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Create a Transact-SQL statement that will find all extract letter characters from the middle of a string field and group by them.

tdeinzer2005
tdeinzer2005 asked
on
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Last Modified: 2008-02-01
Let's say I have a Cust ID field and some of the values look like this:

1235ABC6789
123DEG9876
12345HIJ987
3214ABC5421

I want a SQL statement that returns a list of records grouped by the three letters that exist in all the CustID values.  In other words, the three letters need to be extracted from the string, but the first letter doesn't always start in the same position.  Then I need to group on those 3 letters.
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Top Expert 2012

Commented:
Something like this:
SELECT     SUBSTRING(CustID, PATINDEX('%[^0-9]', CustID, 3)
From YourTableName
GROUP BY SUBSTRING(CustID, PATINDEX('%[^0-9]', CustID, 3)
Scott PletcherSenior DBA
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Most Valuable Expert 2018
Distinguished Expert 2019

Commented:
If the letters are always contiguous (one after the other, not separated):

SELECT ...
FROM ...
WHERE PATINDEX('%[a-zA-z]%', CustID) > 0
GROUP BY SUBTRING(CustID, PATINDEX('%[a-zA-z]%', CustID), 3)
CERTIFIED EXPERT
Top Expert 2012

Commented:
Let's try that again:

SELECT      SUBSTRING(CustID, PATINDEX('%[^0-9]', CustID), 3)
From      YourTableName
GROUP BY
      SUBSTRING(CustID, PATINDEX('%[^0-9]', CustID), 3)
Scott PletcherSenior DBA
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Most Valuable Expert 2018
Distinguished Expert 2019

Commented:
CORRECTION: "SUBSTRING" not "SUBTRING" :-)

Author

Commented:
Scott,

Can you please explain this part for me:  '%[a-zA-z]%'?  I understand how the percent symbols work and I would understand the dash if it were just [a-z], but what's with the 2 dashes?
Senior DBA
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Most Valuable Expert 2018
Distinguished Expert 2019
Commented:
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Author

Commented:
Thanks Scott!  That worked!

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