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$PARTITION (Transact-SQL)

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

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

Transact-SQL syntax conventions

Syntax

[ database_name. ] $PARTITION.partition_function_name(expression)

Arguments

database_name

The name of the database that contains the partition function.

partition_function_name

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

expression

An expression whose data type must either match or be implicitly convertible to the data type of its corresponding partitioning column. This parameter 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.

Note

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

  1. Create a partition scheme, RangePS1, for the partition function RangePF1.

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

  2. Create a table, dbo.PartitionTable, on the RangePS1 partition scheme with col1 as the partitioning column.

    CREATE TABLE dbo.PartitionTable
(
    col1 INT PRIMARY KEY,
    col2 CHAR (20)
) ON RangePS1 (col1);
GO

  3. Insert four rows into the dbo.PartitionTable table. These rows are inserted into partitions based on the partition function RangePF1 definition: 1 and 10 go to partition 1, while 500 and 1000 go to 3.

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

  4. Query 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.

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

Here's the result set.

Partition COUNT
1 2
3 2

Rows aren't returned for partition number 2, which exists but doesn't 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;

Here's the result set.

col1 col2
500 another row
1000 another row

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