I have case in which I want same autoincrement value in new created column for same business code
I have tried below but I am not getting expected result
select *
, rank() over (partition by business_code order by ID)
from table
I am getting same same value in ID column for all business code which is not desired result.
My Output
Expected Output
Can you try the the following SQL statement:
SELECT id
,business_code
,DENSE_RANK() OVER (ORDER BY m) NewColumn
FROM (SELECT id
,business_code
,MIN(id) OVER (PARTITION BY business_code) m
FROM myTable) d
An explanation how it works: because of provided data i thought to get the minimum id of each business_code first. And as second step ranking that minimum id by value.
Use this
select row_number() over (order by (select null)), a.*
from Table a;
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 have a table as shown in the image.In Number column, the values are appeared more than once (for example 63 appeared twice). I would like to keep only one value. Please see my code:
delete from t1 where
(SELECT *,row_number() OVER (
PARTITION BY
Number
ORDER BY
Date) as rn from t1 where rn > 1)
It shows error. Can anyone please assist.
enter image description here
The column created by row_number() was not accessed by your main query, in order to enable that, you can create a quick sub query and use the desired filter
SELECT *
FROM
(
SELECT *,
row_number() OVER (PARTITION BY Number ORDER BY Date) as rn
FROM t1 ) T
where rn = 1;
The partition by determines how row numbers repeat. The row numbers are assigned per group of partition by keys. So, you can get duplicates.
If you want a unique row number over all rows, just leave out the partition by:
select t1.*
from (select t1.*,
row_number() over (order by date) as rn
from t1
) t1
where rn > 1
if you want to keep only one value, rn = 1 instead of "> 1"
I use SQL Server 2012,
I have a following table:
id, name, surname, timestamp, type
type has two possible values: 1 and 2.
Now, I would like to find two rows - for each group (1 and 2) row with maximal value in particular type.
The problem is that I would like to find both name and surname.
I can do it with SELECT TOP 1 - WHERE ORDER BY - UNION approach, but I would like to find antother, better idea.
Can you help me ?
This sounds like you want the most recent for each row, for each type. If that's the case, here is a way with row_number()
with cte as(
select
id
,name
,surname
,timestamp
,type
RN = row_number() over (partition by id,type order by timestamp desc))
select *
from cte
where RN = 1
This is the sample table
What I need to achieve is to get or display only the record of tenant with the highest month value. If ever month is equal, I need to base on the latest date value. Here is the sample desired output
With this, I started by this code using max function and incorporated temp table, but unable to get the desired result.
select tenant, name, date, month
into #sample
from tenant
select *
from #sample
where months = (select max(months)from #sample)
and output to something like this. As I believe, the code is getting the max value in the whole list not considering per tenant filtering.
Any help will be greatly appreciated :)
This can be done with the row_number window function:
select tenant, name, date, months
from (select t.*,
row_number() over (partition by t.tenant, t.name order by t.months desc, t.date desc) as rn
from TableName t) x
where rn = 1
You can use a row_number function.
Query
;with cte as
(
select rn = row_number() over
(
partition by tenant
order by months desc,[date] desc
),*
from table_name
)
select tenant,name,[date],months from cte
where rn = 1;
How can I add a sequential row number to a query that is using order by?
Let say I have a request in this form :
SELECT row_number() over(), data
FROM myTable
ORDER BY data
This will produce the desired result as rows are ordered by "data", but the row numbers are also ordered by data. I understand this is normal as my row number is generated before the order by, but how can I generate this row number after the order by?
I did try to use a subquery like this :
SELECT row_number() over(ORDER BY data), *
FROM
(
SELECT data
FROM myTable
ORDER BY data
) As t1
As shown here, but DB2 doesn't seem to support this syntax SELECT ..., * FROM
Thanks !
You also need to use alaias name before '*'
SELECT row_number() over(ORDER BY data), t1.*
FROM
(
SELECT data
FROM myTable
ORDER BY data
) As t1
You don't need a subquery to do this,
SELECT data , row_number() over(ORDER BY data) as rn
FROM myTable
ORDER BY data