PostgreSQL declarative partitioning
i am trying to understand how can declarative partitioning used in my particular case.
i have a table that look like
CREATE TABLE test
(
created timestamp without time zone NOT NULL,
tx bigint NOT NULL unique
) PARTITION BY RANGE (created);
CREATE TABLE test_part PARTITION OF test
FOR VALUES FROM ('2017-01-01') TO ('2017-02-01');
and, if i try to create it i get
DETAIL: UNIQUE constraint on table "test" lacks column "created" which is part of the partition key.
but, what i would like to do, it's to create a unique key on tx, and not on tx+created.
if i create a unique key (tx,created) then this it's valid, and it's not in my case
insert into test values('2017-01-02',1)
insert into test values('2017-01-01',1)
how can i create a unique index that doesn't contain the partition key?
in that case, i would like to store all the data partitioned based on created, but the txid should be unique on his own.
postgresql database-partitioning postgresql-11
add a comment |
i am trying to understand how can declarative partitioning used in my particular case.
i have a table that look like
CREATE TABLE test
(
created timestamp without time zone NOT NULL,
tx bigint NOT NULL unique
) PARTITION BY RANGE (created);
CREATE TABLE test_part PARTITION OF test
FOR VALUES FROM ('2017-01-01') TO ('2017-02-01');
and, if i try to create it i get
DETAIL: UNIQUE constraint on table "test" lacks column "created" which is part of the partition key.
but, what i would like to do, it's to create a unique key on tx, and not on tx+created.
if i create a unique key (tx,created) then this it's valid, and it's not in my case
insert into test values('2017-01-02',1)
insert into test values('2017-01-01',1)
how can i create a unique index that doesn't contain the partition key?
in that case, i would like to store all the data partitioned based on created, but the txid should be unique on his own.
postgresql database-partitioning postgresql-11
This is currently not possible.
– a_horse_with_no_name
Nov 13 '18 at 11:50
add a comment |
i am trying to understand how can declarative partitioning used in my particular case.
i have a table that look like
CREATE TABLE test
(
created timestamp without time zone NOT NULL,
tx bigint NOT NULL unique
) PARTITION BY RANGE (created);
CREATE TABLE test_part PARTITION OF test
FOR VALUES FROM ('2017-01-01') TO ('2017-02-01');
and, if i try to create it i get
DETAIL: UNIQUE constraint on table "test" lacks column "created" which is part of the partition key.
but, what i would like to do, it's to create a unique key on tx, and not on tx+created.
if i create a unique key (tx,created) then this it's valid, and it's not in my case
insert into test values('2017-01-02',1)
insert into test values('2017-01-01',1)
how can i create a unique index that doesn't contain the partition key?
in that case, i would like to store all the data partitioned based on created, but the txid should be unique on his own.
postgresql database-partitioning postgresql-11
i am trying to understand how can declarative partitioning used in my particular case.
i have a table that look like
CREATE TABLE test
(
created timestamp without time zone NOT NULL,
tx bigint NOT NULL unique
) PARTITION BY RANGE (created);
CREATE TABLE test_part PARTITION OF test
FOR VALUES FROM ('2017-01-01') TO ('2017-02-01');
and, if i try to create it i get
DETAIL: UNIQUE constraint on table "test" lacks column "created" which is part of the partition key.
but, what i would like to do, it's to create a unique key on tx, and not on tx+created.
if i create a unique key (tx,created) then this it's valid, and it's not in my case
insert into test values('2017-01-02',1)
insert into test values('2017-01-01',1)
how can i create a unique index that doesn't contain the partition key?
in that case, i would like to store all the data partitioned based on created, but the txid should be unique on his own.
postgresql database-partitioning postgresql-11
postgresql database-partitioning postgresql-11
edited Nov 13 '18 at 11:51
a_horse_with_no_name
293k46447541
293k46447541
asked Nov 13 '18 at 11:48
user3804873user3804873
2516
2516
This is currently not possible.
– a_horse_with_no_name
Nov 13 '18 at 11:50
add a comment |
This is currently not possible.
– a_horse_with_no_name
Nov 13 '18 at 11:50
This is currently not possible.
– a_horse_with_no_name
Nov 13 '18 at 11:50
This is currently not possible.
– a_horse_with_no_name
Nov 13 '18 at 11:50
add a comment |
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This is currently not possible.
– a_horse_with_no_name
Nov 13 '18 at 11:50