How can I make every first row from my Id? - sql-server

This is my image of database result:

using Row_number you can get first row of your ID
Select * from (
Select *,
Row_number()OVER(PARTITION BY ID ORDER BY (SELECT NULL))RN
From Your Table )T
Where T.RN = 1

Select Id,Label,Score,Flag,AutoGenId from (
Select *,
Row_number()OVER(PARTITION BY ID ORDER BY (SELECT NULL))RN
From [VisionApiResponseData] )T
Where T.RN = 1

Related

Query table and Select latest 2 rows (in SQL Server)

I have a table that logs all updates made to an application. I want to query the table and return the last update by [Timestamp] and the update before that for a different value [ITEM]. I'm struggling to figure out how to get what i need. I'm returning more than one record for each ID and don't want that.
;WITH cte AS
(
SELECT
ID,
LAG(ITEM) OVER (PARTITION BY ID ORDER BY timestamp DESC) AS ITEM,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY timestamp DESC) RN
FROM
MyLoggingTable
WHERE
accountid = 1234
)
SELECT
cte.ID,
dl.ITEM,
DL.timestamp
FROM
cte
JOIN
MyLoggingTable DL ON cte.ID = DL.ID
WHERE
rn = 1
AND cte.ID IN ('id here | Sub select :( ..')
Is ID unique? Because if it is, your code shouldn't return duplicates. If it isn't, you will get duplicates because you are joining back to the MyLoggingTable which isn't needed. You should just move those columns (dl.Item & dl.timestamp) into the cte and return them from the cte like you did cte.ID.
I removed the LAG since you didn't return that column in your final query.
;WITH cte AS
(
SELECT
ID,
ITEM,
[timestamp],
--LAG(ITEM) OVER (PARTITION BY ID ORDER BY timestamp DESC) AS ITEM,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY timestamp DESC) RN
FROM
MyLoggingTable
WHERE
accountid = 1234
)
SELECT
cte.ID,
cte.ITEM,
cte.timestamp
FROM
cte
WHERE
rn = 1
AND cte.ID IN ('id here | Sub select :( ..')
Note, if you wanted the second to the last item, as you stated in your comments, make rn=2

How to get desired result by combining two CTE's with different tables?

I have two CTE's,firstly
;WITH CTE AS (SELECT A.*
, Row_NUMBER() Over (Partition by ID order by Date asc) RN
FROM TABLE A)
SELECT Sum(Weight) as IN_WT
FROM CTE
WHERE RN = 1 and name='dev1'
and then
;WITH CTE AS (SELECT B.*
, Row_NUMBER() Over (Partition by ID order by Date desc) RN1
FROM TABLE B)
SELECT Sum(Weight) AS out_wt
FROM CTE
WHERE RN1 = 1 and name='dev1'
Here we had two tablesTableA,TableB.We are getting In_wt from TableA and Out_wt from TableB.Now I had a requiremnt that the output should be combined and get in_wt,out_wt in single row from different tables and same name.I tried combining both the CTE's but didn't get the desired result.How can we do that?
Try this:
;WITH CTE1 AS (
SELECT name,
Row_NUMBER() Over (Partition by ID order by Date asc) RN
FROM TABLEA
), CTE2 AS (
SELECT name,
Row_NUMBER() Over (Partition by ID order by Date desc) RN
FROM TABLEB
)
SELECT (SELECT Sum(Weight) FROM CTE1 WHERE RN = 1 and name='dev1') AS IN_WT,
(SELECT Sum(Weight) FROM CTE2 WHERE RN = 1 and name='dev1') AS OUT_WT

How do I select the top 1 results of the if the same identifier exist? SQL Server

I have a data table in sql server 2008 that I would like to select the top 1 out of each identifier:
The results shld looks like this during before and after:
Thus it should only select the 1st results if the same identifier do exist. Thanks a lot.
select distinct [Primary Identifier] from tbl
If you have entire records (other columns) instead of that single column, you can row number them and choose one.
select {list of columns}
from
(
select *, rn = row_number over (partition by [Primary Identifier]
order by 1/0)
from tbl
) X
where rn = 1;
order by 1/0 is arbitrary. If you need to choose a specific one from the "duplicates", for example the highest cost, you order by cost descending, i.e.
(partition by [Primary Identifier]
order by [cost] descending)
Just distinct them:
select distinct [primary identifier] from tablename
Or by grouping:
select [primary identifier] from tablename group by [primary identifier]
If more columns exist you can rank rows with window function:
;with cte as(select *, row_number() over(partition by [primary identifier] order by (select null)) rn from tablename)
select * from cte where rn = 1
Change order by (select null) to appropriate ordering column.
i think this will be an appropriate solution to your need-
;WITH cte AS
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY [Primary Identifier] ORDER BY [sort columns]) AS rowid
FROM [table]
)
SELECT *
FROM cte
WHERE rowid = 1

tsq - take the highest int in duplicate row

I have the following rows in a table
name, tagid
-------
test1,1
test1,100
test2,2
test2,200
test3,3
test3,300
There are duplicates in the name.
Is there a way to select unique names by taking the highest tagid of each group?
select name,max(tagid) as highest_tagid
from tbl
group by name
;WITH cte AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY name ORDER BY tagid DESC) AS rn
FROM table_1
)
SELECT *
FROM cte
WHERE rn = 1

select top 1 with a group by

I have two columns:
namecode name
050125 chris
050125 tof
050125 tof
050130 chris
050131 tof
I want to group by namecode, and return only the name with the most number of occurrences. In this instance, the result would be
050125 tof
050130 chris
050131 tof
This is with SQL Server 2000
I usually use ROW_NUMBER() to achieve this. Not sure how it performs against various data sets, but we haven't had any performance issues as a result of using ROW_NUMBER.
The PARTITION BY clause specifies which value to "group" the row numbers by, and the ORDER BY clause specifies how the records within each "group" should be sorted. So partition the data set by NameCode, and get all records with a Row Number of 1 (that is, the first record in each partition, ordered by the ORDER BY clause).
SELECT
i.NameCode,
i.Name
FROM
(
SELECT
RowNumber = ROW_NUMBER() OVER (PARTITION BY t.NameCode ORDER BY t.Name),
t.NameCode,
t.Name
FROM
MyTable t
) i
WHERE
i.RowNumber = 1;
select distinct namecode
, (
select top 1 name from
(
select namecode, name, count(*)
from myTable i
where i.namecode = o.namecode
group by namecode, name
order by count(*) desc
) x
) as name
from myTable o
SELECT max_table.namecode, count_table2.name
FROM
(SELECT namecode, MAX(count_name) AS max_count
FROM
(SELECT namecode, name, COUNT(name) AS count_name
FROM mytable
GROUP BY namecode, name) AS count_table1
GROUP BY namecode) AS max_table
INNER JOIN
(SELECT namecode, COUNT(name) AS count_name, name
FROM mytable
GROUP BY namecode, name) count_table2
ON max_table.namecode = count_table2.namecode AND
count_table2.count_name = max_table.max_count
I did not try but this should work,
select top 1 t2.* from (
select namecode, count(*) count from temp
group by namecode) t1 join temp t2 on t1.namecode = t2.namecode
order by t1.count desc
Here are to examples that you could use but the temp table use is more efficient than the view, but was done on a small data sample. You would want to check your own statistics.
--Creating A View
GO
CREATE VIEW StateStoreSales AS
SELECT t.state,t.stor_id,t.stor_name,SUM(s.qty) 'TotalSales'
,ROW_NUMBER() OVER (PARTITION BY t.state ORDER BY SUM(s.qty) DESC) AS 'Rank'
FROM [dbo].[sales] s
JOIN [dbo].[stores] t ON (s.stor_id = t.stor_id)
GROUP BY t.state,t.stor_id,t.stor_name
GO
SELECT * FROM StateStoreSales
WHERE Rank <= 1
ORDER BY TotalSales Desc
DROP VIEW StateStoreSales
---Using a Temp Table
SELECT t.state,t.stor_id,t.stor_name,SUM(s.qty) 'TotalSales'
,ROW_NUMBER() OVER (PARTITION BY t.state ORDER BY SUM(s.qty) DESC) AS 'Rank' INTO #TEMP
FROM [dbo].[sales] s
JOIN [dbo].[stores] t ON (s.stor_id = t.stor_id)
GROUP BY t.state,t.stor_id,t.stor_name
SELECT * FROM #TEMP
WHERE Rank <= 1
ORDER BY TotalSales Desc
DROP TABLE #TEMP

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