我已经经历了几天的stackoverflow并且google了很多但仍然可以找到解决方案。 我有2个表格和资格 在表格配置文件中
pro_id surname firstname
-------- -------- ------------
1 John James
2 King Fred
3 Paul Smith
4 Cindy Hayes
Qua_id Degree School Year
------ ------ ------ -----
1 MBA Wharton university 2002
1 LLB Yale University 2001
2 BSc Covington University 1998
2 BEd Kellog University 1995
2 Msc MIT 2011
3 MBA Havard Business School 2002
3 MSc Yale University 2012
4 BSc University of Edinburgh 2010
4 BA University of Liverpool 2009
现在我想要实现的是这个
1 John James MBA Wharton university 2002, LLB Yale University 2001
2 King Fred BSc Covington University 1998, BEd Kellog University 1995, Msc MIT 2011
3 Paul Smith MBA Havard Business School 2002, MSc Yale University 2012
4 Cindy Hayes BSc University of Edinburgh 2010, BA University of Liverpool 2009
但我现在拥有的是:
1 John James MBA Wharton university 2002
1 John James LLB Yale University 2001
2 King Fred BSc Covington University 1998 BEd
2 King Fred Kellog University 1995
2 King Fred Msc MIT 2011
3 Paul Smith MBA Havard Business School 2002
3 Paul Smith MSc Yale University 2012
4 Cindy Hayes BSc University of Edinburgh 2010
4 Cindy Hayes BA University of Liverpool 2009
这是我的代码
Select pro_id, surname, firstname concat(degree,school,year) as qual from profile,qualification Where profile.proid=qualification.qua_id
答案 0 :(得分:0)
Select pro_id, surname, firstname, group_concat(degree,school,year) as qual
from profile
left join qualification
on qualification.qua_id = profile.pro_id
group by qua_id
你的归档名称看起来有点奇怪 - qua_id应该是profile_id(我认为shoudl实际上是user_id - 即它们应该是用户ID和用户表的外键)。