好的,我在 Postgres 上为您尝试过这个(我这里没有 MySQL,所以可能有点不同):
select matches.id,
(matches.tfirstname
+ matches.tlastname
+ matches.tschool
+ matches.tcollege
+ matches.tuniversity) as total
from (
select u.id,
(case when u.firstname like '%a%' then 1 else 0 end) as tfirstname,
(case when u.lastname like '%b%' then 1 else 0 end) as tlastname,
sum(e2.nschool) as tschool,
sum(e2.ncollege) as tcollege,
sum(e2.nuniversity) as tuniversity
from tbluser u left outer join (
select e.usr,
(case when e.school like '%c%' then 1 else 0 end) as nschool,
(case when e.college like '%d%' then 1 else 0 end) as ncollege,
(case when e.university like '%e%' then 1 else 0 end) as nuniversity
from tbleduc e
) e2 on u.id=e2.usr
group by u.id, u.firstname, u.lastname
) as matches
我使用这些 DDL 语句来创建表:
create table tbluser (
id int primary key,
firstname varchar(255),
lastname varchar(255)
)
create table tbleduc (
id int primary key,
usr int references tbluser,
school varchar(255),
college varchar(255),
university varchar(255)
)
还有一些示例数据:
insert into tbluser(id, firstname, lastname)
values (1, 'Jason', 'Bourne');
insert into tbleduc(id, usr, school, college, university)
values (1, 1, 'SomeSchool', 'SomeCollege', 'SomeUniversity');
insert into tbleduc(id, usr, school, college, university)
values (2, 1, 'MoreSchool', 'MoreCollege', 'MoreUniversity');
tbluser
如果和之间的关系tbleduc
是1:1 ,则查询可以简化一点。
不要忘记将%a%
, %b
, ... 替换为您的变量(我建议使用准备好的语句)。
我希望这个模板作为一个基本的解决方案有所帮助 - 你可以随意调整它:-) 你也可以删除最外面的 select 语句,以获取单个结果的计数器。