Generate row index for table based on last updated date - sql

I would like to generate an index row in a table that may contain duplicates. the index need to be based on values from the table and last update date.
The data looks like this:
ID Val1 LastUpdateDate
-- ------ -------------
1 0 07.09.2019
1 2.5 12.09.2019
1 2.5 27.09.2019
1 3.5 01.10.2019
1 2.5 24.10.2019
1 0 01.11.2019
I would like to have:
ID Val1 LastUpdateDate index
-- ------ ------------- ----
1 0 07.09.2019 1
1 2.5 12.09.2019 2
1 2.5 27.09.2019 2
1 3.5 01.10.2019 3
1 2.5 24.10.2019 4
1 0 01.11.2019 5
I've tried with the following code but it's not working:
SELECT ID
,Value1
,Value2
,Value3
,LastUpdateDate
,(ROW_NUMBER() OVER (PARTITION BY ID ORDER BY last_update_date) - ROW_NUMBER()OVER(PARTITION BY ID,Value1,Value2,Value3 ORDER BY ID,Value1,Value2,Value3)) AS index
FROM Table1
ORDER BY LastUpdateDate

You can interpret this as a gaps-and-islands problem. However, I think the simplest way is to use LAG() and count the changes:
You seem to want dense_rank():
SELECT t1.*,
SUM(CASE WHEN prev_val1 = val1 THEN 0 ELSE 1 END) OVER (PARTITION BY id ORDER BY last_update_date) as seqnum
FROM (SELECT t1.*,
LAG(val1) OVER (PARTITION BY ID ORDER BY last_update_date) as prev_val1
FROM Table1 t1
) t1
ORDER BY LastUpdateDate;
Note that index is a really bad name for a column, because it is a SQL keyword.

Related

Update a column based on other rows column value

I have a table t which looks like this
key fill store end_date status
1 123 1 2019-04-30 0
2 1234 1 2019-04-30 0
3 123 1 2019-05-01 0
Now I need to update the first record and set status=1 as the third record has same fill, store value and it is latest.
Output:
key fill store end_date status
1 123 1 2019-04-30 1
2 1234 1 2019-04-30 0
3 123 1 2019-05-01 0
I tried calculating row_number and tried to update the column based on it but unable to figure out how to use the result in the update clause.
update t set
status = 1
from (
select *
from (
select *
, row_number() over (partition by fill, store order by end_dt desc) as row_num from t
) a
where row_num = 2
) b
This query is updating all the records, what should change in my query to get the expected result?
I think that you want:
with cte as (
select status, row_number() over(partition by fill, store order by end_date desc) rn
from t
)
update cte set status = 1 where rn > 1
In the common table expression, row_number() ranks records having the same fill and store by descending end_date. Then, the outer query sets status to 1 on rows that were not ranked first.
You can do a correlated subquery:
update my_table a
set status = 1
where exists (
select 1
from my_table b
where b.fill = a.fill
and b.store = a.store
and b.end_date > a.end_date
)

DENSE_RANK() Query

I have something similar to the below dataset...
ID RowNumber
101 1
101 2
101 3
101 4
101 5
101 1
101 2
What I would like to get is an additional column as below...
ID RowNumber New
101 1 1
101 2 1
101 3 1
101 4 1
101 5 1
101 1 2
101 2 2
I have toyed with dense_rank(), but no such luck.
Gordon already mentioned, you required a column to specify the order of data. If i consider ID as order by column, this following logic may help you to get your desired result-
WITH your_table(ID,RowNumber)
AS
(
SELECT 101,1 UNION ALL
SELECT 101,2 UNION ALL
SELECT 101,3 UNION ALL
SELECT 101,4 UNION ALL
SELECT 101,5 UNION ALL
SELECT 101,1 UNION ALL
SELECT 101,2
)
SELECT A.ID,A.RowNumber,
SUM(RN) OVER
(
ORDER BY ID
ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
) +1 New
FROM
(
SELECT *,
CASE
WHEN LAG(RowNumber) OVER(ORDER BY ID) > RowNumber THEN 1
ELSE 0
END RN
FROM your_table
)A
Above will always change the ROW NUMBER if the value in RowNumber decreased than previous one. Alternatively, the same output alsoo can be achieved if you wants to change row number whenever value 1 found. This is bit static option-
SELECT A.ID,A.RowNumber,
SUM(RN) OVER
(
ORDER BY ID
ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
) New
FROM(
SELECT *,
CASE
WHEN RowNumber = 1 THEN 1
ELSE 0
END RN
FROM your_table
)A
Output is-
ID RowNumber New
101 1 1
101 2 1
101 3 1
101 4 1
101 5 1
101 1 2
101 2 2
SQL tables represent unordered sets. There is no ordering unless a column specifies the ordering.
Assuming you have such a column, you can do what you want simply by counting the number of "1" up to each point:
select t.*,
sum(case when rownumber = 1 then 1 else 0 end) over (partition by id order by <ordering column>) as new
from t;
As Gordon alluded to, there is no default order in your example so it's difficult to imagine how to get a deterministic result (e.g. where the same values supplied to the same query always resulting in the exact same answer.)
This sample data includes a sequential PK column which is used to define the order of this set.
DECLARE #tbl TABLE (PK INT IDENTITY, ID INT, RowNumber INT)
INSERT #tbl(ID, RowNumber) VALUES (101,1),(101,2),(101,3),(101,4),(101,5),(101,1),(101,2);
SELECT t.* FROM #tbl AS t;
Returns:
PK ID RowNumber
----- ------ -----------
1 101 1
2 101 2
3 101 3
4 101 4
5 101 5
6 101 1
7 101 2
This query uses DENSE_RANK to get you what you want:
DECLARE #tbl TABLE (PK INT IDENTITY, ID INT, RowNumber INT)
INSERT #tbl(ID, RowNumber) VALUES (101,1),(101,2),(101,3),(101,4),(101,5),(101,1),(101,2);
SELECT t.ID, t.RowNumber, New = DENSE_RANK() OVER (ORDER BY t.PK - RowNumber)
FROM #tbl AS t;
Returns:
ID RowNumber New
----- ----------- ------
101 1 1
101 2 1
101 3 1
101 4 1
101 5 1
101 1 2
101 2 2
Note that ORDER BY New does not affect the plan.
Please try the below
Load the data into Temp table
Select id,RowNumber,Row_number()over(partition by RowNumber Order by id)New from #temp
Order by Row_number()over(partition by RowNumber Order by id),RowNumber

How to Generate Row number Partition by two column match in sql

Tbl1
---------------------------------------------------------
Id Date Qty ReOrder
---------------------------------------------------------
1 1-1-18 1 3
2 2-1-18 0 3
3 3-1-18 2 3
4 4-1-18 3< >3
5 5-1-18 2 3
6 6-1-18 0 3
7 7-1-18 1 3
8 8-1-18 0 3
---------------------------------------------------------
I want the result like below
---------------------------------------------------------
Id Date Qty ReOrder
---------------------------------------------------------
1 1-1-18 1 3
5 5-1-18 2 3
---------------------------------------------------------
if ReOrder not same with Qty then date will be same upto after reorder=Qty
You can use cumulative approach with row_number() function :
select top (1) with ties *
from (select *, max(case when qty = reorder then 'v' end) over (order by id desc) grp
from table
) t
order by row_number() over(partition by grp order by id);
Unfortunately this will require SQL Server, But you can also do:
select *
from (select *, row_number() over(partition by grp order by id) seq
from (select *, max(case when qty = reorder then 'v' end) over (order by id desc) grp
from table
) t
) t
where seq = 1;

How to combine 2 rows into single row in Teradata

I have a resultset in the below-mentioned form returned by a SQL:
ID Key
1 A
2 A
3 A
Now my requirement is to show the data in the below form:
Key ID1 ID2 ID3
A 1 2 3
How to build an SQL for this?
A Windowed Aggregate based solution with a single STATS-step in Explain:
SELECT
key,
-- value from 1st row = current row
ID AS ID1,
-- value from next row, similar to LEAD(ID, 1) Over (PARTITION BY Key ORDER BY ID)
Min(ID)
Over (PARTITION BY Key
ORDER BY ID
ROWS BETWEEN 1 Following AND 1 Following) AS ID2 ,
-- value from 3rd row
Min(ID)
Over (PARTITION BY Key
ORDER BY ID
ROWS BETWEEN 2 Following AND 2 Following) AS ID3
FROM mytable
QUALIFY -- only return the 1st row
Row_Number()
Over (PARTITION BY key
ORDER BY ID) = 1
As teradata 14.10 doesn't have a PIVOT function and assuming that for every unique key, there will be no more than 3 IDs( as mentioned in comments), you can use row_number() and aggregate function as below to get your desired result.
SELECT
key1,
MAX(CASE WHEN rn = 1 THEN ID END) AS ID1,
MAX(CASE WHEN rn = 2 THEN ID END) AS ID2,
MAX(CASE WHEN rn = 3 THEN ID END) AS ID3
FROM
(SELECT
t.*,
ROW_NUMBER() OVER (PARTITION BY key1 ORDER BY ID) AS rn
FROM table1 t) t
GROUP BY key1;
Result:
+------------+-----+-----+-----+
| key1 | id1 | id2 | id3 |
+------------+-----+-----+-----+
| A | 1 | 2 | 3 |
+------------+-----+-----+-----+
DEMO

Any other alternative to write this SQL query

I need to select data base upon three conditions
Find the latest date (StorageDate Column) from the table for each record
See if there is more then one entry for date (StorageDate Column) found in first step for same ID (ID Column)
and then see if DuplicateID is = 2
So if table has following data:
ID |StorageDate | DuplicateTypeID
1 |2014-10-22 | 1
1 |2014-10-22 | 2
1 |2014-10-18 | 1
2 |2014-10-12 | 1
3 |2014-10-11 | 1
4 |2014-09-02 | 1
4 |2014-09-02 | 2
Then I should get following results
ID
1
4
I have written following query but it is really slow, I was wondering if anyone has better way to write it.
SELECT DISTINCT(TD.RecordID)
FROM dbo.MyTable TD
JOIN (
SELECT T1.RecordID, T2.MaxDate,COUNT(*) AS RecordCount
FROM MyTable T1 WITH (nolock)
JOIN (
SELECT RecordID, MAX(StorageDate) AS MaxDate
FROM MyTable WITH (nolock)
GROUP BY RecordID)T2
ON T1.RecordID = T2.RecordID AND T1.StorageDate = T2.MaxDate
GROUP BY T1.RecordID, T2.MaxDate
HAVING COUNT(*) > 1
)PT ON TD.RecordID = PT.RecordID AND TD.StorageDate = PT.MaxDate
WHERE TD.DuplicateTypeID = 2
Try this and see how the performance goes:
;WITH
tmp AS
(
SELECT *,
RANK() OVER (PARTITION BY ID ORDER BY StorageDate DESC) AS StorageDateRank,
COUNT(ID) OVER (PARTITION BY ID, StorageDate) AS StorageDateCount
FROM MyTable
)
SELECT DISTINCT ID
FROM tmp
WHERE StorageDateRank = 1 -- latest date for each ID
AND StorageDateCount > 1 -- more than 1 entry for date
AND DuplicateTypeID = 2 -- DuplicateTypeID = 2
You can use analytic function rank , can you try this query ?
Select recordId from
(
select *, rank() over ( partition by recordId order by [StorageDate] desc) as rn
from mytable
) T
where rn =1
group by recordId
having count(*) >1
and sum( case when duplicatetypeid =2 then 1 else 0 end) >=1