I'm not sure if I understand your issue, but could you output
SELECT
COUNT(*)
FROM rmas
WHERE
id = 1008122437
AND status = 'r';
HTH
Kind regards/met vriendelijke groet,
Serge Fonville
Convince Microsoft!
They need to add TRUNCATE PARTITION in SQL Server
2013/5/23 Keith Fiske <keith@xxxxxxxxxx>
This is running 9.2.4 on CentOS. If anyone can suggest how I can look into this deeper and find what the problem may be, I'd appreciate it. I'm here at PGCon if anyone is available to help IRL as wellClient reported an issue where it appears a foreign key has been violatedAttempting to reinsert this data again causes a violation error, so it doesn't appear to be broken
prod=#\d rma_items
[snip]
rma_items_rma_id_status_fk" FOREIGN KEY (rma_id, rma_status) REFERENCES rmas(id, status) ON UPDATE CASCADE ON DELETE CASCADE
prod=# select i.rma_id, i.rma_status, r.id, r.status from rmas r join rma_items i on i.rma_id = r.id and i.rma_status != r.status;
rma_id | rma_status | id | status
------------+------------+------------+--------
1008122437 | r | 1008122437 | c
(1 row)
prod=# begin;
BEGIN
prod=# insert into rma_items (rma_id, order_item_id, return_reason_id, rma_status) values (1008122437, 1007674099, 9797623, 'r');
ERROR: insert or update on table "rma_items" violates foreign key constraint "rma_items_rma_id_status_fk"
DETAIL: Key (rma_id, rma_status)=(1008122437, r) is not present in table "rmas".
prod=# rollback;
ROLLBACK
