RE: Regular Expression confusion

  • From: William B Ferguson <wbfergus@xxxxxxxx>
  • To: Ron.Reidy@xxxxxxxxxxxxxxxxxx, oracle-l@xxxxxxxxxxxxx
  • Date: Wed, 20 Apr 2005 14:22:02 -0600

Ron,

I tried that before in my application, and it strips out the '%'
(wildcard) operator that I need to allow for, as valid user input. 


------------------------------------------------------------
Bill Ferguson
U.S. Geological Survey - Minerals Information Team
PO Box 25046, MS-750
Denver, Colorado 80225
Voice (303)236-8747 ext. 321 Fax (303)236-4208




-----Original Message-----
From: Reidy, Ron [mailto:Ron.Reidy@xxxxxxxxxxxxxxxxxx] 
Sent: Wednesday, April 20, 2005 2:19 PM
To: William B Ferguson; oracle-l@xxxxxxxxxxxxx
Subject: RE: Regular Expression confusion

Can you say something like "upper(regex_replace(name, '[:punct;]' ..."?

-----------------
Ron Reidy
Lead DBA
Array BioPharma, Inc.
303.386.1480


-----Original Message-----
From: oracle-l-bounce@xxxxxxxxxxxxx
[mailto:oracle-l-bounce@xxxxxxxxxxxxx]On Behalf Of William B Ferguson
Sent: Wednesday, April 20, 2005 2:13 PM
To: oracle-l@xxxxxxxxxxxxx
Subject: Regular Expression confusion


I'm getting confused on regular expressions (Oracle 10.1.0.4). I want to
strip all punctuation from a user supplied input string except for the '%'
wildcard character. To see if my formatting of the syntax was correct, I
wrote the following sql statement:

SQL> set escape off
SQL> select UPPER(regexp_replace(name,'[''''|"|;| |-|,|.]')) test_data
  2  from names
  3  where upper(regexp_replace(name,'[[:punct:][:space:]]'))
  4   like ('%JGO%');

TEST_DATA
----------------------------------------------------------------------
A-JGOLDMININGCO
AJGOLDMINE
AJGOLDMININGCO
AJGOLDMININGCO
AJGOLDMININGCO
EJGOGGINSPROSPECT
JJGORDONQUARRY

7 rows selected.

Not quite what I wanted, as a hyphen still appears.

So, I tried escaping the hyphen:
SQL> select UPPER(regexp_replace(name,'[''''|"|;| |\-|,|.]')) test_data
  2  from names
  3  where upper(regexp_replace(name,'[[:punct:][:space:]]'))
  4   like ('%JGO%');

TEST_DATA
----------------------------------------------------------------------
A-JGMC
AJGM
AJGMC
AJGMC
AGMC
EJGP
JGQ

7 rows selected.

Definitely not what I wanted! So, I figured I'd try adding an extra set of
brackets (like the second regexp_replace uses):

SQL> select UPPER(regexp_replace(name,'[[''''|"|;| |\-|,|.]]')) 
SQL> test_data
  2  from names
  3  where upper(regexp_replace(name,'[[:punct:][:space:]]'))
  4   like ('%JGO%');

TEST_DATA
----------------------------------------------------------------------
A-J GOLD MINING CO
A. J. GOLD MINE
A. J. GOLD MINING CO.
A.J. GOLD MINING CO.
AJ GOLD MINING CO.
E J GOGGINS PROSPECT
JJ GORDON QUARRY

7 rows selected.

That had the effect of negating what I wanted to accomplish in the first
place! I might as well have just selected name.

And if I changed the first regexp_replace to
UPPER(regexp_replace(name,'[''''|"|;| |'-'|,|.]')), Then I get an
ORA-01722 invalid number.

How do I modify my first regexp_replace to also strip out the hyphen while
allowing the '%' wildcard, unlike the :punct: class?


------------------------------------------------------------
Bill Ferguson
U.S. Geological Survey - Minerals Information Team PO Box 25046, MS-750
Denver, Colorado 80225 Voice (303)236-8747 ext. 321 Fax (303)236-4208
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