How to identify the column used to partition a table from the Postgres system catalogs
Given a table created as so...
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (logdate);
How do I identify which column it's been partitioned on? - in this case 'logdate' solely by querying the postgres catalog.
I've looked in the obvious places in the catalog (pg_class, pg_index) but nothing springs out.
(Using version 10.5)
postgresql partitioning postgresql-10
add a comment |
Given a table created as so...
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (logdate);
How do I identify which column it's been partitioned on? - in this case 'logdate' solely by querying the postgres catalog.
I've looked in the obvious places in the catalog (pg_class, pg_index) but nothing springs out.
(Using version 10.5)
postgresql partitioning postgresql-10
@McNets - "select * from pg_class where relkind='p'" gives me those tables that are partitioned, it doesn't tell which column they've been partitioned on (not that I can see anyhow!)
– Hemel
Nov 24 at 16:14
add a comment |
Given a table created as so...
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (logdate);
How do I identify which column it's been partitioned on? - in this case 'logdate' solely by querying the postgres catalog.
I've looked in the obvious places in the catalog (pg_class, pg_index) but nothing springs out.
(Using version 10.5)
postgresql partitioning postgresql-10
Given a table created as so...
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (logdate);
How do I identify which column it's been partitioned on? - in this case 'logdate' solely by querying the postgres catalog.
I've looked in the obvious places in the catalog (pg_class, pg_index) but nothing springs out.
(Using version 10.5)
postgresql partitioning postgresql-10
postgresql partitioning postgresql-10
edited Nov 26 at 7:16
a_horse_with_no_name
38.5k775111
38.5k775111
asked Nov 24 at 15:34
Hemel
21829
21829
@McNets - "select * from pg_class where relkind='p'" gives me those tables that are partitioned, it doesn't tell which column they've been partitioned on (not that I can see anyhow!)
– Hemel
Nov 24 at 16:14
add a comment |
@McNets - "select * from pg_class where relkind='p'" gives me those tables that are partitioned, it doesn't tell which column they've been partitioned on (not that I can see anyhow!)
– Hemel
Nov 24 at 16:14
@McNets - "select * from pg_class where relkind='p'" gives me those tables that are partitioned, it doesn't tell which column they've been partitioned on (not that I can see anyhow!)
– Hemel
Nov 24 at 16:14
@McNets - "select * from pg_class where relkind='p'" gives me those tables that are partitioned, it doesn't tell which column they've been partitioned on (not that I can see anyhow!)
– Hemel
Nov 24 at 16:14
add a comment |
2 Answers
2
active
oldest
votes
I'm not a PostgreSQL pro, but digging a bit I've founded a solution that perhaps can help you. (Works only for version 10 or above)
First I've slightly modified you table by adding two columns to the partition definition (just to show you the final result):
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (city_id,logdate);
This is my solution:
select
par.relnamespace::regnamespace::text as schema,
par.relname as table_name,
partnatts as num_columns,
column_index,
col.column_name
from
(select
partrelid,
partnatts,
case partstrat
when 'l' then 'list'
when 'r' then 'range' end as partition_strategy,
unnest(partattrs) column_index
from
pg_partitioned_table) pt
join
pg_class par
on
par.oid = pt.partrelid
join
information_schema.columns col
on
col.table_schema = par.relnamespace::regnamespace::text
and col.table_name = par.relname
and ordinal_position = pt.column_index;
schema | table_name | num_columns | column_index | column_name
:-------------------------- | :---------- | ----------: | -----------: | :----------
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 1 | city_id
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 2 | logdate
db<>fiddle here
pg_partitioned_table
The catalog pg_partitioned_table stores information about how tables
are partitioned.
Unnesting partattrs
you can get column index of each row involved in the partition. Then you can join information_schema.columns
just to retrieve the name of every column.
+-----------+------------+---------------------+-------------------------------------------------------------|
| partattrs | int2vector | pg_attribute.attnum | This is an array of partnatts values that indicate |
| | | | which table columns are part of the partition key. |
| | | | For example, a value of 1 3 would mean that the first |
| | | | and the third table columns make up the partition key. |
| | | | A zero in this array indicates that the corresponding |
| | | | partition key column is an expression, rather than a simple |
| | | | column reference. |
+-----------+------------+---------------------+-------------------------------------------------------------|
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
I'm glad to help.
– McNets
Nov 24 at 17:22
add a comment |
The answer by McNets already helps, but here is a query that produces slightly prettier output:
select c.relnamespace::regnamespace::text as schema,
c.relname as table_name,
pg_get_partkeydef(c.oid) as partition_key
from pg_class c
where c.relkind = 'p';
Below is output that the above query produces:
schema │ table_name │ partition_key
────────┼─────────────┼──────────────────────────
public │ measurement │ RANGE (city_id, logdate)
(1 row)
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'm not a PostgreSQL pro, but digging a bit I've founded a solution that perhaps can help you. (Works only for version 10 or above)
First I've slightly modified you table by adding two columns to the partition definition (just to show you the final result):
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (city_id,logdate);
This is my solution:
select
par.relnamespace::regnamespace::text as schema,
par.relname as table_name,
partnatts as num_columns,
column_index,
col.column_name
from
(select
partrelid,
partnatts,
case partstrat
when 'l' then 'list'
when 'r' then 'range' end as partition_strategy,
unnest(partattrs) column_index
from
pg_partitioned_table) pt
join
pg_class par
on
par.oid = pt.partrelid
join
information_schema.columns col
on
col.table_schema = par.relnamespace::regnamespace::text
and col.table_name = par.relname
and ordinal_position = pt.column_index;
schema | table_name | num_columns | column_index | column_name
:-------------------------- | :---------- | ----------: | -----------: | :----------
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 1 | city_id
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 2 | logdate
db<>fiddle here
pg_partitioned_table
The catalog pg_partitioned_table stores information about how tables
are partitioned.
Unnesting partattrs
you can get column index of each row involved in the partition. Then you can join information_schema.columns
just to retrieve the name of every column.
+-----------+------------+---------------------+-------------------------------------------------------------|
| partattrs | int2vector | pg_attribute.attnum | This is an array of partnatts values that indicate |
| | | | which table columns are part of the partition key. |
| | | | For example, a value of 1 3 would mean that the first |
| | | | and the third table columns make up the partition key. |
| | | | A zero in this array indicates that the corresponding |
| | | | partition key column is an expression, rather than a simple |
| | | | column reference. |
+-----------+------------+---------------------+-------------------------------------------------------------|
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
I'm glad to help.
– McNets
Nov 24 at 17:22
add a comment |
I'm not a PostgreSQL pro, but digging a bit I've founded a solution that perhaps can help you. (Works only for version 10 or above)
First I've slightly modified you table by adding two columns to the partition definition (just to show you the final result):
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (city_id,logdate);
This is my solution:
select
par.relnamespace::regnamespace::text as schema,
par.relname as table_name,
partnatts as num_columns,
column_index,
col.column_name
from
(select
partrelid,
partnatts,
case partstrat
when 'l' then 'list'
when 'r' then 'range' end as partition_strategy,
unnest(partattrs) column_index
from
pg_partitioned_table) pt
join
pg_class par
on
par.oid = pt.partrelid
join
information_schema.columns col
on
col.table_schema = par.relnamespace::regnamespace::text
and col.table_name = par.relname
and ordinal_position = pt.column_index;
schema | table_name | num_columns | column_index | column_name
:-------------------------- | :---------- | ----------: | -----------: | :----------
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 1 | city_id
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 2 | logdate
db<>fiddle here
pg_partitioned_table
The catalog pg_partitioned_table stores information about how tables
are partitioned.
Unnesting partattrs
you can get column index of each row involved in the partition. Then you can join information_schema.columns
just to retrieve the name of every column.
+-----------+------------+---------------------+-------------------------------------------------------------|
| partattrs | int2vector | pg_attribute.attnum | This is an array of partnatts values that indicate |
| | | | which table columns are part of the partition key. |
| | | | For example, a value of 1 3 would mean that the first |
| | | | and the third table columns make up the partition key. |
| | | | A zero in this array indicates that the corresponding |
| | | | partition key column is an expression, rather than a simple |
| | | | column reference. |
+-----------+------------+---------------------+-------------------------------------------------------------|
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
I'm glad to help.
– McNets
Nov 24 at 17:22
add a comment |
I'm not a PostgreSQL pro, but digging a bit I've founded a solution that perhaps can help you. (Works only for version 10 or above)
First I've slightly modified you table by adding two columns to the partition definition (just to show you the final result):
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (city_id,logdate);
This is my solution:
select
par.relnamespace::regnamespace::text as schema,
par.relname as table_name,
partnatts as num_columns,
column_index,
col.column_name
from
(select
partrelid,
partnatts,
case partstrat
when 'l' then 'list'
when 'r' then 'range' end as partition_strategy,
unnest(partattrs) column_index
from
pg_partitioned_table) pt
join
pg_class par
on
par.oid = pt.partrelid
join
information_schema.columns col
on
col.table_schema = par.relnamespace::regnamespace::text
and col.table_name = par.relname
and ordinal_position = pt.column_index;
schema | table_name | num_columns | column_index | column_name
:-------------------------- | :---------- | ----------: | -----------: | :----------
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 1 | city_id
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 2 | logdate
db<>fiddle here
pg_partitioned_table
The catalog pg_partitioned_table stores information about how tables
are partitioned.
Unnesting partattrs
you can get column index of each row involved in the partition. Then you can join information_schema.columns
just to retrieve the name of every column.
+-----------+------------+---------------------+-------------------------------------------------------------|
| partattrs | int2vector | pg_attribute.attnum | This is an array of partnatts values that indicate |
| | | | which table columns are part of the partition key. |
| | | | For example, a value of 1 3 would mean that the first |
| | | | and the third table columns make up the partition key. |
| | | | A zero in this array indicates that the corresponding |
| | | | partition key column is an expression, rather than a simple |
| | | | column reference. |
+-----------+------------+---------------------+-------------------------------------------------------------|
I'm not a PostgreSQL pro, but digging a bit I've founded a solution that perhaps can help you. (Works only for version 10 or above)
First I've slightly modified you table by adding two columns to the partition definition (just to show you the final result):
CREATE TABLE measurement (
city_id int not null,
logdate date not null,
peaktemp int,
unitsales int
) PARTITION BY RANGE (city_id,logdate);
This is my solution:
select
par.relnamespace::regnamespace::text as schema,
par.relname as table_name,
partnatts as num_columns,
column_index,
col.column_name
from
(select
partrelid,
partnatts,
case partstrat
when 'l' then 'list'
when 'r' then 'range' end as partition_strategy,
unnest(partattrs) column_index
from
pg_partitioned_table) pt
join
pg_class par
on
par.oid = pt.partrelid
join
information_schema.columns col
on
col.table_schema = par.relnamespace::regnamespace::text
and col.table_name = par.relname
and ordinal_position = pt.column_index;
schema | table_name | num_columns | column_index | column_name
:-------------------------- | :---------- | ----------: | -----------: | :----------
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 1 | city_id
fiddle_ctcqwfrzpcyngmgnqkdy | measurement | 2 | 2 | logdate
db<>fiddle here
pg_partitioned_table
The catalog pg_partitioned_table stores information about how tables
are partitioned.
Unnesting partattrs
you can get column index of each row involved in the partition. Then you can join information_schema.columns
just to retrieve the name of every column.
+-----------+------------+---------------------+-------------------------------------------------------------|
| partattrs | int2vector | pg_attribute.attnum | This is an array of partnatts values that indicate |
| | | | which table columns are part of the partition key. |
| | | | For example, a value of 1 3 would mean that the first |
| | | | and the third table columns make up the partition key. |
| | | | A zero in this array indicates that the corresponding |
| | | | partition key column is an expression, rather than a simple |
| | | | column reference. |
+-----------+------------+---------------------+-------------------------------------------------------------|
edited Nov 24 at 17:21
answered Nov 24 at 17:12
McNets
14.6k41757
14.6k41757
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
I'm glad to help.
– McNets
Nov 24 at 17:22
add a comment |
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
I'm glad to help.
– McNets
Nov 24 at 17:22
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
quote "I'm not a PostgreSQL pro" consider yourself promoted!
– Hemel
Nov 24 at 17:21
I'm glad to help.
– McNets
Nov 24 at 17:22
I'm glad to help.
– McNets
Nov 24 at 17:22
add a comment |
The answer by McNets already helps, but here is a query that produces slightly prettier output:
select c.relnamespace::regnamespace::text as schema,
c.relname as table_name,
pg_get_partkeydef(c.oid) as partition_key
from pg_class c
where c.relkind = 'p';
Below is output that the above query produces:
schema │ table_name │ partition_key
────────┼─────────────┼──────────────────────────
public │ measurement │ RANGE (city_id, logdate)
(1 row)
add a comment |
The answer by McNets already helps, but here is a query that produces slightly prettier output:
select c.relnamespace::regnamespace::text as schema,
c.relname as table_name,
pg_get_partkeydef(c.oid) as partition_key
from pg_class c
where c.relkind = 'p';
Below is output that the above query produces:
schema │ table_name │ partition_key
────────┼─────────────┼──────────────────────────
public │ measurement │ RANGE (city_id, logdate)
(1 row)
add a comment |
The answer by McNets already helps, but here is a query that produces slightly prettier output:
select c.relnamespace::regnamespace::text as schema,
c.relname as table_name,
pg_get_partkeydef(c.oid) as partition_key
from pg_class c
where c.relkind = 'p';
Below is output that the above query produces:
schema │ table_name │ partition_key
────────┼─────────────┼──────────────────────────
public │ measurement │ RANGE (city_id, logdate)
(1 row)
The answer by McNets already helps, but here is a query that produces slightly prettier output:
select c.relnamespace::regnamespace::text as schema,
c.relname as table_name,
pg_get_partkeydef(c.oid) as partition_key
from pg_class c
where c.relkind = 'p';
Below is output that the above query produces:
schema │ table_name │ partition_key
────────┼─────────────┼──────────────────────────
public │ measurement │ RANGE (city_id, logdate)
(1 row)
answered Nov 26 at 4:04
Amit L
24125
24125
add a comment |
add a comment |
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@McNets - "select * from pg_class where relkind='p'" gives me those tables that are partitioned, it doesn't tell which column they've been partitioned on (not that I can see anyhow!)
– Hemel
Nov 24 at 16:14