display null value using rank functions in oracle sql

trim(t1.c1) != NULL

trim(t1.c1) IS NOT NULL

SELECT
    CASE WHEN Value IS NULL THEN NULL
         ELSE RANK() OVER (ORDER BY VALUE DESC)
    END AS RANK,
    USER_ID,
    VALUE
FROM yourtable

with q as (
select 10 deptno, 'rrr' empname, 10000.00 sal from dual union all
select 11, 'nnn', 20000.00 from dual union all
select 11, 'mmm', 5000.00 from dual union all
select 12, 'kkk', 30000 from dual union all
select 10, 'fff', 40000 from dual union all
select 10, 'ddd', 40000 from dual union all
select 10, 'bbb', 50000 from dual union all
select 10, 'xxx', null from dual union all
select 10, 'ccc', 50000 from dual)
select empname, deptno, sal
     , rank() over (partition by deptno order by sal nulls first) r
     , dense_rank() over (partition by deptno order by sal nulls first) dr1
     , dense_rank() over (partition by deptno order by sal nulls last) dr2
 from q; 

EMP     DEPTNO        SAL          R        DR1        DR2
--- ---------- ---------- ---------- ---------- ----------
xxx         10                     1          1          4
rrr         10      10000          2          2          1
fff         10      40000          3          3          2
ddd         10      40000          3          3          2
ccc         10      50000          5          4          3
bbb         10      50000          5          4          3
mmm         11       5000          1          1          1
nnn         11      20000          2          2          2
kkk         12      30000          1          1          1

9 rows selected.

select id, inx, max(num) over (partition by id order by inx) as num

select id, inx,
       (case when num is null
             then lag(num) over (partition by id order by inx)
             else num
        end)as null;

select id, inx,
       (case when num is null
             then lag(num ignore nulls) over (partition by id order by inx)
             else num
        end)as null