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SUBSTRING (Transact-SQL)

Returns part of a character, binary, text, or image expression. For more information about the valid SQL Server data types that can be used with this function, see Data Types (Transact-SQL).

Topic link iconTransact-SQL Syntax Conventions


SUBSTRING (value_expression ,start_expression ,length_expression )

value_expression

Is a character, binary, text, ntext, or imageexpression.

start_expression

Is an integer or bigint expression that specifies where the returned characters start. If start_expression is less than 1, the returned expression will begin at the first character that is specifed in value_expression. In this case, the number of characters that are returned is the largest value of either the sum of start_expression and length_expression or 0. If start_expression is greater than the number of characters in the value expression, a zero-length expression is returned.

length_expression

Is a positive integer or bigint expression that specifies how many characters of the value_expression will be returned. If length_expression is negative, an error is generated and the statement is terminated. If the sum of start_expression and length_expression is greater than the number of characters in value_expression, the whole value expression beginning at start_expression is returned.

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.

Specified expression

Return type

char/varchar/text

varchar

nchar/nvarchar/ntext

nvarchar

binary/varbinary/image

varbinary

The values for start_expression and length_expression 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 value_expression must be varchar(max) or varbinary(max) when the start_expression or length_expression contains a value larger than 2147483647.

NoteNote

Compatibility levels can affect return values. For more information about compatibility levels, see sp_dbcmptlevel (Transact-SQL).

A. Using SUBSTRING with a character string

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

USE AdventureWorks;
GO
SELECT LastName, SUBSTRING(FirstName, 1, 1) AS Initial
FROM Person.Contact
WHERE LastName like 'Barl%'
ORDER BY LastName;

Here is the result set.

LastName Initial

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

Barley R

Barlow B

(2 row(s) affected)

Here is how to display the second, third, and fourth characters of the string constant abcdef.

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

Here is the result set.

x

----------

bcd

(1 row(s) affected)

B. Using SUBSTRING with text, ntext, and image data

NoteNote

To run the following examples, you must install the pubs database. For information about how to install the pubs database, see Downloading Northwind and pubs Sample Databases.

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;
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 is the result set.

pub_id logo pr_info

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

1756 0x474946383961E3002500 This is sa

(1 row(s) affected)

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
         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');
INSERT npub_info values('0877', N'üThis is sample text data for Binnet & Hardley, publisher 0877 in the pubs databa');
INSERT npub_info values('1389', N'üThis is sample text data for Algodata Infosystems, publisher 1389 in the pubs da');
INSERT npub_info values('9952', N'üThis is sample text data for Scootney Books, publisher 9952 in the pubs database');
INSERT npub_info values('1622', N'üThis is sample text data for Five Lakes Publishing, publisher 1622 in the pubs d');
INSERT npub_info values('1756', N'üThis is sample text data for Ramona Publishers, publisher 1756 in the pubs datab');
INSERT npub_info values('9901', N'üThis is sample text data for GGG&G, publisher 9901 in the pubs database. GGG&G i');
INSERT npub_info values('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 pr INNER JOIN npub_info npr
   ON pr.pub_id = npr.pub_id
ORDER BY pr.pub_id ASC;

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