开发者

How to retrieve all entries with JOIN?

开发者 https://www.devze.com 2023-04-13 00:04 出处:网络
There is my problem : enter code here I have a table called test that looks like this id|service|sub service| Qt|date

There is my problem :

enter code here

I have a table called test that looks like this

id|  service  |   sub service  | Qt  |   date

1 | service_1 | sub_service_11 |  3  | 2011-12-03
2 | service_1 | sub_service_12 |  6  | 2011-12-03
3 | service_1 | sub_service_13 |  4  | 2011-12-03

Later I have a table called datedim that looks like this

id|   date 

1 | 2011-12-01
2 | 2011-12-02
3 | 2011-12-03
4 | 2011-12-04
5 | 2011-12-05

What I am trying to do is that for each sub_service bring back all the date from datedim even if开发者_开发技巧 there is no match.

So basically something that would look like this

sub_service_11 | 2011-12-01 | NULL
sub_service_11 | 2011-12-02 | NULL
sub_service_11 | 2011-12-03 | 3
sub_service_11 | 2011-12-04 | NULL
sub_service_11 | 2011-12-05 | NULL
sub_service_12 | 2011-12-01 | NULL
sub_service_12 | 2011-12-02 | NULL
sub_service_12 | 2011-12-03 | 6
sub_service_12 | 2011-12-04 | NULL
sub_service_12 | 2011-12-05 | NULL
sub_service_13 | 2011-12-01 | NULL
sub_service_13 | 2011-12-02 | NULL
sub_service_13 | 2011-12-03 | 4
sub_service_13 | 2011-12-04 | NULL
sub_service_13 | 2011-12-05 | NULL

I did try RIGHT JOIN, UNIONS and stuff but I can't figure it out.

Does anyone know how I can accomplish that ?

Thank you,


This is what you need:

select 
  ss.sub_service, 
  dd.date,
  ts.qt
from 
  (datedim dd, (select distinct sub_service from test) ss)
 left join test ts on (dd.date = ts.date and ts.sub_service = ss.sub_service)
order by ss.sub_service, dd.date


As you have a table containing each sub_service (lets call it sub_service_list) you should be able to do the following:

SELECT
s.sub_service, d.date, t.Qt
FROM (sub_service_list s,datedim d) 
LEFT JOIN test t ON s.sub_service=t.sub_service AND d.date=t.date


you can do this:

select t.id, sub_service, case when d.date = t.date then qt else null end as qt
from test t cross join datedim d 
order by sub_service, d.date
0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号