Regexp_replace help

Hello:

I have a string which will be composed of 1 to 3 rooms. The first separator will be 1 or 2 brands of ' / '. I have a regular expression, which works for most of the data, but not in a particular case, and I don't know why. In the case below where the source is simply the string 'A', all 3 parts are returned as 'A'. I want A PT1 and PT2 and PT3 to be null. Why is this not the case, and how do I change the regular expression to manage this source string?

Thank you.

with base as)

Select ' a / / BCD / / XYZ "as double CBC

Union of all the

Select ' a / / BCD/XYZ "as double CBC

Union of all the

Select ' A / / BCD' as double CBC

Union of all the

Select 'A' as double CBC

)

Select the CBC

regexp_replace (RTrim (SRC), ' ([A - Z] +)(/{1,2}) ([A - Z] *)?) () /+)? ([A-Z] *) ? (',' \1') PT1.

regexp_replace (RTrim (SRC), ' ([A - Z] +)(/{1,2}) (([A-Z]*)?)) (/+)? ([A-Z] *) ? (',' \3') PT2.

regexp_replace (RTrim (SRC), ' ([A - Z] +)(/{1,2}) ([A - Z] *)?) () /+)? ([A-Z] *) ? (',' \5') PT3

Basic

What I get is:

CBC	STR	PT2	STP
A / / BCD / / XYZ	A	BCD	XYZ
A / / BCD/XYZ	A	BCD	XYZ
A / / BCD	A	BCD	(null)
A	A	A	A

Hello

You can get what you want more simply using REGEXP_SUBSTR:

SELECT src

, REGEXP_SUBSTR (CBC, ' [^ /] +', 1, 1) AS pt1

, REGEXP_SUBSTR (CBC, ' [^ /] +', 1, 2) AS pt2

, REGEXP_SUBSTR (CBC, ' [^ /] +' 1, 3) AS pt3

BASIC;

Tags: Database

Similar Questions

helps the regexp_replace - converting numbers to strings

I try to use regexp_replace replace all occurrences of the type (number) (in brackets with a number inside) with the expression ('number') (add "before and after the number, and by here - to transform the number to varchar).
I can't find a way to do it properly.
It would be great if you guys could help out me here.
Example-
Before change: abcde (737) (6) s (v) sbsgs37
After change: abcde('737') s sbsgs37('6') (v)
Thank you
Ilya Golosovsky.

("Select REGEXP_REPLACE (' abcde (737) (6) s (v) sbsgs37 ', '$(\d+)$ ', q '[('\1 ')]')

of qry

Result:

ABCDE('737') s sbsgs37('6') (v)

q "[.. '] is called a string q, it allows us to embed single quotes in string literals without having to get away from them."

$(\d+)$
- ------(corresponds to the character (literally
- 1 capturing group (\d+)
  - \d+ matches a digit [0-9]
    - Quantifier: + between one and unlimited times, as many times as possible, giving necessary [greedy]
- \) corresponds to the character) literally
('\1')
- 1 capturing group ("\1")
  - literally "corresponds to the character.
  - \1 matches the text even more recently matched by the capturing group 1
  - literally "corresponds to the character.

Help REGEXP_REPLACE

Hi all
I tried to replace the characters all before X.
For example, in this case, the X is replaced by 0
Select REGEXP_REPLACE (12X99999', '[Aa - Zz]','0 ') of the double
120to 99999
I want to replace 12 X 0, so I want to replace 12X99999 and see 00099999.

Can someone help me?
Thank you.

Select RPAD ('0', REGEXP_INSTR(s,'[A-Za-z]'),'0 '). REGEXP_REPLACE(s,'.+[A-Za-z]') of

(SELECT "12 X 99999" s FROM DUAL ")

Hi all, need help with replacing or Regexp_Replace

Hi all
I have a string with me, i.e. '9 03000200090 R R 9 R 9 9 1 9 2 R R 001535447 R 1 R Y R Y 4 1 100002589 1 9 9 9 9 9 9 00002589'
Now, I would like to remove R-9 characters of my channel, look to the characters below marked "BOLD" - these are those, I want to delete.
"9 03000200090 R R R 9 9 9 1 9 2 R 001535447 R 1 R Y R Y 4 1 100002589 1 9 9 9 9 9 9 00002589' R '.
Want my output end: ' 03000200090 1 2 001535447 1 Y Y 4 1 100002589 100002589'.
Can someone help me please?
Thank you
Amit

Hi, amit,

Here's one way:

SELECT REPLACE (REGEXP_REPLACE (REPLACE (str)

, ' '

, ' '

)

, '(^| ) [R9] ( |$)'

)

, ' '

, ' '

)

FROM table_x

;

With a little more coding, your could do it without regular expressions.

Helps the REGEXP_REPLACE
Hello Experts

I'm unable to analysis at the address below. Please see the part of the desired effect.
Please help solve this problem. I have tried the REGEXP_REPLACE but could not obtain the success.

Is the version of Oracle, I'm working on that

Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - 64 bit Production
With partitioning, OLAP, Data Mining and Real Application Testing options

Thank you

RB

with table1 as
* (by selecting '1805SOUTHETHELAVE' addr1 throughout Union double *)
Select '1165MCALLISTERRD' addr1 Union double all the
Select '2161COMMERCCDR APT206' addr1 Union double all the
Select '1048HARPERLEEDR' addr1 Union double all the
Select '20021NWCR4290' double addr1
*)*

looking for a SOLUTION AS

* 1805 SOUTHETHEL AVE *.
* 1165 MCALLISTER RD *.
* 2161 COMMERCC DR. *
* DR. HARPERLEE 1048 *.
* 20021 NWCR 4290 *.
Hello

Here's one way:
```
SELECT     addr1
,     REGEXP_REPLACE ( addr1
                 , '(\d+)([[:alpha:]]+)(AVE|DR|RD|\d+).*'
                 , '\1 \2 \3'
                 )     AS new_addr1
FROM    table1
;
```
Output:
```
ADDR1                 NEW_ADDR1
--------------------- ----------------------------------------
1805SOUTHETHELAVE     1805 SOUTHETHEL AVE
1165MCALLISTERRD      1165 MCALLISTER RD
2161COMMERCCDR APT206 2161 COMMERCC DR
1048HARPERLEEDR       1048 HARPERLEE DR
20021NWCR4290         20021 NWCR 4290
```
He gets good results of the sample data, but he could get good results for the wrong reasons. I have no way of knowing, since I do not know the reasons why you want to get the results that you do. Explain exactly what you're trying to do. Depending on what you want, you might need nested functions.

need help to get the value using the REGEXP_REPLACE function.
I am trying to extract the value placed between "< listitems >... < / listitems >" but I am unable to get these value using the REGEXP_REPLACE function.
```
select REGEXP_REPLACE('<InputParameters></ParamSet><listitems>14545454</listitems></ParamSet></InputParameters>'
,'(<InputParameters>.*<listitems>)(.*)(</listitems>.*<InputParameters>)'
,'\2') from dual;
```
```
required output:
 1454554
```
Can someone please help me achieve this goal.
Hello

You forgot the / before the last InputParameters. If fix you that, your code will work.

Here is a slightly different way:
```
SELECT  REGEXP_REPLACE ( '14545454'
                 , '^.*?(.*?).*$'
                 , '\1'
                 )     AS listitems
FROM    dual
;
```
Least to type, the less likely you will have to make typos.
If there are 2 (or more) elements of listitmes, 1 will be returned.

From Oracle 11.1, you can also use REGEXP_SUBSTR. I have Oracle 10.2 avaialable now, so I can test the following, but I think you'd say:
```
SELECT  REGEXP_SUBSTR ( '14545454'
                 , '(.*?)'
                 , 1
                 , 1
                 , NULL
                 , 1
                 )     AS listitems
FROM    dual;
```
When it comes with the text that is or looks to, XML, consider using XML functions.

need help to understand REGEXP_REPLACE
Hi all
I'm new to this site so please forgive me for making mistakes. I have a field 'DESCRIPTION of the STUDENT' and they have the following values
2830 ORO - (2011) Rob Miller [6]
2830 ORO - Cathy Ingrid (2011) [7]
2830 ORO - (2011) Sam Sullivan [8]
2650 Robert Lawson
2660 Pat Ortt (2009)
2690 - mark lively
2710 Tim Lacreta
What I want in my desired output is
(2011) Rob Miller [6]
Cathy Ingrid (2011) [7]
(2011) Sam Sullivan [8]
Robert Lawson
Pat Ortt (2009)
Mark lively
Tim Lacreta

Need help please, I know I have to use REGEXP_REPLACE, but I do not understand how I went through the documentation, but has not been of any help.

Thank you
Hello

Welcome to the forum!

Whenever you have a problem, please post CREATE TABLE and INSERT statements for your sample data. Since this is your first post, I'll do it for you:
```
CREATE TABLE     table_x
(       student_description     VARCHAR2 (80)
);

INSERT INTO table_x (student_description) VALUES ('2830-BGC - (2011) Rob Miller [6]');
INSERT INTO table_x (student_description) VALUES ('2830-BGC - (2011) Cathy Ingid [7]');
INSERT INTO table_x (student_description) VALUES ('2830-BGC - (2011) Sam Sullivan [8]');
INSERT INTO table_x (student_description) VALUES ('2650 - Robert Lawson');
INSERT INTO table_x (student_description) VALUES ('2660 - Pat Ortt(2009)');
INSERT INTO table_x (student_description) VALUES ('2690 - Mark Lively');
INSERT INTO table_x (student_description) VALUES ('2710 - Tim Lacreta');
```
Explain how you get the results you want from these data. For example: "I want the student_description part that comes after the substring of characters 3 space-dash-space. Spaces are important, because the first part of student_description, the part I want to delete, can contain a hyphen (for example ' 2830 - ORO - (2011) Rob Miller [6] "). »

I think you want something like this:
```
SELECT     student_description
,     REGEXP_REPLACE ( student_description
                 , '.* - (.*$)'
                 , '\1'
                 )          AS after_dash
FROM    table_x
;
```
«. "*" means "0 or more characters (all).
"-" means exactly what it says: a space, followed by a hyphen, followed by a space. Hyphen has no special meaning outside the brackets.
' $' means the end of the string.

It would be more effective to do this particular job without using regular expressions:
```
SELECT     student_description
,     SUBSTR ( student_description
            , 3 + INSTR ( student_description
                              , ' - '
                  )
            )          AS after_dash
FROM    table_x
;
```
Depending on your needs, you may need to adjust this query if student_description does not always contain ' - '.

Published by: Frank Kulash, December 29, 2011 14:08

Help me about features as regexp_substr, regexp_replace
Hello world
Can someone help me understand these functions such as regexp_substr, regexp_replace...
Will be better if Tunis examples with different situations or it may be links

THX
its here... in the forum itself.

Introduction to regular expressions...

Vivek L

Help needed to add the hyphen in REGEXP_REPLACE
Hi all
I need a help to add the hyphen in my selection criteria so that it does not filter the hyphen.
I use below REGEXP_REPLACE get my result, everything's fine now except the hyphen (-) character is not recognized.

Select REGEXP_REPLACE ('abcd - efgh123 {} $ (), ',' [^ [a - z, A - Z, 0-9, (,), {,}, _, $,., ", [: space:]]] *',") twice;

Can you please help?

Thank you
If you want to keep the hyphen in your string, you can use sth. as
```
select REGEXP_REPLACE('abcd-efgh123{}$(),', '[^a-zA-Z0-9(){}_$.,''[:space:]-]*','') from dual
```
Best regards

Maxim

need help with regexp_replace
Hello

Can someone help me? I'll try to find a regexp for the following model?

In my paper, here is the model I'd like to remove.

< < snip > >
* [News 16]
* [Sport 17]
* [Leisure 18]
* [19] news
* [20] what you say
* [Family 21]
* [Video 22]
* [23] ads
* [Advertise 24]
* [25] buy/sell
* [26] Click2Find
* [Dating [27]
* [28] cars
* [29 houses]
* [30] jobs
< < snip > >

Thank you
Ajay
Maybe
```
regexp_replace(col, '\* \[\d+\][[:alpha:]]+')
```

Help with the Regular Expressions and regexp_replace
Oh great guru Oracle can I can receive assistance

I need to clean the phone numbers that have been entered in the table per_phones of Oracle e-Business. Some of the phone numbers have hyphens, some have spaces and some have tank. I just want to get out all the figures and then re - format the number.

E.g.
914-123-1234... out (914) 123-1234
9141231234... new (914) 123-1234
914 123 1234... (914) 123-1234
MyPhone... just null
(914)-123-1234... (914) 123-1234

I really tried to understand the instructions of regular expressions, but for some reason, I can't understand it.
For example:
```
SQL> with sample_data as (
  2   select '914-123-1234' phone_number from dual union all
  3   select '9141231234'                from dual union all
  4   select '914 123 1234'              from dual union all
  5   select '(914)-123-1234'            from dual
  6  )
  7  select regexp_replace(
  8           regexp_replace(phone_number, '\D')
  9         , '(...)(...)(....)'
 10         , '(\1) \2-\3'
 11         ) as formatted_num
 12  from sample_data
 13  ;

FORMATTED_NUM
--------------------------------------------------------------------------------
(914) 123-1234
(914) 123-1234
(914) 123-1234
(914) 123-1234
 
```

Regexp_replace? help please

Hi all

I get a VARCHAR2 string, which can contain mostly spaces and enything digital alpha also of special characters.
Right now I use a lot of functions which replaces some things, filling in the spaces.

Basically, I want to use regexp_replace to simply remove all the spaces and characters spacial (commas, slashes rear, ambersands e, t, c) and just take the digital alpha data.

He takes their identity card, then concatenates the two first letters of the name, then the first two letters of firtname firdt two letters of the given name... all the missing characters get stuffed with *.

This is the code I have for the moment

   declare
   
   rms_id NUMBER := 4767;
   last_name VARCHAR2(10) := 'fds dgdg';
   first_name VARCHAR2(10) := 'tu/  yu rt';
   middle_name VARCHAR2(10) := '';
   
   answer VARCHAR2(20);
   
   
   BEGIN
   
   SELECT rms_id || rpad(lower(substr(replace(nvl(last_name,'**'),' ',''),1,2)),2,'*') || 
                            rpad(lower(substr(replace(nvl(first_name,'**'),' ',''),1,2)),2,'*') || 
                            rpad(lower(substr(replace(nvl(middle_name,'**'),' ',''),1,2)),2,'*')

                            INTO answer
     

   FROM dual;
   
   dbms_output.put_line(answer);
   
   END;

So, you want to say something like:

SQL> ed
Wrote file afiedt.buf

  1  with t as (select 4767 as rms_id, 'fds dgdg' as last_name, 'tu/  yu rt' as first_name, '' as middle_name from dual)
  2  --
  3  -- END OF TEST DATA
  4  --
  5  select rms_id||
  6         lpad(nvl(substr(regexp_replace(first_name,'[^[:alnum:]]'),1,2),'*'),2,'*')||
  7         lpad(nvl(substr(regexp_replace(middle_name,'[^[:alnum:]]'),1,2),'*'),2,'*')||
  8         lpad(nvl(substr(regexp_replace(last_name,'[^[:alnum:]]'),1,2),'*'),2,'*')
  9* from t
SQL> /

RMS_ID||LP
----------
4767tu**fd

SQL>

need help to get the desired outcome of the SQL using REGEXP_REPLACE

Hi all
I have following query column values.
Select (1-4, 8-9, 7, 11-13') twice;
Here, I need to replace the '-' in the sequence number, as shown in the below expected o/p.
Expected results
-------------------------
1,2,3,4,8,9,7,11,12,13

Hello

dbaKrish wrote:

Hi all

I have following query column values.

Select (1-4, 8-9, 7, 11-13') twice;

Here, I need to replace the '-' in the sequence number, as shown in the below expected o/p.

Expected results

-------------------------

1,2,3,4,8,9,7,11,12,13

Its one-way:

WITH got_str AS

(

SELECT ' 1-4, 8-9, 7, 11-13' STR

OF the double

)

got_comma_delimited_parts AS

(

SELECT LEVEL AS part_num

REGEXP_SUBSTR (str,

, '[^,]+'

1

LEVEL

) IN the framework

OF got_str

CONNECT BY LEVEL<= regexp_count="" (="">

, '[^,]+'

)

)

got_nums AS

(

SELECT part_num

TO_NUMBER (REGEXP_SUBSTR (part

, '\d+'

)

) AS low_num

TO_NUMBER (REGEXP_SUBSTR (part

, '\d+$'

)

) AS high_num

OF got_comma_delimited_parts

)

got_max_range AS

(

SELECT MAX (high_num - low_num) AS max_range

OF got_nums

)

AS cntr

(

SELECT the LEVEL - 1 AS n

OF got_max_range

CONNECT BY LEVEL<= max_range="" +="">

)

SELECT (n.low_num + c.n LISTAGG

, ','

) (ORDER BY n.part_num group

, c.n

), Exit

OF got_nums n

CNTR JOIN c ON c.n<= n.high_num="" -="">

ORDER BY part_num

;

This assumes that each part separated by commas composed entirely unsigned integerss, or exactly 2 integers separated by a hyphen and the integer before the dash is not that the whole greateer after the hyphen. If you can not be sure of these things, you will need to change the query slightly to check and handle exceptions.

REGEXP_REPLACE - replaceing a character not allowed

Hi friends,
Please, help me to replace a character not allowed of a given string. Allowed characters are
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789/ -?: ()., "+ {space} "

If a character in the specified string is not in the list above, which should be replaced by a space. Can you please help me to do with REGEXP or in any other way.

Thanks in advance!

In my previous solution the dash (-) in the second row replaced with space. That shouldn't be the case. This is because the brackets, hyphen is considered operator of range as [0-9]. So I moved the link to the last to solve the problem.
```
SQL> with t
  2  as
  3  (
  4    select 'ka{th}ck' str from dual union all
  5    select 'k$rt-ic)' str from dual union all
  6    select 'k$rthi@k' str from dual union all
  7    select 'KARTHI123' str from dual
  8  )
  9  select str
 10       , regexp_replace(str, '[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789/?:().,''+{}-]', ' ') new_str
 11    from t;

STR       NEW_STR
--------- ----------
ka{th}ck  ka{th}ck
k$rt-ic)  k rt-ic)
k$rthi@k  k rthi k
KARTHI123 KARTHI123
```

Why doesn't my REGEXP_REPLACE?

Select count (*) from PRX. PROX_REQ where status = "ERROR" and REQUEST_BODY like ' % < country > Germany < / country > %';
> 66
update of PRX. PROX_REQ set REQUEST_BODY = REGEXP_REPLACE (REQUEST_BODY, '< country > Germany </country >', '< country > Deutschland/< country >') where status = "ERROR";
> 102 lines updates
which is wrong with my order? It should update the 66 lines and not 102!
I want to replace < country >Germany< / country > with < country >Germany< / country > in the table PRX. PROX_REQ and the request_body column
... < / Street > < Home > 6 number < / house number > < Zip > 21073 < / Zip > < City > Hamburg < / City >< country > Germany < / country >< / ShipToAddress > < HardwareRequired > true < / HardwareRequired > < DeliveryIdentificationType > STANDARD < / DeliveryIdentificationType > < / HardwareDetails > < SendContractDocument > false < / SendContractDocument > < SendConfirmationFax > false < / SendConfirmationFax > < TrustedShopAmount > < amount > 7.49 < / amount > < currencies > EUR < / currency > < / TrustedShopAmount >...
Help, please...
Thanks in advance

Update...

Set request_body = regexp_replace (...), status = 'OK '.

When status = ' ERROR and request_body_like «...» »

Regexp_replace help

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