SQL Server 2019

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select @@version;


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Microsoft SQL Server 2019 (RTM-CU16) (KB5011644) - 15.0.4223.1 (X64) 
 Apr 11 2022 16:24:07 
 Copyright (C) 2019 Microsoft Corporation
 Express Edition (64-bit) on Windows Server 2019 Standard 10.0 <X64> (Build 17763: ) (Hypervisor)



create table t (date date,  price float,  stock INT, item_code INT);
insert into t values('2020-01-01',   17.99,        6,           01);
insert into t values('2020-01-02',   17.99,        null,        01);
insert into t values('2020-01-03',   17.99,        4,           01);
insert into t values('2020-01-04',   17.99,        null,        01);
insert into t values('2020-01-01',   11.59,        null,        02);
insert into t values('2020-01-02',   11.59,        8,           02);
insert into t values('2020-01-03',   11.59,        null,        02);
insert into t values('2020-01-04',   11.59,        4,           02);
8 rows affected

select * from t


date
price
stock
item_code


2020-01-01
17.99
6
1

2020-01-02
17.99
null
1

2020-01-03
17.99
4
1

2020-01-04
17.99
null
1

2020-01-01
11.59
null
2

2020-01-02
11.59
8
2

2020-01-03
11.59
null
2

2020-01-04
11.59
4
2


SELECT 
   date, 
   price,
   CASE
    WHEN stock Is NULL then AVG(stock) OVER(PARTITION BY item_code ORDER BY item_code, date ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING)
    ELSE stock
   END AS stock, 
   item_code
FROM t
ORDER BY item_code, date


date
price
stock
item_code


2020-01-01
17.99
6
1

2020-01-02
17.99
5
1

2020-01-03
17.99
4
1

2020-01-04
17.99
4
1

2020-01-01
11.59
8
2

2020-01-02
11.59
8
2

2020-01-03
11.59
6
2

2020-01-04
11.59
4
2


Warning: Null value is eliminated by an aggregate or other SET operation.

date	price	stock	item_code
2020-01-01	17.99	6	1
2020-01-02	17.99	null	1
2020-01-03	17.99	4	1
2020-01-04	17.99	null	1
2020-01-01	11.59	null	2
2020-01-02	11.59	8	2
2020-01-03	11.59	null	2
2020-01-04	11.59	4	2