Need help understanding range between in SQL window functions

Question

I am trying understand how range clause is working in below case (oracle database)

SELECT
    EMPID,NAME, 
    HIRE_DATE_1, 
    SALARY,
    count(1) over(order by HIRE_DATE_1 range between 1 preceding and 1 preceding) as PREV_MIN_SA
FROM (
    SELECT 
        EMPID,
        NAME,
        (EXTRACT(year from HIRE_DATE)*10000)+(EXTRACT(MONTH FROM HIRE_DATE) * 100) + (extract(DAY from HIRE_DATE)) as HIRE_DATE_1,SALARY
     FROM EMPLOYEE A order by HIRE_DATE,SALARY
) A
ORDER BY HIRE_DATE_1 

Result Set :

EMPID   NAME    HIRE_DATE_1 SALARY  PREV_MIN_SA
100     Ravi    20180101    5000    0
101     Kumar   20180101    7000    0
102     Satish  20180101    13000   0
103     Naresh  20180102    7500    3
105     Lalith  20180104    17300   0
104     Suresh  20180104    40000   0
106     Latha   20180201    16000   0

The inner query is just converting date into numeric YYYYMMDD format. My intention is to get the count of people who joined immediately prior to the date of the employee in each record. I can take the count of rows with same HIRE_DATE and use LAG function but somehow not understand how the sql is returning this result set.

Also, once I am done with the counts I would like to get the MIN(SALARY) of the employees who joined immediately prior to the employee in current row and find the difference in salaries so wondering if somehow I can define the window to only have all records with immediately prior HIRE_DATE.

Thanks

Answer 1

This should get the preceding hire date...

SELECT
    EMPID,NAME, HIRE_DATE, SALARY,
    MAX(HIRE_DATE) OVER (ORDER BY HIRE_DATE
                            RANGE BETWEEN UNBOUNDED PRECEDING
                                      AND INTERVAL '1 DAY' PRECEDING
                        )
                           AS PREV_HIRE_DATE
FROM
    EMPLOYEE

Then I think you need to join back on to the employee table to find the number of employees and their min salary?

SELECT
    *
FROM
(
    SELECT
        EMPID,NAME, HIRE_DATE, SALARY,
        MAX(HIRE_DATE) OVER (ORDER BY HIRE_DATE
                                RANGE BETWEEN UNBOUNDED PRECEDING
                                          AND INTERVAL '1 DAY' PRECEDING
                            )
                               AS PREV_HIRE_DATE
    FROM
        EMPLOYEE
)
    EMPS
LEFT JOIN
(
    SELECT
        HIRE_DATE,
        COUNT(*)     AS COUNT_EMPS,
        MIN(SALARY)  AS MIN_SALARY
    FROM
        EMPLOYEE
    GROUP BY
        HIRE_DATE
)
    PREV_EMPS
        ON PREV_EMPS.HIRE_DATE = EMPS.PREV_HIRE_DATE

EDIT:

Maybe try something like this? (I have to run, good luck!)

WITH
  ranked AS
(
    SELECT
      *,
      DENSE_RANK() OVER (ORDER BY HIRE_DATE)   AS HIRE_SEQ_ID
    FROM
      EMPLOYEE
)
SELECT
  *,
  MIN(SALARY) OVER (ORDER BY HIRE_SEQ_ID
                       RANGE BETWEEN 1 PRECEDING AND 1 PRECEDING
                   )
                      AS PREV_MIN_SALARY,
  COUNT(*)    OVER (ORDER BY HIRE_SEQ_ID
                       RANGE BETWEEN 1 PRECEDING AND 1 PRECEDING
                   )
                      AS COUNT_PREV_EMPS
FROM
  ranked

Answer 2

This does what you're after, I believe (I've left it as an exercise for you to find the difference between the max and min salaries!):

WITH employee AS (SELECT 100 empid, 'Ravi' NAME, to_date('01/01/2018', 'dd/mm/yyyy') hire_date, 5000 salary FROM dual UNION ALL
                  SELECT 101 empid, 'Kumar' NAME, to_date('01/01/2018', 'dd/mm/yyyy') hire_date, 7000 salary FROM dual UNION ALL
                  SELECT 102 empid, 'Satish' NAME, to_date('01/01/2018', 'dd/mm/yyyy') hire_date, 13000 salary FROM dual UNION ALL
                  SELECT 103 empid, 'Naresh' NAME, to_date('02/01/2018', 'dd/mm/yyyy') hire_date, 7500 salary FROM dual UNION ALL
                  SELECT 104 empid, 'Lalith' NAME, to_date('04/01/2018', 'dd/mm/yyyy') hire_date, 17300 salary FROM dual UNION ALL
                  SELECT 105 empid, 'Suresh' NAME, to_date('04/01/2018', 'dd/mm/yyyy') hire_date, 40000 salary FROM dual UNION ALL
                  SELECT 106 empid, 'Latha' NAME, to_date('01/02/2018', 'dd/mm/yyyy') hire_date, 16000 salary FROM dual)
SELECT empid,
       NAME,
       hire_date,
       salary,
       COUNT(*) OVER (ORDER BY hire_date RANGE BETWEEN 1 PRECEDING AND 1 PRECEDING) prev_day_hire_count,
       MIN(salary) OVER (ORDER BY hire_date RANGE BETWEEN 1 PRECEDING AND 1 PRECEDING) prev_day_min_sal,
       MAX(salary) OVER (ORDER BY hire_date RANGE BETWEEN 1 PRECEDING AND 1 PRECEDING) prev_day_max_sal
FROM   employee;

     EMPID NAME   HIRE_DATE       SALARY PREV_DAY_HIRE_COUNT PREV_DAY_MIN_SAL PREV_DAY_MAX_SAL
---------- ------ ----------- ---------- ------------------- ---------------- ----------------
       100 Ravi   01/01/2018        5000                   0                  
       101 Kumar  01/01/2018        7000                   0                  
       102 Satish 01/01/2018       13000                   0                  
       103 Naresh 02/01/2018        7500                   3             5000            13000
       104 Lalith 04/01/2018       17300                   0                  
       105 Suresh 04/01/2018       40000                   0                  
       106 Latha  01/02/2018       16000                   0