Print the sum of even digits of a number in Pl/sql - oracle

Question

I cant seem to resolve this issue

Answer 1

As of the error itself, it is because you're trying to put the whole row (temp, which is declared as a cursor variable - c_nm%rowtype) into an integer variable (i). It won't work, although temp actually contains only one column - num.
So, line #15 on your screenshot:
No : i := temp;
Yes: i := temp.num;
Once you fix it, your code works but - unfortunately, produces wrong result. In input number 121974553, even digits are 2, 9, 4 and 5 whose sum equals 20, not 6 (as your result suggests):
SQL> set serveroutput on;
SQL> declare
2 ad int := 0;
3 i int;
4 b int;
5 cursor c_nm is select * From numb;
6 temp c_nm%rowtype;
7 begin
8 open c_nm;
9 loop
10 fetch c_nm into temp;
11 exit when c_nm%notfound;
12 i := temp.num; --> this
13
14 while i > 0 loop
15 b := mod(i, 10);
16 if mod(b, 2) = 0 then
17 ad := b + ad;
18 end if;
19 i := trunc(i/10);
20 end loop;
21 end loop;
22 dbms_output.put_line('ad = ' ||ad);
23 close c_nm;
24 end;
25 /
ad = 6
PL/SQL procedure successfully completed.
SQL>
A simpler option might be this: split number into digits (each in its separate row) so that you could apply SUM function to its even digits:
SQL> select * from numb;
NUM
----------
121974553 --> sum of even digits = 2 + 9 + 4 + 5 = 20
253412648 --> = 5 + 4 + 2 + 4 = 15
SQL> with
2 split as
3 -- split number into rows
4 (select num,
5 substr(num, column_value, 1) digit,
6 case when mod(column_value, 2) = 0 then 'Y'
7 else 'N'
8 end cb_odd_even
9 from numb cross join table(cast(multiset(select level from dual
10 connect by level <= length(num)
11 ) as sys.odcinumberlist))
12 )
13 -- final result: summary of digits in even rows
14 select num,
15 sum(digit)
16 from split
17 where cb_odd_even = 'Y'
18 group by num;
NUM SUM(DIGIT)
---------- ----------
253412648 15
121974553 20
SQL>
If it must be PL/SQL, slightly rewrite it as
SQL> declare
2 result number;
3 begin
4 for cur_r in (select num from numb) loop
5 with
6 split as
7 -- split number into rows
8 (select substr(cur_r.num, level, 1) digit,
9 case when mod(level, 2) = 0 then 'Y'
10 else 'N'
11 end cb_odd_even
12 from dual
13 connect by level <= length(cur_r.num)
14 )
15 -- final result: summary of digits in even rows
16 select sum(digit)
17 into result
18 from split
19 where cb_odd_even = 'Y';
20
21 dbms_output.put_line(cur_r.num || ' --> ' || result);
22 end loop;
23 end;
24 /
121974553 --> 20
253412648 --> 15
PL/SQL procedure successfully completed.
SQL>

Answer 2

You could do something like this. A few things of note: first, you can use an implicit cursor (the for rec in (select ...) loop construct). You don't need to declare the data type of rec, you don't need to open the cursor - and then remember to close it when you are finished - etc. Implicit cursors are a great convenience, best to learn about it as early as possible. Second, note that rec (in my notation) is just a pointer; to access the value from it, you must reference the table column (as in, rec.num) - Littlefoot has already shown that in his reply. Third, a logic mistake in your attempt: if the table has more than one row, you must initialize ad to 0 for each new value - you can't just initialize it to 0 once, at the beginning of the program.
So, with all that said - here is some sample data, then the procedure and its output.
create table numb (num int);
insert into numb values(121975443);
insert into numb values(100030000);
insert into numb values(null);
insert into numb values(113313311);
insert into numb values(2020);
commit;
. . . . .
declare
ad int;
i int;
begin
for rec in (select num from numb) loop
ad := 0;
i := rec.num;
while i > 0 loop
if mod(i, 2) = 0 then
ad := ad + mod(i, 10);
end if;
i := trunc(i/10);
end loop;
dbms_output.put_line('num: ' || nvl(to_char(rec.num), 'null') ||
' sum of even digits: ' || ad);
end loop;
end;
/
num: 121975443 sum of even digits: 10
num: 100030000 sum of even digits: 0
num: null sum of even digits: 0
num: 113313311 sum of even digits: 0
num: 2020 sum of even digits: 4
PL/SQL procedure successfully completed.