obtain the employee names who made a higher salary in March 2018 than their average monthly salary for all previous months

Question

select * from emp1;

EMP_ID	FIRST_NAME	LAST_NAME
10100	Marcos	Bisset
10101	Kate	Perez
10102	Carlos	Casco

select * from payment history;

EMP_ID	PAY_TYPE	AMT_PAID	PAY_DATED
10100	monthly pay	12000	02-FEB-18
10101	monthly pay	2800	02-FEB-18
10102	monthly pay	1900	02-MAR-18
10101	bonus	1500	08-MAR-18
10102	adjustment	124	10-MAR-18

Please give answer using With Clause ( I got the question from inte.net.. but I am unable to solve this question

Answer 1

Here's one option, step-by-step. Read comments within code.

with 
avg_sal as
  -- average salary per each employee
  (select emp_id, 
          avg(amt_paid) avg_sal
   from payment_history
   group by emp_id
  ),
march_sal as
  -- salary employee earned in March 2018
  (select emp_id, 
          amt_paid
   from payment_history
   where to_char(pay_dated, 'yyyymm') = '201803'
  )
select e.first_name
from emp1 e join avg_sal a on a.emp_id = e.emp_id
            join march_sal m on m.emp_id = e.emp_id
where m.amt_paid > a.avg_sal;

---

WHERE clause in march_sal CTE might need to be rewritten if there's an index on pay_dated column, as applying to_char function to it (the column) won't use that index. In that case, you'd use

WHERE pay_dated >= date '2018-03-01'
  AND pay_dated <  date '2018-04-01'

On the other hand, you might have a function-based index on to_char(pay_dated, 'yyyymm') - in that case, leave it as is .

Answer 2

To find the salary for march and the average previous salary you can sum the salary for each employee and month and then use conditional aggregation:

SELECT MAX(e.first_name) AS first_name,
       MAX(e.last_name) AS last_name,
       SUM(CASE WHEN month = DATE '2018-03-01' THEN pay END) AS march_pay,
       AVG(CASE WHEN month < DATE '2018-03-01' THEN pay END) AS avg_previous_pay
FROM   emp1 e
       INNER JOIN (
         SELECT emp_id,
                TRUNC(pay_dated, 'MM') AS month,
                SUM(amt_paid) AS pay
         FROM   payment_history
         GROUP BY emp_id,
                TRUNC(pay_dated, 'MM')
       ) p
       ON e.emp_id = p.emp_id
GROUP BY e.emp_id

Which, for the sample data:

CREATE TABLE emp1 (EMP_ID, FIRST_NAME, LAST_NAME) AS
SELECT 10100, 'Marcos', 'Bisset' FROM DUAL UNION ALL
SELECT 10101, 'Kate',   'Perez'  FROM DUAL UNION ALL
SELECT 10102, 'Carlos', 'Casco'  FROM DUAL;

CREATE TABLE payment_history (EMP_ID, PAY_TYPE, AMT_PAID, PAY_DATED) AS
SELECT 10100, 'monthly pay', 12000, DATE '2018-02-02' FROM DUAL UNION ALL
SELECT 10101, 'monthly pay',  2800, DATE '2018-02-02' FROM DUAL UNION ALL
SELECT 10102, 'monthly pay',  1900, DATE '2018-03-02' FROM DUAL UNION ALL
SELECT 10101, 'bonus',        1500, DATE '2018-03-08' FROM DUAL UNION ALL
SELECT 10102, 'adjustment',    124, DATE '2018-03-10' FROM DUAL;

Outputs:

FIRST_NAME	LAST_NAME	MARCH_PAY	AVG_PREVIOUS_PAY
Marcos	Bisset	null	12000
Kate	Perez	1500	2800
Carlos	Casco	2024	null

To then find the employee with the higher salary in march than their previous average salary, you can use a HAVING clause:

SELECT MAX(e.first_name) AS first_name,
       MAX(e.last_name) AS last_name
FROM   emp1 e
       INNER JOIN (
         SELECT emp_id,
                TRUNC(pay_dated, 'MM') AS month,
                SUM(amt_paid) AS pay
         FROM   payment_history
         GROUP BY emp_id,
                TRUNC(pay_dated, 'MM')
       ) p
       ON e.emp_id = p.emp_id
GROUP BY e.emp_id
HAVING SUM(CASE WHEN month = DATE '2018-03-01' THEN pay END)
        > AVG(CASE WHEN month < DATE '2018-03-01' THEN pay END)

Which outputs no rows as you can see from the previous data that Employees either have a higher average previous pay or they had no pay in one of the months; to get a result you need a larger data set where there is a row that actually matches the requirements.

---

Requiring a WITH clause is an artificial requirement that is not necessary to solve the problem; if you really must use it ( you do not need to ) then move the sub-query from the INNER JOIN into a sub-query factoring clause:

WITH avg_pay (emp_id, month, pay) AS (
  <sub-query>
)
SELECT ...
FROM   emp1 e
       INNER JOIN avg_pay p
       ON ...

fiddle