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$PARTITION (Transact-SQL)

  • Cikk

Applies to: SQL Server Azure SQL Database Azure SQL Managed Instance

Returns the partition number into which a set of partitioning column values would be mapped for any specified partition function.

Transact-SQL syntax conventions

Syntax

[ database_name. ] $PARTITION.partition_function_name(expression)

Arguments

database_name
Is the name of the database that contains the partition function.

partition_function_name
Is the name of any existing partition function against which a set of partitioning column values are being applied.

expression
Is an expression whose data type must either match or be implicitly convertible to the data type of its corresponding partitioning column. expression can also be the name of a partitioning column that currently participates in partition_function_name.

Return types

int

Remarks

$PARTITION returns an int value between 1 and the number of partitions of the partition function.

$PARTITION returns the partition number for any valid value, regardless of whether the value currently exists in a partitioned table or index that uses the partition function.

Examples

A. Get the partition number for a set of partitioning column values

This example creates a partition function RangePF1 using RANGE LEFT that will partition a table or index into four partitions. $PARTITION is used to determine that the value 10, representing the partitioning column of RangePF1, would be put in partition 1 of the table.

CREATE PARTITION FUNCTION RangePF1 ( INT )  
AS RANGE LEFT FOR VALUES (10, 100, 1000) ;  
GO

SELECT $PARTITION.RangePF1 (10) ;  
GO

B. Get the number of rows in each nonempty partition of a partitioned table or index

This example shows how to use $PARTITION to return the number of rows in each partition of table that contains data.

The example:

  • Creates a partition scheme, RangePS1, for the partition function RangePF1.
  • Creates a table, dbo.PartitionTable, on the RangePS1 partition scheme with col1 as the partitioning column.
  • Inserts four rows into the dbo.PartitionTable table. Based on the partition function definition, these rows will be inserted into partitions 2 and 3. Partitions 1 and 4 will remain empty.
  • Queries the dbo.PartitionTable and uses $PARTITION.RangePF1(col1) in the GROUP BY clause to query the number of rows in each partition that contains data.

Note

To execute this example, you must first create the partition function RangePF1 using the code in the previous example.

CREATE PARTITION SCHEME RangePS1  
    AS PARTITION RangePF1  
    ALL TO ('PRIMARY') ;  
GO  

CREATE TABLE dbo.PartitionTable (col1 int PRIMARY KEY, col2 char(10))  
    ON RangePS1 (col1) ;  
GO

INSERT dbo.PartitionTable (col1, col2)
VALUES ((1,'a row'),(100,'another row'),(500,'another row'),(1000,'another row'))


SELECT 
	$PARTITION.RangePF1(col1) AS Partition,   
	COUNT(*) AS [COUNT] 
FROM dbo.PartitionTable
GROUP BY $PARTITION.RangePF1(col1)  
ORDER BY Partition ;  
GO

The SELECT query should return the following results:

Partition COUNT
2 1
3 3

Rows are not returned for partitions number 1 and 4, which exist but do not contain data.

C. Return all rows from one partition of a partitioned table or index

The following example returns all rows that are in partition 3 of the table PartitionTable.

SELECT col1, col2
FROM dbo.PartitionTable
WHERE $PARTITION.RangePF1(col1) = 3 ;

The query should return the following results:

col1 col2
101 another row
500 a third row
501 a fourth row

Next steps

Learn more about table partitioning in these articles: