Postgres get first and last version for individual vendor - sql

Question

I have mapping table for RFQ(request for quotation) and Vendor's bid amount with version.
Table :
id rfq_id(FK) vendor_id(FK) amount version
-----------------------------------------------
1 1 1 100 1
2 1 1 90 2
3 1 1 80 3
4 1 2 50 1
5 1 7 500 1
6 1 7 495 2
7 1 7 500 3
8 1 7 525 4
9 1 7 450 5
10 1 7 430 6
11 2 1 200 1
12 2 2 300 1
13 2 2 350 2
14 2 3 40 1
15 3 4 70 1
In above table, I want analysis for vendor's first and last bid for particular rfq_id.
Expected Output for rfq_id=1 :
vendor_id first_bid last_bid
---------------------------------
1 100 80
2 50 50
7 500 430
From Postgres : get min and max rows count in many to many relation table I have came to know about window and partition. So I have tried below query.
SELECT
vendor_id,
version,
amount,
first_value(amount) over w as first_bid,
last_value(amount) over w as last_bid,
row_number() over w as rn
FROM
rfq_vendor_version_mapping
where
rfq_id=1
WINDOW w AS (PARTITION BY vendor_id order by version)
ORDER by vendor_id;
With above query, every vendor's maximum rn is my output.
http://sqlfiddle.com/#!15/f19a0/7

Answer 1

Window functions add columns to all the existing rows, instead of grouping input rows into a single output row. Since you are only interested in the bid values, use a DISTINCT clause on the fields of interest.
Note that you need a frame clause for the WINDOW definition to make sure that all rows in the partition are considered. By default, the frame in the partition (the rows that are being used in calculations) runs from the beginning of the partition to the current row. Therefore, the last_value() window function always returns the value of the current row; use a frame of UNBOUNDED PRECEDING TO UNBOUNDED FOLLOWING to extend the frame to the entire partition.
SELECT DISTINCT
vendor_id,
version,
amount,
first_value(amount) OVER w AS first_bid,
last_value(amount) OVER w AS last_bid
row_number() over w as rn
FROM
rfq_vendor_version_mapping
WHERE rfq_id = 1
WINDOW w AS (PARTITION BY vendor_id ORDER BY version
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)
ORDER BY vendor_id;

Answer 2

You have to GROUP BY vendor_id because you want just one row per vendor_id:
SELECT
vendor_id,
MAX(CASE WHEN rn = 1 THEN amount END) AS first_bid,
MAX(CASE WHEN rn2 = 1 THEN amount END) AS last_bid
FROM (
SELECT
vendor_id,
version,
amount,
row_number() over (PARTITION BY vendor_id order BY version) as rn,
row_number() over (PARTITION BY vendor_id order BY version DESC) as rn2
FROM
rfq_vendor_version_mapping
WHERE
rfq_id=1) AS t
GROUP BY vendor_id
ORDER by vendor_id;
The query uses conditional aggregation in order to extract amount values that correspond to first and last bid.
Demo here

Answer 3

Without ORDER BY OLAP-functions default to ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING but with ORDER BY this changes to ROW UNBOUNDED PRECEDING.
You were quite close, but you need two different windows:
select vendor_id, amount as first_bid, last_bid
from
(
SELECT
vendor_id,
version,
amount,
last_value(amount) -- highest version's bid
over (PARTITION BY vendor_id
order by version
rows between unbiunded preceding and unbounded following) as last_bid,
row_number()
over (PARTITION BY vendor_id
order by version) as rn
FROM
rfq_vendor_version_mapping
where
rfq_id=1
) as dt
where rn = 1 -- row with first version/bid
ORDER by vendor_id;