For example, I have a table ordered by column "code". Also, I now exact number of rows of my table (6 for this case).
I need to create one more column with rank using next rules:
The first value has the first code (1)
The second value has the last code (6)
The third value has the second code (2)
The forth value has penultimate code (5) etc.
How can I create this order? Even if you have just an idea without query, share it with me, please.
You could use:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER(ORDER BY code ASC) rn1,
ROW_NUMBER() OVER(ORDER BY code DESC) rn2
FROM tab
)
SELECT *
FROM cte
ORDER BY ABS(rn2 - rn1) DESC, code;
db<>fiddle demo
How it works: two counters based on code, calculate difference so first and last has the same value, when tie prefer code.
I would use row_number() too, but I think the logic you want is more:
select *
from (
select t.*,
row_number() over(order by code asc ) rn_asc,
row_number() over(order by code desc) rn_desc
from tab t
) t
order by case when rn_asc <= rn_desc then rn_asc else rn_desc end, rn_asc;
This ranks records in both directions, and then uses the smallest of the two ranks for ordering. The second sorting criteria ensures that the smallest value of the two consistently comes first.
Related
Hopefully, someone can help me...
I'm trying to get the last two values from a row_number() window function. Let's say my results contain row numbers up to 6, for example. How would it be possible to get the rows where the row number is 5 and 6?
Let me know if it can be done with another window function or in another way.
Kind regards,
Using QUALIFY:
SELECT *
FROM tab
QUALIFY ROW_NUMBER() OVER(ORDER BY ... DESC) <= 2;
This approach could be further extended to get two rows per each partition:
SELECT *
FROM tab
QUALIFY ROW_NUMBER() OVER(PARTITION BY ... ORDER BY ... DESC) <= 2;
You can use top with order by desc like:
select top 2 row_number() over([partition by] [order by]) as rn
from table
order by rn desc
I'd say #Shmiel is the formal and elegant way, just in case, would be the same as :
WITH CTE AS
(SELECT product,
user_id,
ROW_NUMBER() OVER (PARTITION BY user_id order by product desc)
as RN
FROM Mytable)
SELECT product, user_id
FROM CTE
WHERE RN < 3;
You will use order by [order_condition] with "desc". And then you will use RN(row number) to select as many rows as you want
I have a peculiar problem at hand. I need to rank in the following manner:
Each ID gets a new rank.
rank #1 is assigned to the ID with the lowest date. However, the subsequent dates for that particular ID can be higher but they will get the incremental rank w.r.t other IDs.
(E.g. ADF32 series will be considered to be ranked first as it had the lowest date, although it ends with dates 09-Nov, and RT659 starts with 13-Aug it will be ranked subsequently)
For a particular ID, if the days are consecutive then ranks are same, else they add by 1.
For a particular ID, ranks are given in date ASC.
How to formulate a query?
You need two steps:
select
id_col
,dt_col
,dense_rank()
over (order by min_dt, id_col, dt_col - rnk) as part_col
from
(
select
id_col
,dt_col
,min(dt_col)
over (partition by id_col) as min_dt
,rank()
over (partition by id_col
order by dt_col) as rnk
from tab
) as dt
dt_col - rnk caluclates the same result for consecutives dates -> same rank
Try datediff on lead/lag and then perform partitioned ranking
select t.ID_COL,t.dt_col,
rank() over(partition by t.ID_COL, t.date_diff order by t.dt_col desc) as rankk
from ( SELECT ID_COL,dt_col,
DATEDIFF(day, Lag(dt_col, 1) OVER(ORDER BY dt_col),dt_col) as date_diff FROM table1 ) t
One way to think about this problem is "when to add 1 to the rank". Well, that occurs when the previous value on a row with the same id_col differs by more than one day. Or when the row is the earliest day for an id.
This turns the problem into a cumulative sum:
select t.*,
sum(case when prev_dt_col = dt_col - 1 then 0 else 1
end) over
(order by min_dt_col, id_col, dt_col) as ranking
from (select t.*,
lag(dt_col) over (partition by id_col order by dt_col) as prev_dt_col,
min(dt_col) over (partition by id_col) as min_dt_col
from t
) t;
I think I just need a little help with this but is there a way to incrementally count steps in SQL using some type of CTE row partition? I'm using SQL Server 2008 so won't be able to use the LAG function.
In the below, I am trying to find a way to calculate the Step Number as pictured below where for each unique ITEM in my table, in this case G43251, it calculates the process Step_Number based on the Date (timestamp) and the process type. For those with the same timestamp & process_type, it would label them both as the same Step_Number as there other fields that could cause the timestamp to repeat twice.
Right now I am playing around with this below and seeing how maybe I could fit in a DISTINCT timestamp methodology ? So that it doesn't count each row as something new.
WITH cte AS
(
SELECT
*,
ROW_NUMBER() OVER (ORDER BY Timestamp_Posted DESC)
- ROW_NUMBER() OVER (PARTITION BY Item ORDER BY Timestamp_Posted Desc) rn
FROM
#t1
)
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY Item, rn ORDER BY Timestamp_Posted DESC) rn2
FROM
cte
ORDER BY
Timestamp_Posted DESC
Please use dense_rank() instead of row_number()
SELECT *, dense_rank() OVER(Partition By Item ORDER BY Timestamp_Posted, Process_Type ) Step_Number
FROM #t1
ORDER BY Timestamp_Posted DESC
My current data is as follows:
And I want Data to be
When I use the row_number function it is reordering itself and giving me the wrong row_number,as below
If we See "Adjusted conversion COst" value 0.160 is coming top of result and is numbered 1 which is wrong as per the first screenshot it should be numbered 3
Thanks
MYSQL Using Variable
Result - http://www.sqlfiddle.com/#!9/406f64/8/0
select
colo1,f7,
if(colo1='Total Adj. Conversion Spend',#initVal:=#initVal+1,1) as RowNumber
from temp,(select #initVal:=0) vars
MS-SQL Using Rank and Row Number
I've used Row_Number() to preserve the order and then using Rank() inside a case statement
http://www.sqlfiddle.com/#!18/fde9f/15/0
select subquery_1.colo1,subquery_1.f7
,case when subquery_1.colo1='Total Adj. Conversion Spend' then
rank() over (partition by colo1 order by rownum) else 1 end as rnk
from
(select *,row_number() OVER (ORDER BY (Select 0)) as rownum from temp) as subquery_1
order by subquery_1.rownum
I having below data in one table.
And I want to get NEXT out data from OUT column. So used LEAD function in below query.
SELECT ROW_NUMBER,TIMESTAMP,IN,OUT,LEAD(OUT) OVER (PARTITION BY NULL ORDER BY TIMESTAMP) AS NEXT_OUT
FROM MYTABLE;
It gives data as below NEXT_OUT column.
But I need to know the matching next column value in sequential way like DESIRED columns. Please let me know how can i achieve this in Oracle LEAD FUNCTION
THANKS
Assign row number to all INs and OUTs separately, sort the results by placing them in a single column and calculate LEADs:
WITH cte AS (
SELECT t.*
, CASE WHEN "IN" IS NOT NULL THEN COUNT("IN") OVER (ORDER BY "TIMESTAMP") END AS rn1
, CASE WHEN "OUT" IS NOT NULL THEN COUNT("OUT") OVER (ORDER BY "TIMESTAMP") END AS rn2
FROM t
)
SELECT cte.*
, LEAD("OUT") OVER (ORDER BY COALESCE(rn1, rn2), rn1 NULLS LAST) AS NEXT_OUT
FROM cte
ORDER BY COALESCE(rn1, rn2), rn1 NULLS LAST
Demo on db<>fiddle
Enumerate in the "in"s and the "out"s and use that information for matching.
select tin.*, tout.out as next_out
from (select t.*,
count(in) over (order by timestamp) as seqnum_in
from t
) tin left join
(select t.*,
count(out) over (order by timestamp) as seqnum_out
from t
) tout
on tin.in is not null and
tout.out is not null and
tin.seqnum_in = tout.seqnum_out;