Split name into multiple parts in SELECT statement - sql

Question

I cannot seem to find an existing post on splitting a string into the parts I require. I have a database field in SQL Server that contains the "LastName FirstName MI" (no commas just spaces delimiting each part of a person's name). I have the following SQL to get the FirstName and Last, but cannot figure out how to get the Middle Initial or Middle Name.
Ex. Doe John B
SELECT
RTRIM(LEFT([PATIENT_NAME], CHARINDEX(' ', [PATIENT_NAME]))) AS LastName,
SUBSTRING([PATIENT_NAME], CHARINDEX(' ', [PATIENT_NAME]) + 1, LEN([PATIENT_NAME])) AS FirstName
FROM
Clients
Results in:
FirstName = John B
LastName = Doe
How to just return the first name without the middle initial and get the 'B' as middle name from this string in this SELECT statement?

Answer 1

You can either take the right 1 character, or reverse the string the take the first char.
SELECT RIGHT(LTRIM(RTRIM([Patient_Name])), 1) AS Middle_Initial
SELECT LEFT(REVERSE(LTRIM(RTRIM([Patient_Name]))), 1) AS Middle_Initial
As for removing MI from your firstname string, I would either find the length of the string and take the left N-2 chars or I would charindex the space and then take that many chars. To put it all together:
DECLARE #name VARCHAR(100) = 'Smith David M '
--Clean the string of leading/trailing whitespace
SELECT LTRIM(RTRIM(#name)) AS name_cleaned
--Find the first space to parse out the last name
SELECT CHARINDEX(' ', #name) AS first_space
--Select all chars before the first space
SELECT LEFT(LTRIM(RTRIM(#name)), CHARINDEX(' ', #name)-1) AS last_name
--Find the next space, use the starting location as the previous space and add 1
SELECT CHARINDEX(' ', #name, 7) AS second_space
--Select all chars between the spaces
SELECT SUBSTRING(#name, CHARINDEX(' ', #name)+1, CHARINDEX(' ', #name, 7) - CHARINDEX(' ', #name)) AS first_name
--Select the right most char for middle initial
SELECT RIGHT(LTRIM(RTRIM(#name)), 1) AS middle

Answer 2

You can REPLACE the space characters with period characters (.) and use PARSENAME().
Note that this would work for all 3 parts of the name, not just the middle initial.

Answer 3

When using the CHARINDEX on the last name, you'll use it as the length of the substring. Then, on the FirstName, use it again as start position on the substring. Now, the trick on the Middle, on the CHARINDEX, you have to include the start position which will be the LEN minus the LastName CHARINDEX. this would gives you the second space which is the position you want to start with for taking the Middle Name.
See the example below :
DECLARE #tb TABLE (PATIENT_NAME varchar(250));
INSERT INTO #tb VALUES
('Doe John B')
DECLARE
#LastName INT
, #Middle INT
SELECT
#LastName = CHARINDEX(' ', PATIENT_NAME)
, #Middle = CHARINDEX(' ', PATIENT_NAME, LEN(PATIENT_NAME) - CHARINDEX(' ', PATIENT_NAME))
FROM #tb
SELECT
SUBSTRING(PATIENT_NAME, 1, #LastName) LastName
, SUBSTRING(PATIENT_NAME, #LastName, LEN(PATIENT_NAME) - #LastName) FirstName
, SUBSTRING(PATIENT_NAME, #Middle, LEN(PATIENT_NAME) - #Middle + 1 ) Middle
FROM #tb
I have declared some variables to make things much readable, but you can do it without them.
Surely, LEFT and RIGHT are the easier approaches on taking the lastname and Middle Name. Along with using some helper functions such as REVERSE and TRIM, but I would prefer PARSENAME as a simpler and cleaner approach.
Here is an example :
SELECT
PARSENAME(REPLACE(PATIENT_NAME,' ','.'),3) LastName
, PARSENAME(REPLACE(PATIENT_NAME,' ','.'),2) FirstName
, PARSENAME(REPLACE(PATIENT_NAME,' ','.'),1) Middle

Answer 4

Since the number of elements you must extract from your string is fixed(3) you can use XML based split:
DECLARE #clients TABLE (PATIENT_NAME nvarchar(max));
INSERT INTO #clients VALUES
(' Doe John B ')
,(' Doe Jane C ')
,(' Doe Jill ')
;WITH Splitted
AS (
SELECT PATIENT_NAME as ORIGINAL_PATIENT_NAME
,REPLACE(REPLACE(REPLACE(ltrim(rtrim(PATIENT_NAME)),' ','<>'),'><',''),'<>',' ') as PATIENT_NAME
,CAST('<x>' + REPLACE(REPLACE(REPLACE(REPLACE(ltrim(rtrim(PATIENT_NAME)),' ','<>'),'><',''),'<>',' '), ' ', '</x><x>') + '</x>' AS XML) AS Parts
FROM #clients
)
SELECT
ORIGINAL_PATIENT_NAME
,PATIENT_NAME
,Parts.value(N'/x[1]', 'nvarchar(max)') AS LAST_NAME
,Parts.value(N'/x[2]', 'nvarchar(max)') AS FIRST_NAME
,Parts.value(N'/x[3]', 'nvarchar(max)') AS MIDDLE_NAME
FROM Splitted
Results:
As you can see it works even with random-spaced names.