Replace string after a specific index

Question

Replace string after a specific index

sarath.tata

SSCommitted

Points: 1640
More actions
March 23, 2015 at 9:16 am

#305955

I have data like below
Potter, James J
Williams, Ted R
Allen, Gary G
I want to remove Middle Name from the output
Potter, James
Williams, Ted
Allen, Gary
My Query:
SELECT
CASE WHEN CHARINDEX(' ', Supervisor, CHARINDEX(' ', Supervisor, 0) + 1) > 0 THEN
REPLACE(Supervisor, SUBSTRING(Supervisor, CHARINDEX(' ', Supervisor, CHARINDEX(' ', Supervisor, 0) + 1), LEN(Supervisor)), '')
ELSE Supervisor END AS NewSupervisor from data d
However, I stumble when Middle Name exists somewhere in the name as Replace function repalces every occurrence of the string. For ex: "Allen, Gary G" becomes "Allen,ary"
Do we have any way to say sql to replace after certain index?

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sarath.tata SSCommitted Points: 1640 More actions · Answer 1

I think I may better use Left in stead of Replace

SELECT
CASE WHEN CHARINDEX(' ', Supervisor, CHARINDEX(' ', Supervisor, 0) + 1) > 0 THEN
LEFT(Supervisor, CHARINDEX(' ', Supervisor, CHARINDEX(' ', Supervisor, 0) + 1))
ELSE Supervisor END AS NewSupervisor
FROM data d

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 2

CREATE TABLE #Names (Fullname VARCHAR(50))

INSERT INTO #Names (Fullname)

SELECT 'Potter, James J' UNION ALL

SELECT 'Williams, Ted R' UNION ALL

SELECT 'Allen, Gary G' UNION ALL

SELECT 'Allen, Gary Graham' UNION ALL

SELECT 'Allen, Gary ' UNION ALL

SELECT 'Allen, Gary' UNION ALL

SELECT 'Allen, G'

SELECT

OldFullname = Fullname,

x.[Type],

NewFullname = CASE WHEN [Type] = 0 THEN Fullname ELSE LEFT(Fullname, LEN(Fullname)-2) END

FROM #Names

CROSS APPLY (SELECT [Type] = CASE WHEN Fullname LIKE '%, % [A-Z]' THEN 1 ELSE 0 END) x

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 3

sarath.tata (3/23/2015)
I think I may better use Left in stead of Replace
SELECT
CASE WHEN CHARINDEX(' ', Supervisor, CHARINDEX(' ', Supervisor, 0) + 1) > 0 THEN
LEFT(Supervisor, CHARINDEX(' ', Supervisor, CHARINDEX(' ', Supervisor, 0) + 1))
ELSE Supervisor END AS NewSupervisor
FROM data d

-- This too

SELECT

OldSupervisor = Supervisor,

x.[Type],

NewSupervisor = CASE

WHEN [Type] = 0 THEN Supervisor

ELSE LEFT(Supervisor, LEN(Supervisor)-CHARINDEX(' ',REVERSE(Supervisor))) END

FROM #data

CROSS APPLY (SELECT [Type] = CASE WHEN RTRIM(Supervisor) LIKE '%, % %' THEN 1 ELSE 0 END) x

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden