Get the count of a value for each row in SQL - sql-server

Question

I have a SQL question. Below is my data and query
select ID from table
ID
4
4
5
3
5
3
3
What query should I add to my select list so that it gives me the occurrence of a value for each specfic row(I do not want the total count, I want something like a rank of occurrence).
4 1--this is the first time we got a 4 in the list, so 1
4 2--this is the second time we got a 4 in the list, so 2
5 1--this is the first time we got a 5 in the list, so 1
3 1--this is the first time we got a 3 in the list, so 1
5 2--this is the second time we got a 5 in the list, so 2
3 2--this is the second time we got a 3 in the list, so 2
3 3--this is the third time we got a 3 in the list, so 3

Answer 1

You want ROW_NUMBER():
SELECT
[Id],
ROW_NUMBER() OVER (PARTITION BY [Id] ORDER BY [Id])
FROM ...
You'll probably need to change the Order By clause to get the right output.

Answer 2

To generate a numbering from an order of rows, use row_number() or rank(). The problem is, row_number() does not guarantee the original order will be preserved. You could try this:
select
[id],
row_number() over (partition by id, order by (select 1)) as [rank]
from #t
However, you will find the results are not in the original order and somewhat confusing:
id rank
3 1
3 2
3 3
4 1
4 2
5 1
5 2
To preserve the original row order, you could build a temporary table or table variable with an identity column. Select a row_number() partitioned by id from there:
declare #t table ([tkey] int identity(1,1) primary key clustered, [id] int)
insert into #t (id) values (4), (4), (5), (3), (5), (3), (3)
select
[id],
row_number() over (partition by [Id] order by [tkey]) as [rank]
from #t
order by [tkey]
Note that the final order by [tkey] really is necessary. The query has the desired results:
id rank
4 1
4 2
5 1
3 1
5 2
3 2
3 3
Here is a Common Table Expression (CTE) approach. The CTE adds a row_number() to maintain the rows in the original order. (This is the equivalent of the identity column in the previous example.) The actual ranking comes with the second row_number() when it does a partition by id. This causes the 1st 4 gets 1, the 2nd 4 gets 2, etc.
The second row_number() must be ordered by the original order to rank correctly, but this is still not enough to preserve the order in the output. A final order by ensures the ending order is the same.
declare #t table (id int)
insert into #t (id) values (4), (4), (5), (3), (5), (3), (3)
;with [tRows] (rownum, id) as
(
select
row_number() over (order by (select 1)) as [rownum],
[id]
from #t
)
select
[id],
row_number() over (partition by id order by [rownum]) as [rank]
from [tRows]
order by [rownum]
This query also has the desired results:
id rank
4 1
4 2
5 1
3 1
5 2
3 2
3 3
In this example, you could use rank() instead of the second row_number(). The difference between the functions is explained well in this question. rank() would not work the same if, somehow, there were duplicate row numbers generated by the first row_number(), but that cannot happen.