長度是：
mysql> select count(*) from projects join(select project_skills.skill_id from project_skills where project_skills.skill_id = 7) as s on projects.id = s.skill_id;
----------
| count(*) |
----------
| 869 |
----------
另一個長度是：
mysql> select count(*) from projects join(select project_frameworks.framework_id from project_frameworks where project_frameworks.framework_id = 3) as f on projects.id = f.framework_id;
----------
| count(*) |
----------
| 264 |
----------
我需要結合這些長度（結果）。
所以我想出了：
select count(*) from projects join(select project_skills.skill_id from project_skills where project_skills.skill_id = 7) as s on projects.id = s.skill_id join(select project_frameworks.framework_id from project_frameworks where project_frameworks.framework_id = 3) as f on projects.id = f.framework_id;
但長度是0....
我怎樣才能達到我的理想？
■ 表結構
專案 vs project_skills（一對多）
技能 vs project_skills（一對多）
專案 vs project_frameworks（一對多）
框架 vs project_frameworks（一對多）
project_skill（列：id、project_id、skill_id）
project_frameworks（列：id、project_id、framework_id）
uj5u.com熱心網友回復：
您可以使用 union all 進行總結以獲得總數。
SELECT sum(q.c) FROM (
select
count(*) as c
from
projects
join( select project_skills.skill_id from project_skills where project_skills.skill_id = 7 ) as s
on projects.id = s.skill_id
union all
select
count(*) as c
from
projects
join( select project_frameworks.framework_id from project_frameworks where project_frameworks.framework_id = 3 ) as f
on projects.id = f.framework_id
) q;
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標籤：mysqlsql