Select partitions based on matches in other table

Question

Having the following table (conversations):

 id | record_id  |  is_response  |         text         |
 ---+------------+---------------+------------         


        

Answer 1

There is a simple and fast solution:

SELECT record_id, string_agg(text, ' ') As context
FROM  (
   SELECT c.*, count(r.text) OVER (PARTITION BY c.record_id ORDER BY c.id DESC) AS grp
   FROM   conversations  c
   LEFT   JOIN responses r ON r.text = c.text AND c.is_response
   ORDER  BY record_id, id
   ) sub
WHERE  grp > 0  -- ignore conversation part that does not end with a response
GROUP  BY record_id, grp
ORDER  BY record_id, grp;

count() only counts non-null values. r.text is NULL if the LEFT JOIN to responses comes up empty:

• Select rows which are not present in other table

The value in grp (short for "group") is only increased when a new output row is triggered. All rows belonging to the same output row end up with the same grp number. It's then easy to aggregate in the outer SELECT.

The special trick is to count conversation ends in reverse order. Everything after the last end (coming first when starting from the end) gets grp = 0 and is removed in the outer SELECT.

Similar cases with more explanation:

• Row number with reset in PostgreSQL
• Select longest continuous sequence