How to make LAG() ignore NULLS in SQL Server?

Question

Does anyone know how to replace nulls in a column with a string until it hits a new string then that string replaces all null values below it? I have a column that looks like this

Original Column:

PAST_DUE_COL           
91 or more days pastdue        
Null
Null
61-90 days past due          
Null
Null
31-60 days past due
Null
0-30 days past due
Null       
Null
Null            

Expected Result Column:

PAST_DUE_COL           
91 or more days past due        
91 or more days past due
91 or more days past due
61-90 days past due          
61-90 days past due 
61-90 days past due 
31-60 days past due
31-60 days past due
0-30 days past due
0-30 days past due      
0-30 days past due
0-30 days past due

Essentially I want the first string in the column to replace all null values below it until the next string. Then that string will replace all nulls below it until the next string and so on.

Answer 1

SQL Server does not support the ignore nulls option for window functions such as lead() and lag(), for which this question was a nice fit.

We can work around this with some gaps and island technique:

select
    t.*,
    max(past_due_col) over(partition by grp) new_past_due_col
from (
    select 
        t.*,
        sum(case when past_due_col is null then 0 else 1 end)
            over(order by id) grp
    from mytable t
) t

The subquery does a window sum that increments everytime a non null value is found: this defines groups of rows that contain a non-null value followed by null values.

Then, the outer uses a window max() to retrieve the (only) non-null value in each group.

This assumes that a column can be used to order the records (I called it id).

Demo on DB Fiddle:

ID | PAST_DUE_COL            | grp | new_past_due_col       
-: | :---------------------- | --: | :----------------------
 1 | 91 or more days pastdue |   1 | 91 or more days pastdue
 2 | null                    |   1 | 91 or more days pastdue
 3 | null                    |   1 | 91 or more days pastdue
 4 | 61-90 days past due     |   2 | 61-90 days past due    
 5 | null                    |   2 | 61-90 days past due    
 6 | null                    |   2 | 61-90 days past due    
 7 | 31-60 days past due     |   3 | 31-60 days past due    
 8 | null                    |   3 | 31-60 days past due    
 9 | 0-30 days past due      |   4 | 0-30 days past due     
10 | null                    |   4 | 0-30 days past due     
11 | null                    |   4 | 0-30 days past due     
12 | null                    |   4 | 0-30 days past due     

Answer 2

This is a variation on GMBs answer. It is just a bit simpler:

select t.*,
       max(past_due_col) over(partition by grp) as new_past_due_col
from (select t.*,
             count(past_due_col) over (order by id) as grp
      from mytable t
     ) t;

Note that you need an ordering column of some sort for your question to even make sense.

Another approach uses apply:

select t.*, t2.past_due_col
from mytable t outer apply
     (select top (1) t2.*
      from mytable t2
      where t2.id <= t.id and t2.past_due_col is not null
      order by t2.id desc
     ) t2;