I working on a query for SQL Server 2016. I have order by serial_no and group by pay_type and I would like to add row number same example below
row_no | pay_type | serial_no
1 | A | 4000118445
2 | A | 4000118458
3 | A | 4000118461
4 | A | 4000118473
5 | A | 4000118486
1 | B | 4000118499
2 | B | 4000118506
3 | B | 4000118519
4 | B | 4000118521
1 | A | 4000118534
2 | A | 4000118547
3 | A | 4000118550
1 | B | 4000118562
2 | B | 4000118565
3 | B | 4000118570
4 | B | 4000118572
Help me please..
SELECT
ROW_NUMBER() OVER(PARTITION BY paytype ORDER BY serial_no) as row_no,
paytype, serial_no
FROM table
ORDER BY serial_no
You can assign groups to adjacent pay types that are the same and then use row_number(). For this purpose, the difference of row numbers is a good way to determine the groups:
select row_number() over (partition by pay_type, seqnum - seqnum_2 order by serial_no) as row_no,
t.*
from (select t.*,
row_number() over (order by serial_no) as seqnum,
row_number() over (partition by pay_type order by serial_no) as seqnum_2
from t
) t;
This type of problem is one example of a gaps-and-islands problem. Why does the difference of row numbers work? I find that the simplest way to understand is to look at the results of the subquery.
Here is a db<>fiddle.
add this to your select list
ROW_NUMBER() OVER ( ORDER BY (SELECT 1) )
since you already sorting by your stuff, so you don't need to sorting in your windowing function so consuming less CPU,
Related
Given the following table P_PROV
+----+-----------+-----------+
| id | date | person_id |
+----+-----------+-----------+
| 1 |19/06/2019 | 1 |
| 2 |18/07/2010 | 2 |
| 3 |19/06/2020 | 1 |
| 4 |17/06/2020 | 2 |
| 5 |28/06/2020 | 3 |
+----+-----------+-----------+
I want this output
+----+-----------+-----------+
| id | date | person_id |
+----+-----------+-----------+
| 3 |19/06/2020 | 1 |
| 4 |17/06/2020 | 2 |
| 5 |28/06/2020 | 3 |
+----+-----------+-----------+
Putting this in words, I want to return per person the maximum date. I tried something like this
SELECT DISTINCT pp.date, pp.id FROM P_PROV pp
WHERE (SELECT MAX(aa.date)
FROM P_PROV aa) = pp.date;
This one is only returning one row (of course, because the MAX will return the maximum date only), but I really don't know how to approach this issue, any kind of help would be appreciated
ROW_NUMBER provides one way to handle this:
SELECT id, date, person_id
FROM
(
SELECT t.*, ROW_NUMBER() OVER (PARTITION BY person_id ORDER BY date DESC) rn
FROM yourTable t
) t
WHERE rn = 1;
Oracle has a fun way to do this using aggregation:
select max(id) keep (dense_rank first order by date desc) as id,
max(date) as date, person_id
from P_PROV
group by person_id;
Given that your ids are increasing, this probably also does what you want:
select max(id) as id, max(date) as date, person_id
from P_PROV
group by person_id;
My table looks like this, what I'm trying to achieve is to pull out all the records for one user for the product that have the earliest date
product |type_id| user | Date |Desired ROW_NUMBER as output |
-------+--------+------+-------+---------------------
1 | 1 | A | 0101 | 1
1 | 1 | A | 0102 | 1
2 | 3 | A | 0105 | 2
2 | 5 | A | 0105 | 2
3 | 7 | B | 0101 | 1
3 | 8 | B | 0104 | 1
So I want to pull all the records with "1" in the desired row_num column, but I haven't figured out hot to get this without doing another group by. Any helps would be appreciated.
You can use window functions:
select t.*
from (select t.*,
rank() over (partition by user order by min_date) as seqnum
from (select t.*,
min(date) over (partition by user, product) as min_date
from t
) t
) t
where seqnum = 1;
Or, with only one subquery:
select t.*
from (select t.*,
min(date) over (partition by user, product) as min_date_up,
min(date) over (partition by user) as min_date_u
from t
) t
where min_date_u = min_date_up;
You can interpret this as "return all rows where the product has the minimum date for the user".
Here is a db<>fiddle.
SELECT * FROM [tableName] WHERE Desired ROW_NUMBER = 1 ORDER BY Date[DESC, ASC]
Pass the Desired ROW_NUMBER value dynamically as a parameter.
I'm using ROW_NUMBER and I'm trying to compare arr in rn 1 to arr in rn 2,3,4,etc
to see if they overlap. I can do this with a subquery / simple join. Is there a way that AVOIDS a join?
rn | id | job | arr |desired_result
---+----+-----+--------+---------
1 | 1 | 100 | {1,2} | {1,2}
2 | 1 | 101 | {2,3} | {1,2}
3 | 1 | 102 | {5,6,8}| {1,2}
4 | 1 | 103 | {2,7} | {1,2}
I made a dbfiddle
--USING JOIN
WITH a AS (
SELECT
ROW_NUMBER() OVER (PARTITION BY id ORDER by job) as rn
,*
FROM a_table
)
SELECT *
FROM (
SELECT id,arr
FROM a
WHERE rn = 1
) x
JOIN a
ON a.id=x.id
You can use first_value():
SELECT a.*, first_value(arr) over (partition by id order by job)
FROM a_table a;
row_number() does not seem necessary.
I have a table with 4 columns:
AcctNumb | PeriodEndingDate | WaterConsumption | ReadingType
There are multiple records for each AcctNumb, with the date that each record was recorded.
What I want to do is grab the most recent date, consumption reading, and reading type for each account.
I have tried using MAX(PeriodEndingDate) and GROUP BY AcctNumb, but I would need to aggregate all the other values, and none of the aggregate functions help me for the WaterConsumption, etc.
Can anyone point me in the right direction?
Thanks
EDIT
Here is a sample table
+----------+------------------+------------------+-------------+
| AcctNumb | PeriodEndingDate | WaterConsumption | ReadingType |
+----------+------------------+------------------+-------------+
| 1000 | 2018-03-31 | 122230 | A |
| 1001 | 2018-03-31 | 24850 | A |
| 1002 | 2018-03-31 | 88540 | A |
| 1000 | 2017-12-31 | 123800 | A |
| 1001 | 2017-12-31 | 3000 | E |
+----------+------------------+------------------+-------------+
The ReadingType is whether it's an actual (A) reading, or an estimate (E).
Try this
SELECT
AcctNumb,
PeriodEndingDate,
WaterConsumption,
ReadingType
FROM (SELECT
AcctNumb,
PeriodEndingDate,
WaterConsumption,
ReadingType,
ROW_NUMBER() OVER (PARTITION BY AcctNumb ORDER BY PeriodEndingDate DESC) AS MostrecentRecord
FROM <TableName>) dt
WHERE MostrecentRecord= 1
This can be done using ROW_NUMBER. It has been asked an answered thousands of times but the query is easier to write than find a duplicate.
select *
from
(
select *
, RowNum = ROW_NUMBER() over(partition by AcctNumb order by PeriodEndingDate)
from YourTable
) x
where x.RowNum = 1
SELECT DQ.* FROM
(SELECT *,
Row_Number() OVER (PARTITION BY AcctNumb ORDER BY PeriodEndingDate DESC) AS RN
FROM YourTable
) AS DQ
WHERE DQ.RN = 1
i have a table in a postgres DB which has the following structure:
id | date | groupme1 | groupme2 | value
----------------------------------------
1 |
2 |
3 |
Now i want to achieve the following:
Grouping the table after groupme1 and groupme2
Get the value for every group
But only the last entry for each group-compination (odered after date)
Example:
id | date | groupme1 | groupme2 | value
---------------------------------------
| | A | 1 | 4
| | A | 2 | 7
| | A | 3 | 3
| | B | 1 | 9
My current approach looks like this:
SELECT a.*
FROM table AS a
JOIN (SELECT max(id) AS id
FROM table
GROUP BY groupme1, groupme2) AS b
ON a.id = b.id
The Problems of this approach:
it asumes that higher dates have a higher id
it takes long
Is there a faster and better way of doing this? Can windowing function help with this?
I think you just want window functions:
select t.*
from (select t.*,
row_number() over (partition by groupme1, groupme2 order by date desc) as seqnum
from t
) t
where seqnum = 1;
Or, a better way to do this in Postgres uses distinct on:
select distinct on (groupme1, groupme2) t.*
from t
order by groupme1, groupme2, date desc;