Regular Expression confusion 2005-04-20 - By William B Ferguson
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@(protected)]
Sent: Wednesday, April 20, 2005 2:19 PM
To: William B Ferguson; oracle-l@(protected)
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@(protected)
[mailto:oracle-l-bounce@(protected)]On Behalf Of William B Ferguson
Sent: Wednesday, April 20, 2005 2:13 PM
To: oracle-l@(protected)
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 (See ORA-01722.ora-code.com) 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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