a join user_kills b on a.id = b.user_id
join (select user_id,max(kills) as cnt from user_kills group by user_id) c
on b.user_id = c.user_id and b.kills = c.cnt

where子句方法

select a.user_name, b.timestr, b.kills from user1 a join user_kills b on a.id = b.user_id
where (b.user_id,b.kills) in (select user_id,max(kills) as cnt from user_kills group by user_id);

执行结果

kills杀怪个数

3.where语句中的子查询

where型的子查询就是把内层查询的结果当作外层查询的条件

lc_user表

3.1 查询每类理财产品理财金额最高的两个客户

SELECT * FROM lc_user as a
WHERE 2>(SELECT count(*) FROM lc_user WHERE LoanType = a.LoanType and Amount > a.Amount)
order BY a.LoanType, a.Amount DESC

上面的语句相当于依次执行了一遍下面的SQL

#第一次比较A001
SELECT count(*) FROM lc_user a where  LoanType = "A001" and Amount > 100     # 2>2
SELECT count(*) FROM lc_user a where  LoanType = "A001" and Amount > 1000    # 2>1
SELECT count(*) FROM lc_user a where  LoanType = "A001" and Amount > 5000    # 2>0

#第二次比较A002
SELECT count(*) FROM lc_user a where  LoanType = "A002" and Amount > 100     # 2>2
SELECT count(*) FROM lc_user a where  LoanType = "A002" and Amount > 500     # 2>1
SELECT count(*) FROM lc_user a where  LoanType = "A002" and Amount > 1000    # 2>0

#第三次比较A003
SELECT count(*) FROM lc_user a where  a.LoanType = "A003" and a.Amount > 2000    # 2>0

执行结果