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title description author ms.author ms.reviewer ms.date ms.service ms.subservice ms.topic f1_keywords helpviewer_keywords dev_langs monikerRange
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> ] )

[!INCLUDEsql-server-tsql-previous-offline-documentation]

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.

SELECT COUNT(DISTINCT Title)
FROM HumanResources.Employee;
GO

[!INCLUDEssResult]

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

B. Use COUNT(*)

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

SELECT COUNT(*)
FROM HumanResources.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.

SELECT COUNT(*), AVG(Bonus)
FROM Sales.SalesPerson
WHERE 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
FROM HumanResources.EmployeePayHistory AS eph
JOIN HumanResources.EmployeeDepartmentHistory AS edh
    ON eph.BusinessEntityID = edh.BusinessEntityID
JOIN HumanResources.Department AS d
ON d.DepartmentID = edh.DepartmentID
WHERE edh.EndDate IS NULL
ORDER 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;
  
SELECT COUNT(DISTINCT Title)
FROM dbo.DimEmployee;

[!INCLUDEssResult]

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

F. Use COUNT(*)

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

USE ssawPDW;
  
SELECT COUNT(*)
FROM dbo.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;
  
SELECT COUNT(EmployeeKey) AS TotalCount, AVG(SalesAmountQuota) AS [Average Sales Quota]
FROM dbo.FactSalesQuota
WHERE SalesAmountQuota > 500000 AND 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
FROM dbo.DimEmployee
GROUP BY DepartmentName
HAVING COUNT(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 DISTINCT COUNT(ProductKey) OVER(PARTITION BY SalesOrderNumber) AS ProductCount
    , SalesOrderNumber
FROM dbo.FactInternetSales
WHERE SalesOrderNumber IN (N'SO53115',N'SO55981');

[!INCLUDEssResult]

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

See also