Sequence grouping in TSQL - sql

Question

I'm trying to group data in sequence order. Say I have the following table:
| 1 | A |
| 1 | A |
| 1 | B |
| 1 | B |
| 1 | C |
| 1 | B |
I need the SQL query to output the following:
| 1 | A | 1 |
| 1 | A | 1 |
| 1 | B | 2 |
| 1 | B | 2 |
| 1 | C | 3 |
| 1 | B | 4 |
The last column is a group number that is incremented in each group. The important thing to note is that rows 3, 4 and 5 contain the same data which should be grouped into 2 groups not 1.

Answer 1

For MSSQL2008:
Suppose you have a SampleStatuses table:
Status Date
A 2014-06-11
A 2014-06-14
B 2014-06-25
B 2014-07-01
A 2014-07-06
A 2014-07-19
B 2014-07-21
B 2014-08-13
C 2014-08-19
you write the following:
;with
cte as (
select top 1 RowNumber, 1 as GroupNumber, [Status], [Date] from SampleStatuses order by RowNumber
union all
select c1.RowNumber,
case when c2.Status <> c1.Status then c2.GroupNumber + 1 else c2.GroupNumber end as GroupNumber, c1.[Status], c1.[Date]
from cte c2 join SampleStatuses c1 on c1.RowNumber = c2.RowNumber + 1
)
select * from cte;
you get this result:
RowNumber GroupNumber Status Date
1 1 A 2014-06-11
2 1 A 2014-06-14
3 2 B 2014-06-25
4 2 B 2014-07-01
5 3 A 2014-07-06
6 3 A 2014-07-19
7 4 B 2014-07-21
8 4 B 2014-08-13
9 5 C 2014-08-19

Answer 2

The normal way you would do what you want is the dense_rank function:
select key, val,
dense_rank() over (order by key, val)
from t
However, this does not address the problem of separating the last groups.
To handle this, I have to assume there is an "id" column. Tables, in SQL, do not have an ordering, so I need the ordering. If you are using SQL Server 2012, then you can use the lag() function to get what you need. Use the lag to see if the key, val pair is the same on consecutive rows:
with t1 as (
select id, key, val,
(case when key = lead(key, 1) over (order by id) and
val = lead(val, 1) over (order by id)
then 1
else 0
end) as SameAsNext
from t
)
select id, key, val,
sum(SameAsNext) over (order by id) as GroupNum
from t
Without SQL Server 2012 (which has cumulative sums), you have to do a self-join to identify the beginning of each group:
select t.*,
from t left outer join
t tprev
on t.id = t2.id + 1 and t.key = t2.key and t.val = t2.val
where t2.id is null
With this, assign the group as the minimum id using a join:
select t.id, t.key, t.val,
min(tgrp.id) as GroupId
from t left outer join
(select t.*,
from t left outer join
t tprev
on t.id = t2.id + 1 and t.key = t2.key and t.val = t2.val
where t2.id is null
) tgrp
on t.id >= tgrp.id
If you want these to be consecutive numbers, then put them in a subquery and use dense_rank().

Answer 3

This will give you rankings on your columns.
It will not give you 1,2,3 however.
It will give you 1,3,6 etc based on how many in each grouping
select
a,
b,
rank() over (order by a,b)
from
table1
See this SQLFiddle for a clearer idea of what I mean: http://sqlfiddle.com/#!3/0f201/2/0