I have a table as like below
Id RC_CLASS RC_DATE RC_TYPE
14 FI-321619 22-Jan-16 S
14 FI-399481 29-Jan-16 D
14 FI-321619 20-Jan-17 S
Here is what i tried
SELECT *
FROM (SELECT rc.*,
RANK() OVER (PARTITION BY ID,RC_CLASS order by rc__date) AS LATEST_VERSION
FROM table
)
WHERE LATEST_VERSION = 1
ORDER BY rc_vendorid;
Expected output
Id RC_CLASS RC_DATE RC_TYPE
14 FI-399481 29-Jan-16 D
14 FI-321619 20-Jan-17 S
I wanna group by ID and Class and bring top one sort by the RC_DATE. What i am getting is always the top one based on date, partition is not working here. What is missing?
I think you are very close. Basically, you just need a descending sort to get the latest version:
SELECT rc.*
FROM (SELECT rc.*,
RANK() OVER (PARTITION BY ID, RC_CLASS ORDER BY rc_date DESC) AS LATEST_VERSION
FROM table rc
) rc
WHERE LATEST_VERSION = 1
ORDER BY rc_vendorid;
I note that you use RANK() for this. This can return duplicates, if you have two rows on the same date. If that is not desirable, you can use ROW_NUMBER() which would arbitrarily choose one (if all the other keys are the same).
Related
Let's say we have the database table below, called USER_JOBS.
I'd like to write an SQL query that reflects this algorithm:
Divide the whole table in groups of rows defined by a common USER_ID (in the example table, the 2 resulting groups are colored yellow & green)
From each group, select the oldest row (according to SCHEDULE_TIME)
From this example table, the desired SQL query would return these 2 rows:
You can use ranking function (supported in most RDBS):
SELECT *
FROM
(
SELECT *
,ROW_NUMBER() OVER (PARTITION BY USER_ID ORDER BY SCHEDULE_TIME DESC) AS RowID
FROM [table]
)
WHERE RowID = 1
WITH Ranked AS (
SELECT
RANK() OVER (PARTITION BY User_ID ORDER BY ScheduleTime DESC) as Ranking,
*
FROM [table_name]
)
SELECT Status, Sob_Type, User_ID, TimeStamp FROM ranking WHERE Ranks = 1;
I prepared query that select date from table. In table I got: rank, name, citycode as columns. When I am doing something like that:
select name, citycode
from tab20
where rank <= 20
I got resault of first 20 rows that gets rank <= 20. And Everything would be ok, but I have to show results of first 20 rows per every citystate. Is it possible to create in one query ? I was tryin union etc but it doesn't work well.
Thanks
You would use the row_number() function. Based on the rank that would be:
select t.*
from (select t.*,
row_number() over (partition by citycode order by rank) as seqnum
from tab20 t
) t
where seqnum <= 20;
I have a table called TABLE_SCREW where I want to get the latest records for each code.
For example, in the table below you should obtain the records with ids 3 and 7.
I am a newbie in sql and I hope you can help me.
You could use:
SELECT TOP 1 WITH TIES *
FROM TABLE_SCREW
ORDER BY ROW_NUMBER() OVER(PARTITION BY CODE ORDER BY Date DESC);
Another approach(may have better performance):
SELECT * -- here * should be replaced with actual column names
FROM (SELECT *,ROW_NUMBER() OVER(PARTITION BY CODE ORDER BY Date DESC) AS rn
FROM TABLE_SCREW) sub
WHERE sub.rn = 1;
Let's say I have the following table
Sku | Number | Name
11 1 hat
12 1 hat
13 1 hats
22 2 car
33 3 truck
44 4 boat
45 4 boat
Is there an easy way to figure out how to find the differences within each Number. For example, with the table above, I would want the query to output:
13 | 1 | hats
The reason for this is because our program processes the rows as long as the number matches the name. If there is an instance where the name doesn't match but the rest of the names do, it will fail.
You can find the most common value (the "mode") using window functions and aggregation:
select t.*
from (select number, name, count(*) as cnt,
row_number() over (partition by number order by count(*) desc) as seqnum
from t
group by number, name
) t
where seqnum = 1;
You could then find everything that is not the mode using a join. The easier way is just to change the where condition:
select t.*
from (select number, name, count(*) as cnt,
row_number() over (partition by number order by count(*) desc) as seqnum
from t
group by number, name
) t
where seqnum > 1;
Note: If there are ties in frequency for the most common value, then an arbitrary most common value is chosen.
EDIT:
Actually, if you want the original skus, you might as well do the join:
with modes as (
select t.*
from (select number, name, count(*) as cnt,
row_number() over (partition by number order by count(*) desc) as seqnum
from t
group by number, name
) t
where seqnum = 1
)
select t.*
from t join
modes
on t.number = modes.number and t.name <> modes.name;
This will ignore NULL values (but the logic can easily be fixed to accommodate them).
I have table with some data, for example
ID Specified TIN Value
----------------------
1 0 tin1 45
2 1 tin1 34
3 0 tin2 23
4 3 tin2 47
5 3 tin2 12
I need to get rows with all fields by MAX(Specified) column. And if I have few row with MAX column (in example ID 4 and 5) i must take last one (with ID 5)
finally the result must be
ID Specified TIN Value
-----------------------
2 1 tin1 34
5 3 tin2 12
This will give the desired result with using window function:
;with cte as(select *, row_number(partition by tin order by specified desc, id desc) as rn
from tablename)
select * from cte where rn = 1
Edit: Updated query after question edit.
Here is the fiddle
http://sqlfiddle.com/#!9/20e1b/1/0
SELECT * FROM TBL WHERE ID IN (
SELECT max(id) FROM
TBL WHERE SPECIFIED IN
(SELECT MAX(SPECIFIED) FROM TBL
GROUP BY TIN)
group by specified)
I am sure we can simplify it further, but this will work.
select * from tbl where id =(
SELECT MAX(ID) FROM
tbl where specified =(SELECT MAX(SPECIFIED) FROM tbl))
One method is to use window functions, row_number():
select t.*
from (select t.*, row_number() over (partition by tim
order by specified desc, id desc
) as seqnum
from t
) t
where seqnum = 1;
However, if you have an index on tin, specified id and on id, the most efficient method is:
select t.*
from t
where t.id = (select top 1 t2.id
from t t2
where t2.tin = t.tin
order by t2.specified desc, id desc
);
The reason this is better is that the index will be used for the subquery. Then the index will be used for the outer query as well. This is highly efficient. Although the index will be used for the window functions; the resulting execution plan probably requires scanning the entire table.