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Trimming spaces from a field

Posted on 1998-08-21
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Last Modified: 2010-03-19
Basic problem is checking if certain person isn't already in a table. It could be that someone other created this person, and wrote the name slightly different. Soundex is available, but gives a big difference if there are spaces between the characters or not [SOUNDEX('Van Der') <> SOUNDEX('Vander').
So i would like to trim that column first, before using the soundex function. But no internal function is available for trimming 'inside' spaces. How should i solve this problem ?
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Question by:jvh042097
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by:jvh042097
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by:brunchey
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Have you tried patindex?  It returns the starting position of the first occurrence of pattern in the specified expression, or zeros if the pattern is not found. You can use wildcard characters in pattern, as long as the wildcard character % precedes and follows pattern (except when searching for first or last characters). The expression is usually a column name. You can use this function on text, char, and varchar data.
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by:mitek
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PATINDEX would not help in this situation. clearly, 'vander' is not a pattern in 'van der', neither 'van der' will be found in 'vander'

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by:mitek
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Question: do you absolutely need to do it within a single SELECT query or cursor will be ok as well ?
If it's the first, it may be really hard to do (if at all possible)
If second, it's quite feasible ...

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mayhew earned 100 total points
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I believe what brunchey meant was to use patindex to find the space in 'van der'.

It's a little long but something like:

select substring('van der',1,patindex('% %','van der')-1) + substring('van der',patindex('% %','van der')+1,datalength('van der'))

would do the trick.  Of course this only finds one space.  Like mitek said, though, if you can use a stored proc or something, it wouldn't be tough to iterate this so it strips all spaces out.

Let us know if this helps.  :)

Don
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by:mitek
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There is no way to write a generalized solution in one query.
However, here is a solution for names with 0-2 spaces in them:

SELECT * FROM users
 WHERE SOUNDEX('vander') = CASE
         -- first space detection
         WHEN CHARINDEX(' ',last_name) > 0 THEN
           SOUNDEX(SUBSTRING(last_name,1,CHARINDEX(' ',last_name) - 1) +
           SUBSTRING(last_name,CHARINDEX(' ',last_name) + 1,DATALENGTH(last_name)))
         -- second space detection
         WHEN CHARINDEX(' ',SUBSTRING(last_name,CHARINDEX(' ',last_name),DATALENGTH(last_name))) > 0 THEN
           SOUNDEX(SUBSTRING(last_name,1,CHARINDEX(' ',last_name) - 1) +
           SUBSTRING(last_name,CHARINDEX(' ',last_name) + 1,CHARINDEX(' ',SUBSTRING(last_name,CHARINDEX(' ',last_name),DATALENGTH(last_name))) - 1) +
           SUBSTRING(last_name,CHARINDEX(' ',SUBSTRING(last_name,CHARINDEX(' ',last_name),DATALENGTH(last_name))) + 1,DATALENGTH(last_name)))
         -- no spaces detected
         ELSE
           SOUNDEX(last_name)
       END

following the same pattern, one could write a solution for 0-3, and even for 0-4 spaces. however, if more than 3 spaces are expected, it will make sense to write a generalized solution with a cursor and a space-removing routine


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