Padding numbers within a string

Question

Hi Folks,
I have some strings which are all a mix of letters and numbers - no particular rules, but no special characters.
For ordering purposes, I would like to be able to lpad each number string to 5 digits (in reality there will never be more than 3 digits in a row)
For example:
XX01 -> XX00001
XX100 -> XX00100
XX1YY10ZZ1 ->XX00001YY00010ZZ00001
Any thoughts on how this can be achieved?
I am currently trying to get something working with REGEXP_REPLACE but cant quite get it working:
select pc.pcode,REGEXP_REPLACE(pc.pcode, '([[:digit:]]+)',lpad('\1',5,0))
from test_tab pc
order by 2
The actual/desired results are as follows:
ORIG DESIRED RESULT ACTUAL RESULT
=== =========== ==========
XX01 XX00001 XX00001 (correct)
XX10 XX00010 XX00010 (correct)
XX100 XX00100 XX000100 (incorrect)
XX1YY10ZZ1 XX00001YY00010ZZ00001 XX0001YY00010ZZ0001 (incorrect)
Thanks for any input you may have!
Matt
Edited by: Matthew Morgan on Jan 10, 2013 7:59 PM
Edited by: Matthew Morgan on Jan 10, 2013 7:59 PM
Edited by: Matthew Morgan on Jan 10, 2013 8:00 PM

Answer 1

Hi, Matt,
Matthew Morgan wrote:
... I am currently trying to get something working with REGEXP_REPLACE but cant quite get it working:
select pc.pcode,REGEXP_REPLACE(pc.pcode, '([[:digit:]]+)',lpad('\1',5,0))
from test_tab pc
order by 2
/In case you're wondering why that didn't work:
The arguments to REGEXP_REPLACE can be any kind of expressions of the right datatype. In particular, the 3rd argument is supposed to be a string, so it's perfectly okay to use a function that returns a string, such as
LPAD ('\1', 5, '0')in that place. (The 3rd argument to LPAD is supposed to be string. You passed a NUMBER instead, but Oracle implicitly converted that to a string for you, and no harm was done this time.)
When you pass an expression like that as an argument to REGEXP_REPLACE, the expression is evaluated first, and then it's results are passed back to REGEXP_REPLACE. What does
LPAD ('\1', 5, '0')evaluated to? It's the 5-character string:
000\1'\1' has no special meaning to LPAD; it's just a 2-character string. '\1' has special meaning in REGEXP_REPLACE, but this is not being done in the context of REGEXP_REPLACE, in fact, this has to be done before REGEXP_REPLACE can do its job. So, regardless of what's in pcode, what you wrote is equvalent to
select    pc.pcode
,       REGEXP_REPLACE ( pc.pcode
                   , '([[:digit:]]+)'
                   , '000\1'
from      test_tab pc
order by  2which adds exactly 3 0's before every number, regardless of how many digits are in that number. So when you happened to have 2-digit numbers (as in 'XX01') it happened to return the results you wanted.
chris227 wrote:
A more cautious version of frank solution could be to replace only the kind of digits added previously
SELECT       str
,       REGEXP_REPLACE ( REGEXP_REPLACE ( str
                                   , '(\d+)'
                           , '0000\1'
                   , '0+(\d{5})' --changed \d to 0
                   , '\1'
                   )     AS padded_str
FROM       str
ORDER BY  padded_str
Good idea!
Edited by: Frank Kulash on Jan 11, 2013 10:28 AM