One common task performed in database applications is given a full name, how can
this be split into the first name and last name. In this article, three different methods are discussed to perform this task using three different functions in SQL Server.
It is assumed in this article that the full name is just made up of one first name and one last name.
Using SUBSTRING Function
The first method is the use of the SUBSTRING string function, as can be seen
from the following script:
DECLARE @FullName VARCHAR(100) SET @FullName = 'John Doe'
SELECT SUBSTRING(@FullName, 1, CHARINDEX(' ', @FullName) - 1)
AS [FirstName],
SUBSTRING(@FullName, CHARINDEX(' ', @FullName) + 1, LEN(@FullName))
AS [LastName]
Given the full name of "John Doe", the script will generate the following output:
FirstName LastName ------------ ---------------- John Doe
Aside from the SUBSTRING string function, it also used the CHARINDEX string
function. The CHARINDEX function returns the starting position of a specified
expression in a character string. The CHARINDEX function was used to look for
the space that separates the first name from the last name (CHARINDEX(' ', @FullName)).
If the full name only contains the last name, the following error message
will be encountered:
Server: Msg 536, Level 16, State 3, Line 4 Invalid length paramete
r passed to the substring function.
To avoid this error message, the script above needs to be modified to check
if there's a space in the full name. If there's no space in the full name, it is
assumed that it only contains the last name. The following script overcomes
this problem:
DECLARE @FullName VARCHAR(100) SET @FullName = 'JohnDoe'
SELECT SUBSTRING(@FullName, 1,
NULLIF(CHARINDEX(' ', @FullName) - 1, -1))
AS [FirstName],
SUBSTRING(@FullName, CHARINDEX(' ', @FullName) + 1,
LEN(@FullName)) AS [LastName]
The only difference between this script and the previous one is the addition of the
NULLIF function. The NULLIF function returns a null value if the two specified
expressions are equivalent. In the CHARINDEX function, if the space is not
found in the full name, it returns a value of 0. Then subtracting 1 from this gives
a value of -1. Using the NULLIF function, we are changing the length parameter
passed to the SUBSTRING function to NULL if the value returned by the
CHARINDEX minus 1 is -1. In the SUBSTRING function if the length parameter
is NULL, the string returned is NULL as well. Therefore, if there's no space in the
full name, the first name will have a value of NULL, as shown in the following result:
FirstName LastName ------------ ---------------- NULL JohnDoe
Using LEFT and RIGHT Functions
The second method of getting the first name and last name from a full name is
using the LEFT and RIGHT functions. The LEFT and RIGHT functions are basically
the same as the SUBSTRING function. The LEFT function returns the part of a character
string starting at a specified number of characters from the left while the RIGHT
function returns the part of a character string starting a specified number of characters
from the right.
The script to split the full name string into the corresponding first name and last name using the LEFT and RIGHT string functions is as follows:
DECLARE @FullName VARCHAR(100) SET @FullName = 'John Doe'
SELECT LEFT(@FullName, CHARINDEX(' ', @FullName) - 1) AS [FirstName],
RIGHT(@FullName, CHARINDEX(' ', REVERSE(@FullName)) - 1) AS [LastName]
In getting the first name, it is basically the same as the first version using the
SUBSTRING function. The only difference is that the LEFT function doesn't need the
starting position and it automatically starts at the first position.
In getting the last name, the RIGHT string function is used. The second parameter of the RIGHT function is the number of characters to extract from the full name starting from the right. To get the number of characters to extract, the CHARINDEX is again used. The only difference between the CHARINDEX used in the LEFT string function and RIGHT string function is the addition of the use of the REVERSE string function. Since we want to extract all characters starting from the space until the end of the string, we have to REVERSE the string and count the number of characters from there (CHARINDEX(' ', REVERSE(@FullName)).
To avoid the same error message discussed on the first version when there's no space in the full name, the following script can be used:
DECLARE @FullName VARCHAR(100) SET @FullName = 'JohnDoe'
SELECT LEFT(@FullName, NULLIF(CHARINDEX(' ', @FullName) - 1, -1)) AS [FirstName],
RIGHT(@FullName, ISNULL(NULLIF(CHARINDEX(' ', REVERSE(@FullName)) - 1, -1),
LEN(@FullName))) AS [LastName]
It uses the same logic used in the first version which is the use of the NULLIF function to check if there's a space in the full name and return a NULL value for the first name if there's no space. As for the last name, it uses the NULLIF and ISNULL functions to return the full name as the last name if there's no space in it. This is done by passing the length of the full name as the second parameter to the RIGHT string function if there's no space in the full name.
Using PARSENAME Function
The last method in splitting a full name into its corresponding first name and last name is the use of the PARSENAME string function, as can be seen from the following script:
DECLARE @FullName VARCHAR(100) SET @FullName = 'John Doe'
SELECT PARSENAME(REPLACE(@FullName, ' ', '.'), 2) AS [FirstName],
PARSENAME(REPLACE(@FullName, ' ', '.'), 1) AS [LastName]
The PARSENAME string function returns the specified part of an object name. Parts of an object that can be retrieved are the object name, owner name, database name, and server name. In using the PARSENAME string function, we are tricking SQL Server into thinking that the value being passed to the first parameter, which is the object name, is an object within SQL Server since the function does not validate whether or not an object by the specified name exists.
The name of an object in SQL Server is composed of the following parts, separated by a period (.):
[Server Name].[Database Name].[Owner Name].[Object Name]
The second parameter of the PARSENAME function relates to each part of the
object name, each of which having the following values:
- Object Name - 1
- Owner Name - 2
- Database Name - 3
- Server Name - 4
In using the PARSENAME function to split the full name, we need to replace the space with a period (REPLACE(@FullName, ' ', '.')). Then to get the first name, we pass a value
of 2 to the PARSENAME string function, which corresponds to the [Owner Name]. To get the last name, we pass a value of 1 to the function, which corresponds to the [Object Name]. Even if there's no space separating the first name and last name, no error message will be encountered and a NULL value will be returned for the first name while the whole full name will be returned as the last name.
