add partition left boundary in an existing range right partitioned table Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Data science time! April 2019 and salary with experience Should we burninate the [wrap] tag? The Ask Question Wizard is Live!Add a column with a default value to an existing table in SQL ServerHow to check if a column exists in a SQL Server table?Check if table exists in SQL ServerSQLite - UPSERT *not* INSERT or REPLACEWhat is the difference between Left, Right, Outer and Inner Joins?Check if a temporary table exists and delete if it exists before creating a temporary tableInsert into a MySQL table or update if existsWhat's the difference between INNER JOIN, LEFT JOIN, RIGHT JOIN and FULL JOIN?How to drop a table if it exists in SQL Server?SQL Table Partition Error - Cannot Add A New Partition To Existing Partitioned Table
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add partition left boundary in an existing range right partitioned table
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?
The Ask Question Wizard is Live!Add a column with a default value to an existing table in SQL ServerHow to check if a column exists in a SQL Server table?Check if table exists in SQL ServerSQLite - UPSERT *not* INSERT or REPLACEWhat is the difference between Left, Right, Outer and Inner Joins?Check if a temporary table exists and delete if it exists before creating a temporary tableInsert into a MySQL table or update if existsWhat's the difference between INNER JOIN, LEFT JOIN, RIGHT JOIN and FULL JOIN?How to drop a table if it exists in SQL Server?SQL Table Partition Error - Cannot Add A New Partition To Existing Partitioned Table
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My partition function creates right range type partitions. It turns out after the data migration that in partition_number one there are boundary values that are lesser than the boundary declared in my partition function. So for example, if my minimum partition key is 5, i found values 1,2,3 and 4 in partition number 1. What i need to do is alter my partition function adding boundaries 1,2,3 and 4.How am i going to do this? Does split range work in this case? How is sql server going to re arrange my data in the new partitions. Will it do the job just by altering the table? Do i need to do something extra? Do i need to take a backup in case something goes wrong?
sql sql-server sql-server-2005
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My partition function creates right range type partitions. It turns out after the data migration that in partition_number one there are boundary values that are lesser than the boundary declared in my partition function. So for example, if my minimum partition key is 5, i found values 1,2,3 and 4 in partition number 1. What i need to do is alter my partition function adding boundaries 1,2,3 and 4.How am i going to do this? Does split range work in this case? How is sql server going to re arrange my data in the new partitions. Will it do the job just by altering the table? Do i need to do something extra? Do i need to take a backup in case something goes wrong?
sql sql-server sql-server-2005
add a comment |
My partition function creates right range type partitions. It turns out after the data migration that in partition_number one there are boundary values that are lesser than the boundary declared in my partition function. So for example, if my minimum partition key is 5, i found values 1,2,3 and 4 in partition number 1. What i need to do is alter my partition function adding boundaries 1,2,3 and 4.How am i going to do this? Does split range work in this case? How is sql server going to re arrange my data in the new partitions. Will it do the job just by altering the table? Do i need to do something extra? Do i need to take a backup in case something goes wrong?
sql sql-server sql-server-2005
My partition function creates right range type partitions. It turns out after the data migration that in partition_number one there are boundary values that are lesser than the boundary declared in my partition function. So for example, if my minimum partition key is 5, i found values 1,2,3 and 4 in partition number 1. What i need to do is alter my partition function adding boundaries 1,2,3 and 4.How am i going to do this? Does split range work in this case? How is sql server going to re arrange my data in the new partitions. Will it do the job just by altering the table? Do i need to do something extra? Do i need to take a backup in case something goes wrong?
sql sql-server sql-server-2005
sql sql-server sql-server-2005
edited Mar 22 at 9:10
Bonzay
asked Mar 22 at 9:02
BonzayBonzay
17514
17514
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1 Answer
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I guess your situation is the following: you have a right range partition with the lowest boundary value being 5, and a table that is being partitioned that way, e.g.:
create partition function pf (int) as range right for values (5)
create partition scheme ps as partition pf to ([PRIMARY], [PRIMARY])
create table T (part_key int constraint PK_T primary key)
on ps (part_key)
Now, if you enter the values 1, 2, 3, 4, and 5 into table T and check the distribution of values within each partition, you'll find 1, 2, 3 and 4 in partition 1, and 5 in partition 2:
insert T values (1), (2), (3), (4), (5)
select part_key, $partition.pf(part_key) as partition from T
What you need to do to get each value in its own partition is:
- for each new partition add a new destination filegroup, and
- split the partition range starting at highest value
This could look like:
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (4)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (3)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (2)
Now, if you check again your value distribution, you'll see that all distinct values end up in a separate partition:
select part_key, $partition.pf(part_key) as partition from T
But, be aware that this goes along with data movement, i.e. all rows with partition key values 1-4 need to be physically moved from original partition 2 to their new destination partition. So, if there are millions of such rows, this'll take some time and will blow up your transaction log.
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
add a comment |
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1 Answer
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I guess your situation is the following: you have a right range partition with the lowest boundary value being 5, and a table that is being partitioned that way, e.g.:
create partition function pf (int) as range right for values (5)
create partition scheme ps as partition pf to ([PRIMARY], [PRIMARY])
create table T (part_key int constraint PK_T primary key)
on ps (part_key)
Now, if you enter the values 1, 2, 3, 4, and 5 into table T and check the distribution of values within each partition, you'll find 1, 2, 3 and 4 in partition 1, and 5 in partition 2:
insert T values (1), (2), (3), (4), (5)
select part_key, $partition.pf(part_key) as partition from T
What you need to do to get each value in its own partition is:
- for each new partition add a new destination filegroup, and
- split the partition range starting at highest value
This could look like:
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (4)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (3)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (2)
Now, if you check again your value distribution, you'll see that all distinct values end up in a separate partition:
select part_key, $partition.pf(part_key) as partition from T
But, be aware that this goes along with data movement, i.e. all rows with partition key values 1-4 need to be physically moved from original partition 2 to their new destination partition. So, if there are millions of such rows, this'll take some time and will blow up your transaction log.
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
add a comment |
I guess your situation is the following: you have a right range partition with the lowest boundary value being 5, and a table that is being partitioned that way, e.g.:
create partition function pf (int) as range right for values (5)
create partition scheme ps as partition pf to ([PRIMARY], [PRIMARY])
create table T (part_key int constraint PK_T primary key)
on ps (part_key)
Now, if you enter the values 1, 2, 3, 4, and 5 into table T and check the distribution of values within each partition, you'll find 1, 2, 3 and 4 in partition 1, and 5 in partition 2:
insert T values (1), (2), (3), (4), (5)
select part_key, $partition.pf(part_key) as partition from T
What you need to do to get each value in its own partition is:
- for each new partition add a new destination filegroup, and
- split the partition range starting at highest value
This could look like:
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (4)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (3)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (2)
Now, if you check again your value distribution, you'll see that all distinct values end up in a separate partition:
select part_key, $partition.pf(part_key) as partition from T
But, be aware that this goes along with data movement, i.e. all rows with partition key values 1-4 need to be physically moved from original partition 2 to their new destination partition. So, if there are millions of such rows, this'll take some time and will blow up your transaction log.
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
add a comment |
I guess your situation is the following: you have a right range partition with the lowest boundary value being 5, and a table that is being partitioned that way, e.g.:
create partition function pf (int) as range right for values (5)
create partition scheme ps as partition pf to ([PRIMARY], [PRIMARY])
create table T (part_key int constraint PK_T primary key)
on ps (part_key)
Now, if you enter the values 1, 2, 3, 4, and 5 into table T and check the distribution of values within each partition, you'll find 1, 2, 3 and 4 in partition 1, and 5 in partition 2:
insert T values (1), (2), (3), (4), (5)
select part_key, $partition.pf(part_key) as partition from T
What you need to do to get each value in its own partition is:
- for each new partition add a new destination filegroup, and
- split the partition range starting at highest value
This could look like:
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (4)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (3)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (2)
Now, if you check again your value distribution, you'll see that all distinct values end up in a separate partition:
select part_key, $partition.pf(part_key) as partition from T
But, be aware that this goes along with data movement, i.e. all rows with partition key values 1-4 need to be physically moved from original partition 2 to their new destination partition. So, if there are millions of such rows, this'll take some time and will blow up your transaction log.
I guess your situation is the following: you have a right range partition with the lowest boundary value being 5, and a table that is being partitioned that way, e.g.:
create partition function pf (int) as range right for values (5)
create partition scheme ps as partition pf to ([PRIMARY], [PRIMARY])
create table T (part_key int constraint PK_T primary key)
on ps (part_key)
Now, if you enter the values 1, 2, 3, 4, and 5 into table T and check the distribution of values within each partition, you'll find 1, 2, 3 and 4 in partition 1, and 5 in partition 2:
insert T values (1), (2), (3), (4), (5)
select part_key, $partition.pf(part_key) as partition from T
What you need to do to get each value in its own partition is:
- for each new partition add a new destination filegroup, and
- split the partition range starting at highest value
This could look like:
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (4)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (3)
alter partition scheme ps next used [PRIMARY]
alter partition function pf() split range (2)
Now, if you check again your value distribution, you'll see that all distinct values end up in a separate partition:
select part_key, $partition.pf(part_key) as partition from T
But, be aware that this goes along with data movement, i.e. all rows with partition key values 1-4 need to be physically moved from original partition 2 to their new destination partition. So, if there are millions of such rows, this'll take some time and will blow up your transaction log.
answered Mar 22 at 10:23
hanspohanspo
16
16
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
add a comment |
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
This is perfectly clear. Just one question, how do i choose the next used filegroup. (my partitions are day based and the filegroups month based)
– Bonzay
Mar 22 at 10:42
add a comment |
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