LEN (Transact-SQL)

**APPLIES TO:** ![yes](media/yes.png)SQL Server (starting with 2008) ![yes](media/yes.png)Azure SQL Database ![yes](media/yes.png)Azure SQL Data Warehouse ![yes](media/yes.png)Parallel Data Warehouse

Returns the number of characters of the specified string expression, excluding trailing blanks.

[!NOTE]
To return the number of bytes used to represent an expression, use the DATALENGTH function.

Topic link icon Transact-SQL Syntax Conventions

Syntax

LEN ( string_expression )

Arguments

string_expression
Is the string expression to be evaluated. string_expression can be a constant, variable, or column of either character or binary data.

Return Types

bigint if expression is of the varchar(max), nvarchar(max) or varbinary(max) data types; otherwise, int.

If you are using SC collations, the returned integer value counts UTF-16 surrogate pairs as a single character. For more information, see Collation and Unicode Support.

Remarks

LEN excludes trailing blanks. If that is a problem, consider using the DATALENGTH (Transact-SQL) function which does not trim the string. If processing a unicode string, DATALENGTH will return twice the number of characters. The following example demonstrates LEN and DATALENGTH with a trailing space.

DECLARE @v1 varchar(40),  
    @v2 nvarchar(40);  
SELECT   
@v1 = 'Test of 22 characters ',   
@v2 = 'Test of 22 characters ';  
SELECT LEN(@v1) AS [varchar LEN] , DATALENGTH(@v1) AS [varchar DATALENGTH];  
SELECT LEN(@v2) AS [nvarchar LEN], DATALENGTH(@v2) AS [nvarchar DATALENGTH];

Examples

The following example selects the number of characters and the data in FirstName for people located in Australia. This example uses the AdventureWorks database.

SELECT LEN(FirstName) AS Length, FirstName, LastName   
FROM Sales.vIndividualCustomer  
WHERE CountryRegionName = 'Australia';  
GO

Examples: [!INCLUDEssSDWfull] and [!INCLUDEssPDW]

The following example returns the number of characters in the column FirstName and the first and last names of employees located in Australia.

-- Uses AdventureWorks  
  
SELECT DISTINCT LEN(FirstName) AS FNameLength, FirstName, LastName   
FROM dbo.DimEmployee AS e  
INNER JOIN dbo.DimGeography AS g   
    ON e.SalesTerritoryKey = g.SalesTerritoryKey   
WHERE EnglishCountryRegionName = 'Australia';

Here is the result set.

FNameLength  FirstName  LastName  
-----------  ---------  ---------------  
4            Lynn       Tsoflias