找出連續的資料 如1,2,3,48,50,51,52,53,67,68
找出連續的數字的起點和重點
1,348,48
51,53
67,68
create table test.range_problem(
a int not null,
primary key (a));
insert into test.range_problem values(1);
insert into test.range_problem values(2);
insert into test.range_problem values(3);
insert into test.range_problem values(48);
insert into test.range_problem values(50);
insert into test.range_problem values(51);
insert into test.range_problem values(52);
insert into test.range_problem values(53);
insert into test.range_problem values(66);
insert into test.range_problem values(67);
--思路1 找最後乙個不連續的數
select t1.a
from test.range_problem t1
where not exists
(select a
from test.range_problem t2
where t2.a -1 = t1.a
)--思路2 找所有行對應的結束值
select row_number()over(order by tbase.a) id
,tbase.a
,(select min(a)
from test.range_problem t1
where not exists
(select a
from test.range_problem t2
where t2.a -1 = t1.a
)and t1.a >= tbase.a --t1.a 是3,48,53,67 tbase.a 是每行的值
) a_end
from test.range_problem tbase
--思路3 分組後找出所有連續的起始值和結束值
select min(a) a_start, a_end
from(
select row_number()over(order by tbase.a) id
,tbase.a
,(select min(a)
from test.range_problem t1
where not exists
(select a
from test.range_problem t2
where t2.a -1 = t1.a
)and t1.a >= tbase.a --t1.a 是3,48,53,67 tbase.a 是每行的值
) a_end
from test.range_problem tbase
) tgroup by t.a_end
--對上面語句的優化
with pot as(
select a
from test.range_problem t
where not exists
( select a
from test.range_problem tc
where tc.a -1 = t.a)
)select tbase.a a_start,(select min(a)
from pot t
where tbase.a <= t.a
) a_end
from test.range_problem tbase
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