sql查询基于连续性获取oracle中一行的唯一id

Question

I have a problem that needs to be solved using sql in oracle.

I have a dataset like given below:

value | date
------------- 
1 | 01/01/2017 
2 | 02/01/2017 
3 | 03/01/2017 
3 | 04/01/2017
2 | 05/01/2017 
2 | 06/01/2017 
4 | 07/01/2017 
5 | 08/01/2017

I need to show the result in the below format:

value    | date     |   Group
1   |   01/01/2017  |   1
2   |   02/01/2017  |   2
3   |   03/01/2017  |   3
3   |   04/01/2017  |   3
2   |   05/01/2017  |   4
2   |   06/01/2017  |   4
4   |   07/01/2017  |   5
5   |   08/01/2017  |   6

The logic is whenever value changes over date , it gets assigned a new group/id , but if its the same as the previous one , then its part of the same group .

Answer 1

Here is one method using lag() and cumulative sum:

select t.*,
       sum(case when value = prev_value then 0 else 1 end) over (order by date) as grp
from (select t.*,
             lag(value) over (order by date) as prev_value
      from t
     ) t;

The logic here is to simply count the number of times that the value changes from one month to the next.

This assumes that date is actually stored as a date and not a string. If it is a string, then the ordering will not be correct. Either convert to a date or use a column that specifies the correct ordering.

Answer 2

Here is a solution using the MATCH_RECOGNIZE clause, introduced in Oracle 12.*

select value, dt, mn as grp
from   inputs
match_recognize (
  order by dt
  measures match_number() as mn
  all rows per match
  pattern  ( a b* )
  define   b as value = prev(value)
)
order by dt   --  if needed
;

Here is how this works: Other than SELECT, FROM and ORDER BY, the query has only one clause, MATCH_RECOGNIZE. What this clause does is: it takes the rows from inputs and it orders them by dt . Then it searches for patterns: one row, marked as a , with no constraints, followed by zero or more rows b , where b is defined by the condition that the value is the same as for the prev[ious] row. What the clause calculates or measures is the match_number() - first "match" of the pattern, second match etc. We use this match number as the group number ( grp ) in the outer query - that's all we needed!

*Notes : The existence of solutions like this shows why it is important for posters to state their Oracle version. (Run the statement select * from v$version to find out.) Also: date and group are reserved words in Oracle and shouldn't be used as column names. Not even for posting made-up sample data. (There are workarounds but they aren't needed in this case.) Also, whenever using dates like 03/01/2017 in a post, please indicate whether that is March 1 or January 3, there's no way for "us" to tell. (It wasn't important in this case, but it is in the vast majority of cases.)