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Following / copying computhomas's question, but adding some twists...
I have the following table in MSSQL2008
id | business_key | result | date
1 | 1 | 0 | 9
2 | 1 | 1 | 8
3 | 2 | 1 | 7
4 | 3 | n | 6
5 | 4 | 1 | 5
6 | 4 | 0 | 4
And now i want to group based on the business_key returning the complete entry with the newest date. So my expected result is:
id | business_key | result | date
1 | 1 | 0 | 9
3 | 2 | 1 | 7
4 | 3 | n | 6
5 | 4 | 1 | 5
I also bet that there is a way to achieve that, i just can't find / see / think of it at the moment.
edit: sorry about this, I actually meant something else from original question I did. I felt like editing this might be better than accepting a solution and making another question. my original problem was that I am not filtering by id.
748
Retrieve Last Record for each Group in SQL Server Example 1 First, partition the data by Occupation and assign the rank number using the yearly income. Next, it is going to select the last record from each SQL Server group.
-- Select First Row in each SQL Group By group USE [SQL Tutorial] GO SELECT * FROM ( SELECT [FirstName] , [LastName] , [Education] , [Occupation] , [YearlyIncome] ,ROW_NUMBER () OVER ( PARTITION BY [Occupation] ORDER BY [YearlyIncome] DESC ) AS [ROW NUMBER] FROM [Customers] ) groups WHERE groups.
The plain SQL solution is to divide and conquer. We already have a query to get the current balance of an account. If we write another query to get the credit for each account, we can join the two together and get the complete state of an account. To get the last event for each account in PostgreSQL we can use DISTINCT ON:
SQL GROUP BY and DISTINCT If you use the GROUP BY clause without an aggregate function, the GROUP BY clause behaves like the DISTINCT operator. The following gets the phone numbers of employees and also group rows by the phone numbers. SELECT phone_number FROM employees GROUP BY phone_number;
SELECT t.*
FROM
(
SELECT *, ROW_NUMBER() OVER
(
PARTITION BY [business_key]
ORDER BY [date] DESC
) AS [RowNum]
FROM yourTable
) AS t
WHERE t.[RowNum] = 1
SELECT
*
FROM
mytable
WHERE
ID IN (SELECT MAX(ID) FROM mytable GROUP BY business_key)
SELECT
MAX(T1.id) AS [id],
T1.business_key,
T1.result
FROM
dbo.My_Table T1
LEFT OUTER JOIN dbo.My_Table T2 ON
T2.business_key = T1.business_key AND
T2.id > T1.id
WHERE
T2.id IS NULL
GROUP BY T1.business_key,
T1.result
ORDER BY MAX(T1.id)
Edited based on clarifications
SELECT M1.*
FROM My_Table M1
INNER JOIN
(
SELECT [business_key], MAX([date]) as MaxDate
FROM My_Table
GROUP BY [business_key]
) M2 ON M1.business_key = M2.business_key AND M1.[date] = M2.MaxDate
ORDER BY M1.[id]
Assuming the combination of business_key & date is unique then....
Working example (3rd time is a charm):
declare @src as table(id int, business_key int,result int,[date] int)
insert into @src
SELECT 1,1,0,9
UNION SELECT 2,1,1,8
UNION SELECT 3,2,1,7
UNION SELECT 4,3,1,6
UNION SELECT 5,4,1,5
UNION SELECT 6,4,0,4
;with bkdate(business_key,[date])
AS
(
select business_key,MAX([date])
from @src
group by business_key
)
select src.* from @src src
inner join bkdate
ON src.[date] = bkdate.date
and src.business_key = bkdate.business_key
order by id
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