Question

我有两张桌子：

第一个表格是tbl_ccompany

db_id   db_cname
 1         xxx
 2         yyy

第二张表tbl_marketing

db_mid db_name db_ccompany
1       ggg     1,2

在tbl_marketing db_ccompany中包含第一个表的id

我想选择要打印的db_cname 输出将是这样的

1  ggg xxx,yyy

我在这两个表之间尝试左连接，但只有我收到第一个名字

select 
marketing.*,
ccompany.db_cname
from tbl_marketing as marketing
left join tbl_ccompany as ccompany
on
marketing.db_ccompany=ccompany.db_id

我如何加入表格并获得我想要的输出？

Answer 1

SELECT db_mid,db_name,GROUP_CONCAT(db_cname) FROM tbl_marketing LEFT JOIN tbl_ccompany on db_id=db_ccompany WHERE db_ccompany in(1,2)

Answer 2

我希望这会有所帮助。

数据库结构：

create table tbl_ccompany (
db_id int,
db_cname text );

insert into tbl_ccompany values (1,'xxx');
insert into tbl_ccompany values (2,'yyy');

create table tbl_marketing (
db_mid int ,
db_name text,
db_ccompany text
);

insert into tbl_marketing values (1,'ggg','1,2');

查询：

select db_mid, db_name, group_concat(db_cname) 
from tbl_ccompany tc , tbl_marketing tm
where find_in_set(tc.db_id,tm.db_ccompany)

参考： http://amitbrothers.blogspot.in/2014/03/mysql-split-comma-separated-list-into.html

MySQL从两个表中选择，其中一列包含2个变量

2 个答案: