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select sum£¨a.t)from ta a group by ta.a,ÕâÌõÓï¾ä½«±ítaÖеÄÊý¾Ý°´ÕÕa×ֶηÖ×é»ã×Üt×ֶΡ£
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3 30
4 50
ÎÒÏëµÃµ½Õâ¸ö½á¹û£º
1 0
2 0
3 30
4 50
5 0
Õâ¸ö½á¹û£¬ÇëÎÊsqlÓï¾äÔõôд£¿¶àл

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nobody knows£¿

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ÔÙdecode(sum(a.t),null,0,sum(a.t))

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ÔÙdecode(sum(a.t),null,0,sum(a.t))

up

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SQL code:
select ROWNUM,0 from ÓÐÊý¾ÝµÄ±í(±í¼Ç¼>ÄãÏëÒªµÄÐòÁÐN,ÄãµÄÀý×ÓÀïÊÇ5) where ROWNUM <5
ÀýÈç SELECT ROWNUM,0 AS AGE from EMP where ROWNUM <5
ROWNUM AGE
1 0
2 0
3 0
4 0
5 0
SELECT * from A;
ID AGE
1 30
3 40
select id,max(AGE) from
(SELECT ROWNUM ID,0 AS AGE from EMP where ROWNUM <=5
union
SELECT * from A)
group by id

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select t1.rn, nvl(sum(t2.t),0) from
(select rownum rn from dual connect by rownum <= 5) t1,
ta t2
where t1.rn = t2.a(+)
group by t1.rn;

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select t.l, nvl(sum(t2.t),0) from 
(
select level l from dual connect by level <=5) t  
left join ta t2 on (t.l = t2.a)
group by t.l

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case when sum(columnname) = 0 then 0
else then sum(c