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titledescriptionauthorms.authorms.reviewerms.datems.servicems.subservicems.topicf1_keywordshelpviewer_keywordsdev_langsmonikerRange
COUNT (Transact-SQL)
COUNT (Transact-SQL)
markingmyname
maghan
randolphwest
11/21/2022
sql
t-sql
reference
COUNT_TSQL
COUNT
totals [SQL Server], COUNT function
totals [SQL Server]
counting items in group
groups [SQL Server], number of items in
number of group items
COUNT function [Transact-SQL]
TSQL
>= aps-pdw-2016 || = azuresqldb-current || = azure-sqldw-latest || >= sql-server-2016 || >= sql-server-linux-2017 || = azuresqldb-mi-current||=fabric

COUNT (Transact-SQL)

[!INCLUDE sql-asdb-asdbmi-asa-pdw-fabricse-fabricdw]

This function returns the number of items found in a group. COUNT operates like the COUNT_BIG function. These functions differ only in the data types of their return values. COUNT always returns an int data type value. COUNT_BIG always returns a bigint data type value.

:::image type="icon" source="../../includes/media/topic-link-icon.svg" border="false"::: Transact-SQL syntax conventions

Syntax

Aggregation function syntax

COUNT ( { [ [ ALL | DISTINCT ] expression ] | * } )

Analytic function syntax

COUNT ( [ ALL ] { expression | * } ) OVER ( [ <partition_by_clause> ] )

Arguments

ALL

Applies the aggregate function to all values. ALL serves as the default.

DISTINCT

Specifies that COUNT returns the number of unique nonnull values.

expression

An expression of any type, except image, ntext, or text. COUNT doesn't support aggregate functions or subqueries in an expression.

*

Specifies that COUNT should count all rows to determine the total table row count to return. COUNT(*) takes no parameters and doesn't support the use of DISTINCT. COUNT(*) doesn't require an expression parameter because by definition, it doesn't use information about any particular column. COUNT(*) returns the number of rows in a specified table, and it preserves duplicate rows. It counts each row separately. This includes rows that contain null values.

OVER ( [ partition_by_clause ] [ order_by_clause ] [ ROW_or_RANGE_clause ] )

The partition_by_clause divides the result set produced by the FROM clause into partitions to which the COUNT function is applied. If not specified, the function treats all rows of the query result set as a single group. The order_by_clause determines the logical order of the operation. See OVER clause (Transact-SQL) for more information.

Return types

  • int NOT NULL when ANSI_WARNINGS is ON, however SQL Server will always treat COUNT expressions as int NULL in metadata, unless wrapped in ISNULL.

  • int NULL when ANSI_WARNINGS is OFF.

Remarks

  • COUNT(*) without GROUP BY returns the cardinality (number of rows) in the resultset. This includes rows comprised of all-NULL values and duplicates.
  • COUNT(*) with GROUP BY returns the number of rows in each group. This includes NULL values and duplicates.
  • COUNT(ALL <expression>) evaluates expression for each row in a group, and returns the number of nonnull values.
  • COUNT(DISTINCT *expression*) evaluates expression for each row in a group, and returns the number of unique, nonnull values.

COUNT is a deterministic function when used without the OVER and ORDER BY clauses. It is nondeterministic when used with the OVER and ORDER BY clauses. For more information, see Deterministic and nondeterministic functions.

ARITHABORT and ANSI_WARNINGS

When COUNT has a return value exceeding the maximum value of int (that is, 231-1 or 2,147,483,647), the COUNT function will fail due to an integer overflow. When COUNT overflows and both the ARITHABORT and ANSI_WARNINGS options are OFF, COUNT will return NULL. Otherwise, when either of ARITHABORT or ANSI_WARNINGS are ON, the query will abort and the arithmetic overflow error Msg 8115, Level 16, State 2; Arithmetic overflow error converting expression to data type int. will be raised. To correctly handle these large results, use COUNT_BIG instead, which returns bigint.

When both ARITHABORT and ANSI_WARNINGS are ON, you can safely wrap COUNT call-sites in ISNULL( <count-expr>, 0 ) to coerce the expression's type to int NOT NULL instead of int NULL. Wrapping COUNT in ISNULL means any overflow error will be silently suppressed, which should be considered for correctness.

Examples

A. Use COUNT and DISTINCT

This example returns the number of different titles that an [!INCLUDEssSampleDBCoFull] employee can hold.

SELECTCOUNT(DISTINCT Title) FROMHumanResources.Employee; GO

[!INCLUDEssResult]

----------- 67 (1 row(s) affected)

B. Use COUNT(*)

This example returns the total number of [!INCLUDEssSampleDBCoFull] employees.

SELECTCOUNT(*) FROMHumanResources.Employee; GO

[!INCLUDEssResult]

----------- 290 (1 row(s) affected)

C. Use COUNT(*) with other aggregates

This example shows that COUNT(*) works with other aggregate functions in the SELECT list. The example uses the [!INCLUDEssSampleDBnormal] database.

SELECTCOUNT(*), AVG(Bonus) FROMSales.SalesPersonWHERE SalesQuota >25000; GO

[!INCLUDEssResult]

----------- --------------------- 14 3472.1428 (1 row(s) affected)

D. Use the OVER clause

This example uses the MIN, MAX, AVG and COUNT functions with the OVER clause, to return aggregated values for each department in the [!INCLUDEssSampleDBnormal] database HumanResources.Department table.

SELECT DISTINCT Name , MIN(Rate) OVER (PARTITION BY edh.DepartmentID) AS MinSalary , MAX(Rate) OVER (PARTITION BY edh.DepartmentID) AS MaxSalary , AVG(Rate) OVER (PARTITION BY edh.DepartmentID) AS AvgSalary , COUNT(edh.BusinessEntityID) OVER (PARTITION BY edh.DepartmentID) AS EmployeesPerDept FROMHumanResources.EmployeePayHistoryAS eph JOINHumanResources.EmployeeDepartmentHistoryAS edh ONeph.BusinessEntityID=edh.BusinessEntityIDJOINHumanResources.DepartmentAS d ONd.DepartmentID=edh.DepartmentIDWHEREedh.EndDate IS NULLORDER BY Name;

[!INCLUDEssResult]

Name MinSalary MaxSalary AvgSalary EmployeesPerDept ----------------------------- --------------------- --------------------- --------------------- ---------------- Document Control 10.25 17.7885 14.3884 5 Engineering 32.6923 63.4615 40.1442 6 Executive 39.06 125.50 68.3034 4 Facilities and Maintenance 9.25 24.0385 13.0316 7 Finance 13.4615 43.2692 23.935 10 Human Resources 13.9423 27.1394 18.0248 6 Information Services 27.4038 50.4808 34.1586 10 Marketing 13.4615 37.50 18.4318 11 Production 6.50 84.1346 13.5537 195 Production Control 8.62 24.5192 16.7746 8 Purchasing 9.86 30.00 18.0202 14 Quality Assurance 10.5769 28.8462 15.4647 6 Research and Development 40.8654 50.4808 43.6731 4 Sales 23.0769 72.1154 29.9719 18 Shipping and Receiving 9.00 19.2308 10.8718 6 Tool Design 8.62 29.8462 23.5054 6 (16 row(s) affected)

Examples: [!INCLUDEssazuresynapse-md] and [!INCLUDEssPDW]

E. Use COUNT and DISTINCT

This example returns the number of different titles that an employee of a specific company can hold.

USE ssawPDW; SELECTCOUNT(DISTINCT Title) FROMdbo.DimEmployee;

[!INCLUDEssResult]

----------- 67

F. Use COUNT(*)

This example returns the total number of rows in the dbo.DimEmployee table.

USE ssawPDW; SELECTCOUNT(*) FROMdbo.DimEmployee;

[!INCLUDEssResult]

------------- 296

G. Use COUNT(*) with other aggregates

This example combines COUNT(*) with other aggregate functions in the SELECT list. It returns the number of sales representatives with an annual sales quota greater than $500,000, and the average sales quota of those sales representatives.

USE ssawPDW; SELECTCOUNT(EmployeeKey) AS TotalCount, AVG(SalesAmountQuota) AS [Average Sales Quota] FROMdbo.FactSalesQuotaWHERE SalesAmountQuota >500000AND CalendarYear =2001;

[!INCLUDEssResult]

TotalCount Average Sales Quota ---------- ------------------- 10 683800.0000

H. Use COUNT with HAVING

This example uses COUNT with the HAVING clause to return the departments of a company, each of which has more than 15 employees.

USE ssawPDW; SELECT DepartmentName, COUNT(EmployeeKey)AS EmployeesInDept FROMdbo.DimEmployeeGROUP BY DepartmentName HAVINGCOUNT(EmployeeKey) >15;

[!INCLUDEssResult]

DepartmentName EmployeesInDept -------------- --------------- Sales 18 Production 179

I. Use COUNT with OVER

This example uses COUNT with the OVER clause, to return the number of products contained in each of the specified sales orders.

USE ssawPDW; SELECT DISTINCTCOUNT(ProductKey) OVER(PARTITION BY SalesOrderNumber) AS ProductCount , SalesOrderNumber FROMdbo.FactInternetSalesWHERE SalesOrderNumber IN (N'SO53115',N'SO55981');

[!INCLUDEssResult]

ProductCount SalesOrderID ------------ ----------------- 3 SO53115 1 SO55981

See also

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