Count consecutive duplicate values in SQL

Question

I have a table like so

ID     OrdID     Value
1      1          0     
2      2          0
3      1          1
4      2          1
5      1          1
6              


        

Answer 1

I am going to presume that id is unique and increasing. You can get counts of consecutive values by using the different of row numbers. The following counts all sequences:

select grp, value, min(id), max(id), count(*) as cnt
from (select t.*,
             (row_number() over (order by id) - row_number() over (partition by value order by id)
             ) as grp
      from table t
     ) t
group by grp, value;

If you want the longest sequence of 0s:

select top 1 grp, value, min(id), max(id), count(*) as cnt
from (select t.*,
             (row_number() over (order by id) - row_number() over (partition by value order by id)
             ) as grp
      from table t
     ) t
group by grp, value
having value = 0
order by count(*) desc

Answer 2

A query using not exists to find consecutive 0s

select top 1 min(t2.id), max(t2.id), count(*)
from mytable t
join mytable t2 on t2.id <= t.id
where not exists (
    select 1 from mytable t3
    where t3.id between t2.id and t.id
    and t3.value <> 0
)
group by t.id
order by count(*) desc 

http://sqlfiddle.com/#!3/52989/3