Writing Oracle Queries to Retrieve Latest Values and Min File Code

Step 1: Understand the problem and identify the goal

The problem is to write an Oracle query that retrieves the latest values from a table, separated by a specific column. The goal is to find the minimum file_code for each subscriber_id or filter by property_id of 289 with the latest graph_registration_date.

Step 2: Determine the approach for finding the latest value

To solve this problem, we need to use Oracle’s analytic functions, such as RANK() or ROW_NUMBER(), to rank rows within a partition and then select the top row based on that ranking.

Step 3: Write the query for the minimum file_code

First, let’s write the query that finds the minimum file_code for each subscriber_id:

SELECT 
  S.FILE_CODE,
  SP.SUBSCRIBER_ID,
  COALESCE(RC.FIRST_NAME || RC.LAST_NAME, LC.LEGAL_NAME) AS CUSTOMER_NAME,
  PROPERTY_ID,
  PROPERTY_CONTENT,
  RANK() OVER (PARTITION BY subscriber_id ORDER BY file_code) AS rnk
FROM 
  CUSTOMERS_TCI.A_SUBSCRIBER_PROPERTIES  SP
  INNER JOIN CUSTOMERS_TCI.SUBSCRIBERS S ON S.SUBSCRIBER_ID = SP.SUBSCRIBER_ID
  INNER JOIN CUSTOMERS_TCI.CUSTOMERS_INFO CI ON CI.CUSTOMER_ID = S.CUSTOMER_ID
  LEFT JOIN CUSTOMERS_TCI.REAL_CUSTOMER RC ON RC.CUSTOMER_ID = CI.CUSTOMER_ID
  LEFT JOIN CUSTOMERS_TCI.LEGAL_CUSTOMER LC ON LC.CUSTOMER_ID = CI.CUSTOMER_ID
WHERE 
  PROPERTY_ID IN (277, 289, 290, 291)
AND    SP.PROVINCE_ID = 22

Step 4: Write the query with PIVOT for the specific property_id

Next, let’s write the query that pivots the output for the property_id of 289 and filters to return only the row with the latest graph_registration_date:

SELECT *
FROM (
  SELECT p.*,
         ROW_NUMBER() OVER (
           PARTITION BY subscriber_id
           ORDER BY TO_DATE(GRAPH_REGISTRATION_DATE, 'MM/DD/YYYY') DESC
         ) AS rn
  FROM 
    SELECT S.FILE_CODE,
           SP.SUBSCRIBER_ID,
           COALESCE(RC.FIRST_NAME || RC.LAST_NAME, LC.LEGAL_NAME) AS CUSTOMER_NAME,
           PROPERTY_ID,
           PROPERTY_CONTENT
    FROM 
      CUSTOMERS_TCI.A_SUBSCRIBER_PROPERTIES  SP
        INNER JOIN CUSTOMERS_TCI.SUBSCRIBERS S ON S.SUBSCRIBER_ID = SP.SUBSCRIBER_ID
        INNER JOIN CUSTOMERS_TCI.CUSTOMERS_INFO CI ON CI.CUSTOMER_ID = S.CUSTOMER_ID
        LEFT JOIN CUSTOMERS_TCI.REAL_CUSTOMER RC ON RC.CUSTOMER_ID = CI.CUSTOMER_ID
        LEFT JOIN CUSTOMERS_TCI.LEGAL_CUSTOMER LC ON LC.CUSTOMER_ID = CI.CUSTOMER_ID
    WHERE 
      PROPERTY_ID IN (277, 289, 290, 291)
    AND    SP.PROVINCE_ID = 22
) p
PIVOT (
  MAX(PROPERTY_CONTENT)
  FOR PROPERTY_ID IN (
    277 AS GRAPH,
    289 AS GRAPH_REGISTRATION_DATE,
    290 AS GRAPH_YEAR,
    291 AS GRAPH_MONTH
  )
)
WHERE 
  rn = 1
AND   GRAPH_YEAR = '1402'
ORDER BY CUSTOMER_NAME;

The final answer is: There is no single number that solves this problem as it involves writing an Oracle query.

Last modified on 2023-10-13