Re: Patch: Global Unique Index - Mailing list pgsql-hackers

Hi Bruce,

Thank you for helping review the patches in such detail.

Looking at the patch, I am unclear how the the patch prevents concurrent
duplicate value insertion during the partitioned index checking.  I am
actually not sure how that can be done without locking all indexes or
inserting placeholder entries in all indexes.  (Yeah, that sounds bad,
unless I am missing something.)

For the uniqueness check cross all partitions, we tried to follow the implementation of uniqueness check on a single partition, and added a loop to check uniqueness on other partitions after the index tuple has been inserted to current index partition but before this index tuple has been made visible. The uniqueness check will wait `XactLockTableWait` if there is a valid transaction in process, and performs the uniqueness check again after the in-process transaction finished.

We tried to simulate this duplicate value case in blow steps:

1) prepare the partitioned table,
CREATE TABLE gidx_part (a int, b int, c text) PARTITION BY RANGE (a);
CREATE TABLE gidx_part1 partition of gidx_part FOR VALUES FROM (0) TO (10);
CREATE TABLE gidx_part2 partition of gidx_part FOR VALUES FROM (10) TO (20);

2) having two psql consoles hooked up with gdbs and set break points after _bt_doinsert

result = _bt_doinsert(rel, itup, checkUnique, indexUnchanged, heapRel);

inside btinsert function in nbtree.c file.

3) first, execute `INSERT INTO gidx_part values(1, 1, 'test');` on console-1, and then execute `INSERT INTO gidx_part values(11, 1, 'test');` on console-2 (expect duplicated value '1' in the 2nd column to be detected),

The test results is that: console-2 query will have to wait until either console-1 committed or aborted. If console-1 committed, then console-2 reports duplicated value already exists; if console-1 aborted, then console-2 will report insert successfully. If there is a deadlock, then the one detected this deadlock will error out to allow the other one continue.

I am not quite sure if this is a proper way to deal with a deadlock in this case. It would be so grateful if someone could help provide some cases/methods to verify this cross all partitions uniqueness.

Best regards,

David