自联结

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数据库:ASE12.5
表:
ID                     COL
1                      2434
2                      454
3                      342
4                      67
5                      [NULL]
6                      [NULL]
7                      23432
8                      32
9                      [NULL]
10                     23
11                     [NULL]
12                     1232
13                     [NULL]
14                     [NULL]
15                     [NULL]

要求得到这个结果:
ID                     COL
4                      67
5                      [NULL]
6                      [NULL]
8                      32
9                      [NULL]
10                     23
11                     [NULL]
12                     1232
13                     [NULL]
14                     [NULL]
15                     [NULL]
即空字段及它前面的那个非空字段保留
-----------------------------------
create table F(ID numeric(10,0) identity,                    COL int null)
go
insert into F values(2434)
insert into F values(454)
insert into F values(342)
insert into F values(67)
insert into F values(NULL)
insert into F values(NULL)
insert into F values(23432)
insert into F values(32)
insert into F values(NULL)
insert into F values(23)
insert into F values(NULL)
insert into F values(1232)
insert into F values(NULL)
insert into F values(NULL)
insert into F values(NULL)
go
select * into #tmp from F a where a.COL=NULL or exists (select 1 from F b where  a.COL<>NULL and b.ID=a.ID+1 and b.COL=NULL)
select * from #tmp