COUNT (Transact-SQL)

COUNT (Transact-SQL)

 

Updated: January 8, 2016

THIS TOPIC APPLIES TO:yesSQL Server (starting with 2008) yesAzure SQL Database yesAzure SQL Data Warehouse yesParallel Data Warehouse

Returns the number of items in a group. COUNT works like the COUNT_BIG function. The only difference between the two functions is their return values. COUNT always returns an int data type value. COUNT_BIG always returns a bigint data type value.

Topic link icon Transact-SQL Syntax Conventions


COUNT ( { [ [ ALL | DISTINCT ] expression ] | * } ) 
    [ OVER ( 
        [ partition_by_clause ] 
        [ order_by_clause ] 
        [ ROW_or_RANGE_clause ]
    ) ]
-- Azure SQL Data Warehouse and Parallel Data Warehouse
-- Aggregation Function Syntax
COUNT ( { [ [ ALL | DISTINCT ] expression ] | * } )
-- Azure SQL Data Warehouse and Parallel Data Warehouse
-- Analytic Function Syntax
COUNT ( { expression | * } ) OVER ( [ <partition_by_clause> ] )

ALL

Applies the aggregate function to all values. ALL is the default.

DISTINCT

Specifies that COUNT returns the number of unique nonnull values.

expression

Is an expression of any type except text, image, or ntext. Aggregate functions and subqueries are not permitted.

*

Specifies that all rows should be counted to return the total number of rows in a table. COUNT(*) takes no parameters and cannot be used with DISTINCT. COUNT(*) does not require an expression parameter because, by definition, it does not use information about any particular column. COUNT(*) returns the number of rows in a specified table without getting rid of duplicates. It counts each row separately. This includes rows that contain null values.

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

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

COUNT(*) returns the number of items in a 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.

For return values greater than 2^31-1, COUNT produces an error. Use COUNT_BIG instead.

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

The following example lists the number of different titles that an employee who works at Adventure Works Cycles can hold.

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

Here is the result set.

-----------

67

(1 row(s) affected)

The following example finds the total number of employees who work at Adventure Works Cycles.

SELECT COUNT(*)
FROM HumanResources.Employee;
GO

Here is the result set.

            

-----------

290

(1 row(s) affected)

The following example shows that COUNT(*) can be combined with other aggregate functions in the select list. The example uses the AdventureWorks2012 database.

SELECT COUNT(*), AVG(Bonus)
FROM Sales.SalesPerson
WHERE SalesQuota > 25000;
GO

Here is the result set.

                                 

----------- ---------------------

14 3472.1428

(1 row(s) affected)

The following example uses the MIN, MAX, AVG and COUNT functions with the OVER clause to provide aggregated values for each department in the HumanResources.Department table in the AdventureWorks2012 database.

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;

Here is the result set.

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)

The following example lists the number of different titles that an employee who works at a specific company can hold.

USE AdventureWorksPDW2012;

SELECT COUNT(DISTINCT Title)
FROM dbo.DimEmployee;

Here is the result set.

-----------

67

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

USE AdventureWorksPDW2012;

SELECT COUNT(*)
FROM dbo.DimEmployee;

Here is the result set.

-------------

296

The following example combines COUNT(*) with other aggregate functions in the SELECT list. The query returns the number of sales representatives with a annual sales quota greater than $500,000 and the average sales quota.

USE AdventureWorksPDW2012;

SELECT COUNT(EmployeeKey) AS TotalCount, AVG(SalesAmountQuota) AS [Average Sales Quota]
FROM dbo.FactSalesQuota
WHERE SalesAmountQuota > 500000 AND CalendarYear = 2001;

Here is the result set.

TotalCount  Average Sales Quota

----------  -------------------

10          683800.0000

The following example uses COUNT with the HAVING clause to return the departments in a company that have more than 15 employees.

USE AdventureWorksPDW2012;

SELECT DepartmentName, 
       COUNT(EmployeeKey)AS EmployeesInDept
FROM dbo.DimEmployee
GROUP BY DepartmentName
HAVING COUNT(EmployeeKey) > 15;

Here is the result set.

DepartmentName  EmployeesInDept

--------------  ---------------

Sales           18

Production      179

The following example uses COUNT with the OVER clause to return the number of products that are contained in each of the specified sales orders.

USE AdventureWorksPDW2012;

SELECT DISTINCT COUNT(ProductKey) OVER(PARTITION BY SalesOrderNumber) AS ProductCount
    ,SalesOrderNumber
FROM dbo.FactInternetSales
WHERE SalesOrderNumber IN (N'SO53115',N'SO55981');

Here is the result set.

ProductCount   SalesOrderID

------------   -----------------

3              SO53115

1              SO55981

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