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SQL pull numeric from string

Posted on 2011-03-14
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Last Modified: 2012-06-21
I need to be able to pull the zip code out for either example

'5220 michaux rd,greensboro,n.c.27410' or '5220 michaux rd,greensboro,n.c.27410-9203'

How?
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Question by:cheryl9063
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15 Comments
 
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Expert Comment

by:8080_Diver
Comment Utility
Do you know for sure that the ZIP Code will be the characters after the last space in the string?
Do you only want the 5 digit Zip Code or do you want the full 9 digit if it is available?
-- SQL_1 - as many Zip Code Characters as are available
SELECT RIGHT(yourstring, CHARINDEX(' ', REVERSE(yourstring) - 1)) AS ZipCode
FROM yourtable;

-- SQL_2 - only the 5 digit Zip Code
SELECT LEFT(RIGHT(yourstring, CHARINDEX(' ', REVERSE(yourstring) - 1)), 5) AS ZipCode
FROM yourtable;

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Expert Comment

by:sshah254
Comment Utility
Let's assume that this data is in field1

reverse(field1)
find the first "." (or comma, or space, whatever tells that the zip code is starting) in there
charindex(reverse(field1))
right(field1, len(field1) - charindex(reverse(field1))) should give you what you want

You may need to add +1 or subtract -1 ... don't remember if MSSQL does it with 0-index or not.

Ss

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Accepted Solution

by:
Lowfatspread earned 125 total points
Comment Utility
like this....

select case when zipstr like '%-%' then left(zipstr,charindex('-',zipstr)-1) else zipstr end as zip
  from (
select reverse(left(rcol,charindex('.',rcol)-1)) as zipstr
  from (select yourcolumn as col,reverse(yourcolumn) as rcol
          from ....) as x
       ) as y
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Author Comment

by:cheryl9063
Comment Utility
LowfatSpread..What is rcol and what is "Your column?" This is not coming from a table.. This is a variable passed in that I'm having to unparse..
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Expert Comment

by:Sharath
Comment Utility
Create a function like this.
CREATE FUNCTION [dbo].[fn_Split](@text nvarchar(max), @delimiter char(1) = ' ')
RETURNS @Strings TABLE (position int IDENTITY PRIMARY KEY, value nvarchar(max)) AS 
  BEGIN 
DECLARE @index int
    SET @index = -1
  WHILE (LEN(@text) > 0)
  BEGIN
    SET @index = CHARINDEX(@delimiter , @text)
     IF (@index = 0) AND (LEN(@text) > 0)
  BEGIN 
 INSERT INTO @Strings VALUES (@text)
  BREAK 
    END
     IF (@index > 1)
  BEGIN
 INSERT INTO @Strings VALUES (LEFT(@text, @index - 1))
    SET @text = RIGHT(@text, (LEN(@text) - @index))
    END 
   ELSE 
    SET @text = RIGHT(@text, (LEN(@text) - @index))
    END
 RETURN
    END
 GO

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Use this function in your code as below.Replace your_table with the actual table name and PostalAddress column with actual column name.
SELECT PostalAddress,value zip  
  FROM (SELECT *, 
               ROW_NUMBER() 
                 OVER(PARTITION BY PostalAddress ORDER BY position DESC) rn 
          FROM your_table t 
               CROSS APPLY dbo.fn_Split(PostalAddress,'.')) t1 
 WHERE rn = 1

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I have tested like this.
DECLARE  @table  TABLE( 
                       PostalAddress VARCHAR(100) 
                       ) 

INSERT @table 
VALUES('5220 michaux rd,greensboro,n.c.27410'), 
      ('5220 michaux rd,greensboro,n.c.27410-9203') 

SELECT PostalAddress,value zip  
  FROM (SELECT *, 
               ROW_NUMBER() 
                 OVER(PARTITION BY PostalAddress ORDER BY position DESC) rn 
          FROM @table t 
               CROSS APPLY dbo.fn_Split(PostalAddress,'.')) t1 
 WHERE rn = 1
/*
PostalAddress	zip
5220 michaux rd,greensboro,n.c.27410	27410
5220 michaux rd,greensboro,n.c.27410-9203	27410-9203
*/

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Author Comment

by:cheryl9063
Comment Utility
There is no table... Will any of these work for a variable input such as below..

@Location = 5220 michaux rd greensboro n.c.27410

@NEEdZip =  
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Assisted Solution

by:Sharath
Sharath earned 125 total points
Comment Utility
Yes, you can try like this.

SELECT @NEEdZip = value  
  FROM (SELECT *, 
               ROW_NUMBER() 
                 OVER(PARTITION BY PostalAddress ORDER BY position DESC) rn 
          FROM (select @Location as PostalAddress) t 
               CROSS APPLY dbo.fn_Split(PostalAddress,'.')) t1 
 WHERE rn = 1
select @NEEdZip

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Expert Comment

by:awking00
Comment Utility
Can address also be like this?
'5220 Michaux Rd., Greensboro, NC 27410'
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Expert Comment

by:8080_Diver
Comment Utility
Yes,

Create a function and, in the function, set the output/result to "RIGHT(yourstring, CHARINDEX(' ', REVERSE(yourstring) - 1)) " where yourstring is the string passed into the function as the parameter.
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Assisted Solution

by:8080_Diver
8080_Diver earned 125 total points
Comment Utility
Using the variables you previously indicated:

@NEEdZip = RIGHT(@Location, CHARINDEX(' ', REVERSE(@Location) - 1))

You don't have to create the function (but, just so I could use it elsewhere ;-) I would. ;-)
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Expert Comment

by:awking00
Comment Utility
Without knowing the data, I'm not sure I would count on either a period introducing the zip code or the last space introducing the zip code. Are there any other examples for the input?
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Author Comment

by:cheryl9063
Comment Utility
awKing..that is the problem.. Could be any kind of address with either a 5 digit of the extended american 9 digit with dash.. The state could be spelled out or not, could be dots.. I'm having to just "TRY" each and pass a successful one to an outside process that validates.. If it does not validate I come back and "try" again 5 times.. My problem is the one I mentioned at the top with n.c...
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Expert Comment

by:Sharath
Comment Utility
<< Could be any kind of address with either a 5 digit of the extended american 9 digit with dash.. The state could be spelled out or not, could be dots.. >>

Provide sample data for all these combinations. BTW, did you try my last query?
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Assisted Solution

by:awking00
awking00 earned 125 total points
Comment Utility
Try this -
substr(address,regexp_instr(addr,'[0-9]{5}|[0-9]{5}-[0-9]{4}'))
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Author Closing Comment

by:cheryl9063
Comment Utility
Thanks
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