SUBSTRING

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

Syntax

SUBSTRING ( expression , start , length )

Arguments

expression

Is a character string, binary string, text, image, a column, or an expression that includes a column. Do not use expressions that include aggregate functions.

start

Is an integer that specifies where the substring begins.

length

Is an integer that specifies the length of the substring (the number of characters or bytes to return).

Note  Because start and length specify the number of bytes when SUBSTRING is used on text data, DBCS data, such as Kanji, may result in split characters at the beginning or end of the result. This behavior is consistent with the way in which READTEXT handles DBCS. However, because of the occasional strange result, it is advisable to use ntext instead of text for DBCS characters.

Return Types

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 given expression with the exceptions shown in the table.

Remarks

Offsets (start and length) using the ntext, char, or varchar data types must be specified in number of characters. Offsets using the text, image, binary, or varbinary data types must be specified in number of bytes.

Note  Compatibility levels can affect return values. For more information about compatibility levels, see sp_dbcmptlevel.

Examples

A. Use SUBSTRING with a character string

This example shows how to return only a portion of a character string. From the authors table, this query returns the last name in one column with only the first initial in the second column.

USE pubs
SELECT au_lname, SUBSTRING(au_fname, 1, 1)
FROM authors
ORDER BY au_lname

Here is the result set:

au_lname                                   
---------------------------------------- - 
Bennet                                   A 
Blotchet-Halls                           R 
Carson                                   C 
DeFrance                                 M 
del Castillo                             I 
...
Yokomoto                                 A 

(23 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. Use SUBSTRING with text, ntext, and image data

This example shows how to return the first 200 characters from each of a text and image data column in the publishers 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)

This example shows the effect of SUBSTRING on both text and ntext data. First, this example creates a new table in the pubs database named npr_info. Second, the example creates the pr_info column in the npr_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

See Also

String Functions