Regular Expression/Replace - Oracle 7.3

Hi!
I am trying the regular expression SQL functions of 10g to Oracle 7.3 and it seems the older version does not cover this feature yet.
"Aaaa,Bbbb" --> "Aaaa, Bbbb"
REPLACE *",[0-9A-Za-z]"* WITH *", "*
The string pattern is to look for comma-punctuations that is not followed immediately by whitespacess so I can replace this with a comma followed by a whitespace.
Any workaround for this?

Hi,
Welcome to the forum!
kitsune wrote:
Hi!
I am trying the regular expression SQL functions of 10g to Oracle 7.3 and it seems the older version does not cover this feature yet.You're right; regular expressions only work in Oracle 10.1 and higher.
>
>
"Aaaa,Bbbb" --> "Aaaa, Bbbb"
REPLACE *",[0-9A-Za-z]"* WITH *", "*
The string pattern is to look for comma-punctuations that is not followed immediately by whitespacess so I can replace this with a comma followed by a whitespace.
Any workaround for this?You're best bet in Oracle 7.3 would be a user-defined function. That's a very old version; don't expect much.
Do you know anything else about the string? For example, is there some character (say ~) that never occurs in the string? Will there ever be two (or more) whitespace characters after punctuation? What characters do you consider to be whitespace? Which are punctuation? Depending on the answers, you might be able to do something with nested REPLACE and/or TRANSLATE functions.

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Regular Expressions in Oracle

Hello All,
I come from Perl scripting language background. Perl's regular expressions are rock solid, robust and very fast.
Now I am planning to master Regular Expressions in Oracle.
Could someone please point the correct place to start with it like official Oracle documentation on Regular Expressions, any good book on Regex or may be any online link etc.
Cheers,
Parag
Edited by: Parag Kalra on Dec 19, 2009 11:03 AM

Hi, Parag,
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Sorry, I don't know a good tutorial or introduction.
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Hi All,
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884476 wrote:
Could you please explain what does that pattern mean? We can look at your string as a sequence of substrings where each substring is set of characters (part A) followed by word oracle or, in last substring, by end-of-line (part B). Each of such substrings be want to replace with part B thus removing part A.
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SY.

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Hello,
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Roger22 wrote:
Hello,
Does oracle 9i support regular expressions?
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Regular expression in oracle for hypen

Hi,
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Hi Prasad,
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according to:
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hi,
I need to replace the below string, but i couldnt able to do if we use the special charaters '+', '$' . can anyone suggest a way to do this?
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you should use escape character \.
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Dear All,
This is not a business requirement, just trying to practice regexp.
Suppose we want to replace a part of a regular expression from a string. As an example, in the string 'THIS Number 124356 Is to Change.This Number 5 Also', I am trying to replace the last digit of each number with $ sign.
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MichaelS wrote:
I am trying to achieve this using a SINGLE regexp_replace.Here we go:
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select 'THIS Number 124356 Is to Change.This Number 5 Also' str from dual
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Nice..
Learning for me ..
Never thought of usage like {0,} ... kepping the end value OPEN.
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(<PartName>([a-z]))(.*)(</PartName>)
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I've been on this for a day now and it's getting to the point where I could have gone through and done it manually in less time.
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I gave up and just exported out from my xml into Excel and did a formula there...

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Can anybody tell me the regular expression to validate date in 'DD-MON-YYYY'.
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If it must be a regexp, this is a starter for you, note it carries the caveats mentioned by both posters above and in the linked thread
mkr02@ORA11GMK> with data as (select '10-jan-2012' dt from dual
2 union all select '10-111-2012' from dual
3 union all select 'mm-jan-2012' from dual
4 union all select '10-jan-12' from dual)
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8 from data
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10-jan-2012          1
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mm-jan-2012          0
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Without using a regular expressions in oracle 9i

dear all;
I have the following test data below
insert into p
      (id)
    values
      ('\G1\G2');
insert into p
      (id)
    values
      ('\A1\');
   insert into p
      (id)
    values
      ('\B1\B2\B3');
       insert into p
      (id)
    values
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ID
G1
null
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J3the output is gotten by looking at the second to the last entry and extracting it....
so for \G1\G2 the second to the last entry was G1 hence we have G1
for \A1\ there is no second to the last entry and that is why it is null
for \B1\B2\B3 the second to the last entry was B2

Hi,
user13328581 wrote:
... the output is gotten by looking at the second to the last entry and extracting it....
so for \G1\G2 the second to the last entry was G1 hence we have G1
for \A1\ there is no second to the last entry and that is why it is nullSo backslashes at the end of id don't matter. Whether id is
'\AI',
'\A1\' or
'\A1\\\\\\\\\\\\'
there's only entry in id, and so you want to return NULL. Is that right?
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WITH    got_pos          AS
     SELECT     id
     ,     INSTR ( RTRIM (id, '\')
                , -1
                , 2) + 1     AS pos_from
     ,     INSTR ( RTRIM (id, '\')
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                , 1)     AS pos_to
     FROM     p
SELECT       id
,       CASE
           WHEN pos_from     > 1
           THEN SUBSTR ( id
                            , pos_from
                  , pos_to - pos_from
       END     AS penult
FROM       got_pos
ORDER BY id
;

Regular Expression/Replace - Oracle 7.3

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