Count odd and even digits in a number in PL/SQL

We are given a positive integer of digits and the task is to calculate the count of odd and even digits in a number using PL/SQL.

PL/SQL is a combination of SQL along with the procedural features of programming languages. It was developed by Oracle Corporation in the early 90's to enhance the capabilities of SQL.

PL/SQL is one of three key programming languages embedded in the Oracle Database, along with SQL itself and Java.

Input

int number = 23146579

Output 

count of odd digits in a number are : 5
count of even digits in a number are : 3

Explanation − In the given number, we have 2, 4, 6 as an even digits therefore count of even digits in a number are 3 and we have 3, 1, 5, 7 and 9 as an odd digits therefore count of odd digits in a number are 5.

Input

int number = 4567228

Output 

count of odd digits in a number are : 2
count of even digits in a number are : 5

Explanation − In the given number, we have 5 and 7 as an odd digits therefore count of odd digits in a number are 2 and we have 4, 6, 2, 2 and 8 as an even digits therefore count of even digits in a number are 5.

Approach used in the below program is as follows

• Input a number in an integer type variable of datatype NUMBER used in PL/SQL.

• Take a length of type VARCHAR(50) which describes the maximum size length can store.

• Take two variables as count for odd digits and count for even digits and initially set them to 0

• Start Loop For from 1 till the length while pass a number to it

• Inside the loop, set length as substr(number, i, 1)

• Now, check IF mod of length by 2 is not equals to 0 then increase the count for odd digits in a number

• Else, increase the count of even digits in a number

• Print the result.

Example

DECLARE
   digits NUMBER := 23146579;
   length VARCHAR2(50);
   count_odd NUMBER(10) := 0;
   count_even NUMBER(10) := 0;
BEGIN
   FOR i IN 1..Length(digits)
   LOOP
      length := Substr(digits, i, 1);
      IF mod(length, 2) != 0 THEN
         count_odd := count_odd + 1;
      ELSE
         count_even := count_even + 1;
      END IF;
   END LOOP;
   dbms_output.Put_line('count of odd digits in a number are : ' || count_odd);
   dbms_output.Put_line('count of even digits in a number are : ' || count_even);
END;

Output

If we run the above code it will generate the following output −

count of odd digits in a number are : 5
count of even digits in a number are : 3