我的数据库中有以下表格
项目
+----+-------------------------------------------+
| id | name |
+----+-------------------------------------------+
| 1 | YANNONALI COURT |
| 2 | UNIVERSITY OF COLORARDO DENVER RESEARCH 2 |
| 3 | G.R.E.A.T PROGRAM DESALTER BUILDING |
| 4 | MONARCH CLUB |
| 5 | LAFAYETTE MERCANTILE |
| 6 | CAMELBACK VILLAGE RAQUET AND HEALTH CLUB |
| 7 | BACK COUNTRY |
| 8 | URBAN CRASHPAD |
| 9 | PRIVATE RESIDENCE |
| 10 | EATON RESIDENCE |
+----+-------------------------------------------+
PROJECT_ASSIGNMENTS(WHERE projects.id = project_assignment.target_id)
+-------+-----------+-------------+
| id | target_id | property_id |
+-------+-----------+-------------+
| 19178 | 1 | 48 |
| 19192 | 1 | 39 |
| 19391 | 1 | 3 |
| 19412 | 2 | 3 |
| 19591 | 2 | 34 |
| 19610 | 2 | 34 |
| 21013 | 3 | 2 |
| 21032 | 3 | 2 |
| 30876 | 4 | 2433 |
| 38424 | 5 | 2580 |
+-------+-----------+-------------+
PROPERTIES(WHERE properties.id = project_assignment.property_id)
+----+------------------+
| id | name |
+----+------------------+
| 2 | Residential |
| 3 | Multi Family |
| 34 | New Construction |
| 39 | Contemporary |
| 48 | Southwest |
+----+------------------+
我想通过列表中的no.of项目订购O / P ......
Residential(177) //12 - total no.of projects which is having this property
Multi Family(15)
New Construction(13)
Contemporary(11)
请给我一些MySQL查询
Thank You
答案 0 :(得分:0)
这应该可以解决问题:
select
c.name,
count(c.id) as CountOfProperties
from
projects a,
project_assignments b,
properties c
where
a.ID=b.target_id
and b.property_id=c.ID
group by
c.name
order by
count(c.id) desc;
答案 1 :(得分:0)
试试这个::
select
prop.name,
count(prop.id) as CountOfProperties
from
projects p
inner join project_assignments pa on (p.ID=pa.target_id)
inner join properties prop on (pa.property_id=prop.ID)
group by
prop.name
order by
count(prop.id) desc;