RE: row filtering for logical replication

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От tanghy.fnst@fujitsu.com
Тема RE: row filtering for logical replication
Дата
Msg-id OS0PR01MB6113BB510435B16E9F0B2A59FB519@OS0PR01MB6113.jpnprd01.prod.outlook.com
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Ответ на RE: row filtering for logical replication  ("houzj.fnst@fujitsu.com" <houzj.fnst@fujitsu.com>)
Ответы Re: row filtering for logical replication  (Amit Kapila <amit.kapila16@gmail.com>)
Список pgsql-hackers
On Tuesday, January 11, 2022 10:16 AM houzj.fnst@fujitsu.com <houzj.fnst@fujitsu.com> wrote:
> 
> Attach the v62 patch set which address the above comments and slightly
> adjust the commit message in 0002 patch.
> 

I saw a possible problem about Row-Filter tablesync SQL, which is related
to partition table.

If a parent table is published with publish_via_partition_root off, its child
table should be taken as no row filter when combining the row filters with OR.
But when using the current SQL, this publication is ignored.

For example:
create table parent (a int) partition by range (a);
create table child partition of parent default;
create publication puba for table parent with (publish_via_partition_root=false);
create publication pubb for table child where(a>10);

Using current SQL in patch:
(table child oid is 16387)
SELECT DISTINCT pg_get_expr(prqual, prrelid) FROM pg_publication p
INNER JOIN pg_publication_rel pr ON (p.oid = pr.prpubid)
WHERE pr.prrelid = 16387 AND p.pubname IN ( 'puba', 'pubb' )
AND NOT (select bool_or(puballtables)
FROM pg_publication
WHERE pubname in ( 'puba', 'pubb' ))
AND NOT EXISTS (SELECT 1
FROM pg_publication_namespace pn, pg_class c, pg_publication p
WHERE c.oid = 16387 AND c.relnamespace = pn.pnnspid AND p.oid = pn.pnpubid AND p.pubname IN ( 'puba', 'pubb' ));
pg_get_expr
-------------
 (a > 10)
(1 row)


I think there should be no filter in this case, because "puba" publish table child
without row filter. Thoughts?

To fix this problem, we could use pg_get_publication_tables function in
tablesync SQL to filter which publications the table belongs to. How about the
following SQL, it would return NULL for "puba".

SELECT DISTINCT pg_get_expr(pr.prqual, pr.prrelid)
FROM pg_publication p
LEFT OUTER JOIN pg_publication_rel pr
    ON (p.oid = pr.prpubid AND pr.prrelid = 16387),
LATERAL pg_get_publication_tables(p.pubname) GPT
WHERE GPT.relid = 16387 AND p.pubname IN ( 'puba', 'pubb' );
 pg_get_expr
-------------
 (a > 10)

(2 rows)

Regards,
Tang

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