1樓：賓士
主要過程如下：
1、查詢t2表中按照【field001】分組的【field002】和
select field001,sum(field002) as field002 from t2 group by field001;
2、查詢t3表中按照【field001】分組的【field002】和
select field001,sum(field002) as field002 from t3 group by field001;
3、將上面兩個查詢結果作為結果集和 t1 相關聯，求出最終結果：
select t1.field001,t2.field002,t3.field002
from t1,
(select field001,sum(field002) as field002 from t2 group by field001) t2,
(select field001,sum(field002) as field002 from t3 group by field001) t3
where t1.field001=t2.field001
and t1.field001=t3.field001;
說明：這個問題很多人很可能會想把三個表關聯起來，但是請注意，由於t2表t3表中的【field001】是【多對多關係】，直接關聯會產生【笛卡爾積】從而不會達到預期的結果。
---以上，希望對你有所幫助。
2樓：
我要統計出以下結果
a 5 2
b 2 7
---------
--在t2\t3不存在時用0填充
select
a.field002 as t1_field002,
isnull(b.field002,0) as t2_field002,
isnull(c.field002,0) as t3_field002
from t1 as a
left join (select field001,sum(field002) as field002 from t2 group by field001)b on a.field001=b.field002
left join (select field001,sum(field002) as field002 from t3 group by field001)c on a.field001=c.field002
3樓：匿名使用者
select t1.field002,test1.sum1,test2.sum2 from t1
inner join
(select field001,sum(field002) as sum1 from t2 group by field001) test1
on t1.field001=test1.field001
inner join
(select field001,sum(field002) as sum2 from t3 group by field001) test2
on t1.field001=test2.field001;
補充一下：試驗情況
sql@kokooa>select t1.field002,test1.sum1,test2.sum2 from t1
2 inner join
3 (select field001,sum(field002) as sum1 from t2 group by field001) test1
4 on t1.field001=test1.field001
5 inner join
6 (select field001,sum(field002) as sum2 from t3 group by field001) test2
7 on t1.field001=test2.field001;
field002 sum1 sum2
---------- ---------- ----------
101 5 2
102 2 7
為了方便 a,bjiu用101，102代替了
4樓：
mysql下估計能運動。。 orcal肯定不行。...差不多這意思。我也沒仔細去寫
(select distinct(field001) as field001 from t1) a ((select distinct(field001) as field001 from t1) b left join
(select t2.field001,sum(field002) from t1,t2 where t1.field001=t2.
field001 group by t2.field001) c where b.field001 =c.
field001 ) left join on
(select t2.field001,sum(field002) from t1,t3 where t1.field001=t3.
field001 group by t3.field001) d where a.field001 =d.
field001 )
5樓：匿名使用者
select a.field002,sum(b.field002) ,sum(c.field002) from t1 a,t2 b,t3 c
where a.field001=b.field001 and c.field001=a.field001
group by a.field002
三表聯查一下就行了.
6樓：
select a.field002, b.sum1, c.sum2from t1 a
join (select field001, sum(field002) as sum1 from t2) b on a.field001=b.field001
join (select field001, sum(field002) as sum2 from t3) c on a.field001=c.field001
7樓：匿名使用者
select t1.field002,a.sum ,b.
sumfrom t1,(select field001,sum(field002) sum from t3 groupb by field001) a,
(select field001,sum(field002) sum from t2 groupb by field001) b
where a.field001=b.field001 and t1.field001=a.field001;
8樓：盒子裡的房間
select field002,(select isnull(sum(field002),0) from t2 where field001 = t1.field001),select isnull(sum(field002),0) from t3 where field001 = t1.field001)
from t1
9樓：
select a.field001,b.s1,c.s2from t1 a,(select field001,sun(field002) s1
from t2 group by field001) b,(select field001,sun(field002) s2
from t3 group by field001) cwhere a.field001=b.field001 and b.field001=c.field001
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