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Fun SQL string replace challenge

Posted on 2011-03-24
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Last Modified: 2012-05-11
When I use this code

@stlocation= 3713 Yanceyville st Greensboro NC
 @UseST = NC
REPLACE(@stlocation, @UseST, '')

I want to get
3713 Yanceyville st Greensboro

However-- I get

3713 Yaeyville st Greensboro


It removes the NC not only from the back of the string but also from the STREET!!

Can someone tell me how to remove only from the back of the string? This same thing could happen if a street name is Virginia ST and the state entered is Virginia.. It would take virginia off of both the state and the street!!

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Question by:cheryl9063
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Expert Comment

by:ewangoya
ID: 35209785


select LEFT(@stlocation, DATALENGTH(@stlocation) - DATALENGTH(@UseST))
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Assisted Solution

by:sdstuber
sdstuber earned 50 total points
ID: 35209788
prepend a space before NC

@UseST = ' ' + 'NC'


even more robust would be to use regular expressions but you'll need udf for that
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Assisted Solution

by:bitref
bitref earned 100 total points
ID: 35209896
@UseST = ' NC'
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Author Comment

by:cheryl9063
ID: 35209970
ewangoya--you were close but when I enter

'201 S Elm Street Greensboro North Carolina'

I get this

201 S Elm Stre

To the others.. This has to be native SQL.. I'm a developer and have no control over add-ins or using regular expressions... Also-- these address can be entered in any form imaginable... commas, no commas, spaces etc.. This is a huge website and I have no control over how the data comes in to me.. I'm tasked with tearing it apart and putting it back together again in a form we can use.. Thanks all!!
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Expert Comment

by:sdstuber
ID: 35210066
what about the first part of my suggestion?

prepending a space to whatever value you get

@UseST = ' ' + 'NC'
@UseST = ' ' + 'North Carolina'

@UseST = ' ' + <whatever your input is>

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Accepted Solution

by:
ewangoya earned 300 total points
ID: 35210123
try this

declare @stlocation varchar(100), @UseST varchar(50)

set @stlocation= '201 S Elm Street Greensboro North Carolina'
 set @UseST = 'North Carolina'
 
 select LEFT(@stlocation, DATALENGTH(@stlocation) - DATALENGTH(RTRIM(LTRIM(@UseST))))
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Assisted Solution

by:bitref
bitref earned 100 total points
ID: 35210130
Then, you may try:
SELECT LEFT(@stlocation, LEN(@stlocation) - LEN(@UseST)) 

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Expert Comment

by:Sharath
ID: 35211086
Do you always have state name at the end? Can you provide more sample data with expectd result?
Is @UseST always NC?
If @stLocation = '201 S Elm Street Greensboro North Carolina', what is the value of @UseST and expected result?
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Assisted Solution

by:Anthony Perkins
Anthony Perkins earned 50 total points
ID: 35214997
>>This has to be native SQL.. I'm a developer and have no control over add-ins or using regular expressions... <<
Regular expressions are part of "native SQL" in some dialects that is and since you posted in SQL Query Syntax (as well as SQL Server 2008, SQL Server 2005) the responses you are getting reflect that.
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