for SQL Server 2008 R2
I have a resultset that looks like this (note [price] is numeric, NULL below represents a NULL value, the result set is ordered by product_id and timestamp)
product timestamp price
------- ---------------- -----
5678 2008-01-01 12:00 12.34
5678 2008-01-01 12:01 NULL
5678 2008-01-01 12:02 NULL
5678 2008-01-01 12:03 23.45
5678 2008-01-01 12:04 NULL
I want to transform that to a result set that (essentially) copies a non-null value from the latest preceding row, to produce a resultset that looks like this:
product timestamp price
------- ---------------- -----
5678 2008-01-01 12:00 12.34
5678 2008-01-01 12:01 12.34
5678 2008-01-01 12:02 12.34
5678 2008-01-01 12:03 23.45
5678 2008-01-01 12:04 23.45
I don't find any aggregate/windowing function that will allow me to do this (again this ONLY needed for SQL Server 2008 R2.)
I was hoping to find an analytic aggregate function that do this for me, something like...
LAST_VALUE(price) OVER (PARTITION BY product_id ORDER BY timestamp)
But I don't seem to find any way to do a "cumulative latest non-null value" in the window (to bound the window to the preceding rows, rather than the entire partition)
Aside from creating a table-valued user defined function, is there any builtin that would accomplish this?
UPDATE:
Apparently, this functionality is available in the 'Denali' CTP, but not in SQL Server 2008 R2.
LAST_VALUE http://msdn.microsoft.com/en-us/library/hh231517%28v=SQL.110%29.aspx
I just expected it to be available in SQL Server 2008. It's available in Oracle (since 10gR2 at least), and I can do something similar in MySQL 5.1, using a local variable.
http://download.oracle.com/docs/cd/E14072_01/server.112/e10592/functions083.htm
ISNULL Function in SQL Server To use this function, all you need to do is pass the column name in the first parameter and in the second parameter pass the value with which you want to replace the null value. So, now all the null values are replaced with No Name in the Name column.
UPDATE [table] SET [column]=0 WHERE [column] IS NULL; Null Values can be replaced in SQL by using UPDATE, SET, and WHERE to search a column in a table for nulls and replace them. In the example above it replaces them with 0. Cleaning data is important for analytics because messy data can lead to incorrect analysis.
You can use COALESCE() to instruct using the value of another column if the target column is NULL and if that is also null then use the third column and so on. That's all about how to replace NULL with empty String or blank in SQL SERVER. You can use ISNULL() or COALESCE() to replace NULL with blanks.
Using the SQL COALESCE function, it is easy to replace missing or NULL values in SELECT statements. Specific values can be set directly with COALESCE and the mean, median or mode can be used by combining COALESCE with WINDOW functions.
You can try the following:
* Updated **
-- Test Data
DECLARE @YourTable TABLE(Product INT, Timestamp DATETIME, Price NUMERIC(16,4))
INSERT INTO @YourTable
SELECT 5678, '20080101 12:00:00', 12.34
UNION ALL
SELECT 5678, '20080101 12:01:00', NULL
UNION ALL
SELECT 5678, '20080101 12:02:00', NULL
UNION ALL
SELECT 5678, '20080101 12:03:00', 23.45
UNION ALL
SELECT 5678, '20080101 12:04:00', NULL
;WITH CTE AS
(
SELECT *
FROM @YourTable
)
-- Query
SELECT A.Product, A.Timestamp, ISNULL(A.Price,B.Price) Price
FROM CTE A
OUTER APPLY ( SELECT TOP 1 *
FROM CTE
WHERE Product = A.Product AND Timestamp < A.Timestamp
AND Price IS NOT NULL
ORDER BY Product, Timestamp DESC) B
--Results
Product Timestamp Price
5678 2008-01-01 12:00:00.000 12.3400
5678 2008-01-01 12:01:00.000 12.3400
5678 2008-01-01 12:02:00.000 12.3400
5678 2008-01-01 12:03:00.000 23.4500
5678 2008-01-01 12:04:00.000 23.4500
I have a table containing the following data. I want to update all nulls in salary columns with previous value without taking null value.
Table:
id name salary
1 A 4000
2 B
3 C
4 C
5 D 2000
6 E
7 E
8 F 1000
9 G 2000
10 G 3000
11 G 5000
12 G
here is the query that works for me.
select a.*,first_value(a.salary)over(partition by a.value order by a.id) as abc from
(
select *,sum(case when salary is null then 0 else 1 end)over(order by id) as value from test)a
output:
id name salary Value abc
1 A 4000 1 4000
2 B 1 4000
3 C 1 4000
4 C 1 4000
5 D 2000 2 2000
6 E 2 2000
7 E 2 2000
8 F 1000 3 1000
9 G 2000 4 2000
10 G 3000 5 3000
11 G 5000 6 5000
12 G 6 5000
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With