select
isnull (a.nday,b.nday) as nday,
isnull (b.num,0 ) as num(
select nday from lm group by nday) a full join (
SELECT nday,count(*) as num FROM lm GROUP BY nday
) b
on a.nday=b.nday
解决方案 »
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create #tmp
(
[time] datetime
)--输入连续的日期
declare @i int
set @i = 100
while @i > 0
begin
insert into #tmp select dateadd(day,@i,'2004/05/06')
end --输出
select [time],count(id) from ( select [time],id from 表 union all select [time],null as ID from #tmp ) tdrop table #tmp哈哈,方法比较笨~~。
select @dt=min(nday),@i=datediff(day,max(nday),min(nday))+1
from lm
if @@rowcount=0 returnset rowcount @i
select id=identity(int,0,1),a=0 into #t from syscolumns a,syscolumns b
set @i=@i-@@rowcount
while @i>0
begin
set rowcount @i
insert #t select 0 from syscolumns a,syscolumns b
set @i=@i-@@rowcount
end
set rowcount 0select nday=dateadd(day,a.id,@dt),[count]=isnull(b.[count],0)
from #t a left join(
SELECT nday,[count]=count(*) FROM lm GROUP BY nday
)b on a.id=datediff(day,@dt,b.nday)drop table #t
create #tmp
(
[time] datetime
)--输入连续的日期
declare @i int
set @i = 100
while @i > 0
begin
insert into #tmp select dateadd(day,@i,'2004/05/06')
set @i = @i - 1
end --输出
select [time],count(id) from ( select [time],id from 表 union all select [time],null as ID from #tmp ) t group by [time]drop table #tmp哈哈,方法比较笨~~。