| From: | Michael Paquier <michael(at)paquier(dot)xyz> | 
|---|---|
| To: | David Rowley <david(dot)rowley(at)2ndquadrant(dot)com> | 
| Cc: | Robert Haas <robertmhaas(at)gmail(dot)com>, PostgreSQL Hackers <pgsql-hackers(at)lists(dot)postgresql(dot)org> | 
| Subject: | Re: Should new partitions inherit their tablespace from their parent? | 
| Date: | 2018-12-07 07:15:08 | 
| Message-ID: | 20181207071508.GS2407@paquier.xyz | 
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| Thread: | |
| Lists: | pgsql-hackers | 
On Sat, Nov 24, 2018 at 10:18:17AM +0900, Michael Paquier wrote:
> On Sat, Nov 24, 2018 at 12:32:36PM +1300, David Rowley wrote:
>> On Fri, 23 Nov 2018 at 17:03, David Rowley <david(dot)rowley(at)2ndquadrant(dot)com> wrote:
>> > Here's a patch for that.  Parking here until January's commitfest.
>> 
>> And another, now rebased atop of de38ce1b8 (which I probably should
>> have waited for).
Thanks for the patch, David.  I can see that this patch makes the code
more consistent for partitioned tables and partitioned indexes when it
comes to tablespace handling, which is a very good thing.
-ATExecPartedIdxSetTableSpace(Relation rel, Oid newTableSpace)
+ATExecSetTableSpaceNoStorage(Relation rel, Oid newTableSpace)
NoStorage looks strange as routine name for this case.  Would something
like ATExecPartedRelSetTableSpace be more adapted perhaps?
+   else if (stmt->partbound)
+   {
+       RangeVar   *parent;
+       Relation    parentrel;
+
+       /*
+        * For partitions, when no other tablespace is specified, we default
+        * the tablespace to the parent partitioned table's.
+        */
Okay, so the direct parent is what counts, and not the top-most parent.
Could you add a test with multiple level of partitioned tables, like:
- parent is in tablespace 1.
- child is created in tablespace 2.
- grandchild is created, which should inherit tablespace 2 from the
child, but not tablespace 1 from the parent.  In the existing example,
as one partition is used to test the top-most parent and another for the
new default, it looks cleaner to create a third partition which would be
itself a partitioned table.
--
Michael
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