SQL QUERY replace NULL value in a row with a value from the previous known value

Admin 1
Viewed 129375

I have 2 columns

date   number       
----   ------
1      3           
2      NULL        
3      5           
4      NULL        
5      NULL        
6      2          
.......

I need to replace the NULL values with new values takes on the value from the last known value in the previous date in the date column eg: date=2 number = 3, date 4 and 5 number = 5 and 5. The NULL values appear randomly.

null sql
14 Answers

I know it is a very old forum, but I came across this while troubleshooting my problem :) just realised that the other guys have given bit complex solution to the above problem. Please see my solution below:

DECLARE @A TABLE(ID INT, Val INT)

INSERT INTO @A(ID,Val) SELECT 1, 3
INSERT INTO @A(ID,Val) SELECT 2, NULL
INSERT INTO @A(ID,Val) SELECT 3, 5
INSERT INTO @A(ID,Val) SELECT 4, NULL
INSERT INTO @A(ID,Val) SELECT 5, NULL
INSERT INTO @A(ID,Val) SELECT 6, 2

UPDATE D
    SET D.VAL = E.VAL
    FROM (SELECT A.ID C_ID, MAX(B.ID) P_ID
          FROM  @A AS A
           JOIN @A AS B ON A.ID > B.ID
          WHERE A.Val IS NULL
            AND B.Val IS NOT NULL
          GROUP BY A.ID) AS C
    JOIN @A AS D ON C.C_ID = D.ID
    JOIN @A AS E ON C.P_ID = E.ID

SELECT * FROM @A

Hope this may help someone:)

user_12999200

If you're looking for a solution for Redshift, this will work with the frame clause:

SELECT date, 
       last_value(columnName ignore nulls) 
                   over (order by date
                         rows between unbounded preceding and current row) as columnName 
 from tbl
user_43791600

This will work on Snowflake (credit to Darren Gardner):

create temp table ss (id int, val int);
insert into ss (id,val) select 1, 3;
insert into ss (id,val) select 2, null;
insert into ss (id,val) select 3, 5;
insert into ss (id,val) select 4, null;
insert into ss (id,val) select 5, null;
insert into ss (id,val) select 6, 2;

select *
      ,last_value(val ignore nulls) over 
       (order by id rows between unbounded preceding and current row) as val2
  from ss;
user_15432170

In case you have one identity (Id) and one common (Type) columns:

UPDATE #Table1 
SET [Type] = (SELECT TOP 1 [Type]
              FROM #Table1 t              
              WHERE t.[Type] IS NOT NULL AND 
              b.[Id] > t.[Id]
              ORDER BY t.[Id] DESC)
FROM #Table1 b
WHERE b.[Type] IS NULL
user_43914010
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