SUBSTRING (Transact-SQL)

Applies to: SQL Server Azure SQL Database Azure SQL Managed Instance Azure Synapse Analytics Analytics Platform System (PDW) SQL analytics endpoint in Microsoft Fabric Warehouse in Microsoft Fabric

Returns part of a character, binary, text, or image expression in SQL Server.

Transact-SQL syntax conventions

Syntax for SQL Server.

SUBSTRING ( expression , start , length )

Syntax for Azure SQL Database, Azure SQL Managed Instance, Azure Synapse Analytics, Analytics Platform System (PDW), and Warehouse and SQL analytics endpoint in Microsoft Fabric.

SUBSTRING ( expression , start [ , length ] )

A character, binary, text, ntext, or image expression.

An integer or bigint expression that specifies where the returned characters start. (The numbering is 1 based, meaning that the first character in the expression is 1). If start is less than 1, the returned expression begins at the first character that is specified in expression. In this case, the number of characters that are returned is the largest value of either the sum of start + length - 1, or 0. If start is greater than the number of characters in the value expression, a zero-length expression is returned.

A positive integer or bigint expression that specifies how many characters of the expression are returned. If length is negative, an error is generated and the statement is terminated. If the sum of start and length is greater than the number of characters in expression, the whole value expression beginning at start is returned. If length is omitted, all characters from the start position to the end of the expression is returned.

You can use substring with an optional length argument in Azure SQL Database, Azure SQL Managed Instance, Azure Synapse Analytics, Analytics Platform System (PDW), and Warehouse and SQL analytics endpoint in Microsoft Fabric. However, if you use NULL for length, SUBSTRING returns NULL. Review E. Use SUBSTRING with optional length argument for an example.

Returns character data if expression is one of the supported character data types. Returns binary data if expression is one of the supported binary data types. The returned string is the same type as the specified expression with the exceptions shown in the table.

The values for start and length must be specified in number of characters for ntext, char, or varchar data types and bytes for text, image, binary, or varbinary data types.

The expression must be varchar(max) or varbinary(max) when the start or length contains a value larger than 2,147,483,647.

When you use supplementary character (SC) collations, both start and length count each surrogate pair in expression as a single character. For more information, see Collation and Unicode support.

The following example shows how to return only a part of a character string. From the sys.databases table, this query returns the system database names in the first column, the first letter of the database in the second column, and the third and fourth characters in the final column.

SELECT name,
       SUBSTRING(name, 1, 1) AS Initial,
       SUBSTRING(name, 3, 2) AS ThirdAndFourthCharacters
FROM sys.databases
WHERE database_id < 5;

Here's the result set.

To display the second, third, and fourth characters of the string constant abcdef, use the following query.

SELECT SUBSTRING('abcdef', 2, 3) AS x;

Here's the result set.

x
----------
bcd

The following example shows how to return the first 10 characters from each of a text and image data column in the pub_info table of the pubs database. text data is returned as varchar, and image data is returned as varbinary.

USE pubs;
GO

SELECT pub_id,
       SUBSTRING(logo, 1, 10) AS logo,
       SUBSTRING(pr_info, 1, 10) AS pr_info
FROM pub_info
WHERE pub_id = '1756';

Here's the result set.

 pub_id logo    pr_info
------ ---------------------- ----------
1756   0x474946383961E3002500 This is sa

The following example shows the effect of SUBSTRING on both text and ntext data. First, this example creates a new table in the pubs database named npub_info. Second, the example creates the pr_info column in the npub_info table from the first 80 characters of the pub_info.pr_info column and adds an ü as the first character. Lastly, an INNER JOIN retrieves all publisher identification numbers and the SUBSTRING of both the text and ntext publisher information columns.

IF EXISTS (SELECT table_name
           FROM INFORMATION_SCHEMA.TABLES
           WHERE table_name = 'npub_info')
    DROP TABLE npub_info;
GO

-- Create npub_info table in pubs database. Borrowed from instpubs.sql.
USE pubs;
GO

CREATE TABLE npub_info
(
    pub_id CHAR (4) NOT NULL FOREIGN KEY
        REFERENCES publishers (pub_id)
        CONSTRAINT UPKCL_npubinfo PRIMARY KEY CLUSTERED,
    pr_info NTEXT NULL
);
GO

-- Fill the pr_info column in npub_info with international data.
RAISERROR ('Now at the inserts to pub_info...', 0, 1);
GO

INSERT npub_info
VALUES ('0736', N'üThis is sample text data for New Moon Books, publisher 0736 in the pubs database'),
    ('0877', N'üThis is sample text data for Binnet & Hardley, publisher 0877 in the pubs databa'),
    ('1389', N'üThis is sample text data for Algodata Infosystems, publisher 1389 in the pubs da'),
    ('9952', N'üThis is sample text data for Scootney Books, publisher 9952 in the pubs database'),
    ('1622', N'üThis is sample text data for Five Lakes Publishing, publisher 1622 in the pubs d'),
    ('1756', N'üThis is sample text data for Ramona Publishers, publisher 1756 in the pubs datab'),
    ('9901', N'üThis is sample text data for GGG&G, publisher 9901 in the pubs database. GGG&G i'),
    ('9999', N'üThis is sample text data for Lucerne Publishing, publisher 9999 in the pubs data');
GO

-- Join between npub_info and pub_info on pub_id.
SELECT pr.pub_id,
       SUBSTRING(pr.pr_info, 1, 35) AS pr_info,
       SUBSTRING(npr.pr_info, 1, 35) AS npr_info
FROM pub_info AS pr
     INNER JOIN npub_info AS npr
         ON pr.pub_id = npr.pub_id
ORDER BY pr.pub_id ASC;

The following example shows how to return only a part of a character string. From the dbo.DimEmployee table, this query returns the family name in one column with only the first initial in the second column.

-- Uses AdventureWorks
SELECT LastName,
       SUBSTRING(FirstName, 1, 1) AS Initial
FROM dbo.DimEmployee
WHERE LastName LIKE 'Bar%'
ORDER BY LastName;

Here's the result set.

LastName             Initial
-------------------- -------
Barbariol            A
Barber               D
Barreto de Mattos    P

The following example shows how to return the second, third, and fourth characters of the string constant abcdef.

USE ssawPDW;

SELECT TOP 1 SUBSTRING('abcdef', 2, 3) AS x
FROM dbo.DimCustomer;

Here's the result set.

x
-----
bcd

SELECT SUBSTRING('123abc', 4, NULL) AS [NULL length];

Here's the result set.

NULL length
-----------
NULL

Applies to: Azure SQL Database, Azure SQL Managed Instance, Azure Synapse Analytics, Analytics Platform System (PDW), and Warehouse and SQL analytics endpoint in Microsoft Fabric

The following example shows how to return only a part of a character string from a given start position. Since the length argument isn't provided, the length defaults to return the remaining characters in the string.

SELECT SUBSTRING('123abc', 4) AS y;

Here's the result set.

y
-----
abc

USE AdventureWorks2022;
GO

SELECT [ProductDescriptionID],
       [Description],
       SUBSTRING([Description], LEN('Replacement') + 1) AS [Replacement-Part]
FROM [Production].[ProductDescription]
WHERE [Description] LIKE 'Replacement%';

Here's the result set.

• LEFT (Transact-SQL)
• LTRIM (Transact-SQL)
• RIGHT (Transact-SQL)
• RTRIM (Transact-SQL)
• STRING_SPLIT (Transact-SQL)
• TRIM (Transact-SQL)
• String Functions (Transact-SQL)