Regexp_substr help

Similar Questions

• Regexp_substr helps.

  Hello
  

  with t as
  (select 12857854 id from dual union all
  select 1573321 id from dual union all
  select 95634321 id from dual)
  select * from t

  Could you people please help me at first characters little except the last 3 digits of the data using the ordinary exprassion.
  
  Output

  12857
  1573
  95634

  Thank you.

  Like this

  with t as
  (select 12857854 id from dual union all
  select 1573321 id from dual union all
  select 95634321 id from dual)
  select id, regexp_replace(id, '[[:digit:]]{3}$') from t
  

• helps the regexp_substr

  Hi all
  
  I've got statement
  

  with example as 
  (select '4578, WILD STORM, S.TURK-DELIGHT, 44' as str from dual)
  select case when regexp_substr(str,'S.[[:alpha:]''-]*,') is not null
                   then regexp_substr(str,'S.[[:alpha:]''-]*,') 
                   else '' 
           end as str 
  from example

  It returns 'STORM', while I need 'S.TURK - DELIGHT.
  
  Help me to get the desired result.

  Don't forget '. ' is a wildcard character, so you must escape a------.

  HTH

• 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

• Regexp_substr | starting the regex need help

  Hello

  I need where to go a single string that contains the full qualified table name [schemaname.tablename] I need to break schemaname then tablename

  Schema_name	Table-name
  SCHEMA1	TABLE1:
  SCHEMA2	TABLE2
  SCHEMA3	TABLE 3
  SCHEMA4	TABLE 4

  : SCHEMA_NAME = ' SCHEMA1. TABLE1, SCHEMA2. TABLE 2, SCHEMA3. TABLE3, SCHEMA4. TABLE 4 '

  So far I've written it, Yes, there is much to learn through.

  Please comment to let me know what I missed and how to overcome these errors

  SELECT RTRIM (REGEXP_SUBSTR (: SCHEMA_NAME, '\w+\.', 1, LEVEL),'.) ') AS SCHEMA_NAME

  , RTRIM (REGEXP_SUBSTR (: SCHEMA_NAME, '\w+\,+', 1, LEVEL), ',') AS TABLE_NAME

  OF THE DOUBLE

  CONNECT REGEXP_SUBSTR (: SCHEMA_NAME, ' [^,] +', 1, LEVEL) IS NOT NULL

  SELECT REGEXP_SUBSTR (: SCHEMA_NAME, '\w+', 1, LEVEL * 2-1) AS SCHEMA_NAME

  , REGEXP_SUBSTR (: SCHEMA_NAME, '\w+', 1, LEVEL * 2) AS TABLE_NAME

  OF THE DOUBLE

  CONNECT REGEXP_SUBSTR (: SCHEMA_NAME, ' [^,] +', 1, LEVEL) IS NOT NULL

  You were close. Just take the fitting level calculation

• help required in regexp_substr

  Hi all

  "In fact I want to cut the BHK' 1' of ' villa 1 BHK.

  I can separate the 'villa' of ' 1 BHK Villa"in

  SQL > select REGEXP_SUBSTR ('1 BHK Villa ',' [^ ""] +' 1, 3) from DUAL;

  REGEX

  -----

  Villa

  "" but please could you me how can we separate ' 1 BHK ' of ' villa 1 BHK.

  Concerning

  Rajat

  GregV wrote:

  Hello

  If you want to take the last piece of text (of the last space at the end of the string):

  SELECT REGEXP_SUBSTR ('1 BHK Villa', '(^.+) [[: space:]]. + $', 1, 1, 'i', 1) FROM DUAL;

  Two ways, I prefer this:

  1. SELECT

  2 REGEXP_replace (1 BHK 'Villa'

  3                 ,' [^ ]+$'

  4                        ) s

  5 * OF THE DOUBLE

  >/

  S

  -----

  1 BHK

  1. SELECT

  2 substr ('1 BHK Villa'

  3          ,1

  4, instr ("1 BHK Villa ',' ', - 1")-1

  5              )

  6              s

  7 * OF THE DOUBLE

  >/

  S

  -----

  1 BHK

• How to remove the REGEXP_SUBSTR space

  Choose the length (TRIM (REGEXP_SUBSTR ('V, 370498, 658561477, 20150917, 4.37

  (((' [^,] +' 1, 5))) of the double

  output

  ---

  9

  Select TRIM (REGEXP_SUBSTR ('V, 370498, 658561477, 20150917, 4.37

  ((' [^,] +' 1, 5)) of the double

  output

  -----

  20150917

  
  

  last value actual length 8 only

  can help me please for these

  Thank you

  Thanks for the help

  to replace suggested used

  fix

  Choose the length (regexp_substr (regexp_replace ('V, 370498, 658561477, 20150917, 4.37

  (' [[: space:]] *', ")," [^,] (+' 1, 5)) solution, "

  length (rtrim (regexp_substr ('V, 370498, 658561477, 20150917, 4.37

  ', '[^,]+', 1, 5)

  10) has contributed to

  Double;

  Hi odie_63

  I replied to your previous one (csv to xml using ult_file pl/sql conversion ), it is also very good

  Thank you...

• REGEXP_SUBSTR returns the second group

  (occuranHi! Need help. I'm trying to use regexp_substr to get the channel number. But Oracle returns me NULL when I choose the second game.

  INPUT STRING:

  Line 255 #: 05:59:51.050639000: startup process filling 188234978

  When I choose the first (default) game - it returns:

  start the process of filling 188234978

  SQL:

  Select REGEXP_SUBSTR (' #255 line: 05:59:51.050639000: start filling out the process 188234978', '(start process fill) (\w+)',1,2( )

  de double

  
  

  Why?

  
  

  Database Oracle 11.2.0.4.0 x 64
  

  In my opinion, there are two groups:

  1 # start process filling 188234978

  2 # 188234978

  Yes, there are two groups, but that one game.

  Do not confuse the number of occurrence of a pattern for the number of a corresponding model group.

  It looks more like what you are looking for:

  select regexp_substr(
           'Line #255: 05:59:51.050639000 : start process fill 188234978'
         , 'start process fill (\d+)'
         , 1
         , 1     --< first occurrence of the pattern
         , null
         , 1     --< there you can choose the group in the matching pattern
         )
  from dual ;
  

  However, there are probably more effcient ways of doing, without regex.

• Please help me find the solution for the query

  Hi Experts,

  Please help build a sql query. Thank you

  Examples of data

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

  create the table Material_tb

  (

  Detail varchar2 (20).

  Description varchar2 (200)

  )

  /

  Start

  Insert into material_tb values('Color','Red,Blue,Black,Green,White');

  Insert into material_tb values ('Material','Gold, Silver, Platinum');

  end;

  /

  Select * from material_tb;

  DETAIL DESCRIPTION

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

  Color red, blue, black, green, white

  Material gold, silver, Platinum

  I want that output voltage

  DETAIL DESCRIPTION

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

  Red color

  Blue color

  Black color

  Green color

  White color

  Material gold

  Silver material

  Platinum material

  You can try under sql

  select distinct detail,regexp_substr(description,'[^,]+',1,LEVEL)
   from material_tb
  connect by regexp_substr(description,'[^,]+',1,LEVEL) is not null
  order by 1
  

• regexp_substr: retrieve a block of text between delimiters

  Hello

  regexp_substr is very powerful, but not so easy to understand. I'll try to do my best, but I need your help.

  My question is this: I have a multiline text, stored in a database field. The text is structured as a "windows.ini" file: there are sections in square brackets and the lines of text in the section.

  I want to retrieve all of the text in a given section, and I think it's possible using regexp_substr.

  Here is the text of the field (in bold the part I want back):

  ---

  [ARTICLE 1]

  AAA = 123

  BBB = 456

  CCC = 789

  [ARTICLE 2]

  DDD = 987

  EEE = 654

  ---

  In other words, is there a way to retrieve the text between [ARTICLE 1] and [ARTICLE 2]?

  Thank you in advance.

  Concerning

  Select

  regexp_substr (txt, ' [de] [^] +', 1, level * 2) s

  t

  connect by level<= regexp_count(txt,="">

• regexp help

  Hello

  I need to extract deposited number. Can please help me this extract using regexp.

  for ex:

  at the bottom of the message, I need to extract the value 434234234235435345634345 using REGEXP

  ORA-20102: cw_upload_queue_item_meta | ec3_unique_file_ref:434234234235435345634345 | ORA-00001: unique constraint (IBIS. CW_UPLOAD_QUEUE_ITEM_META_PK) violated

  user575115 wrote:

  Thx.It works in 11g, 10g too error throw

  That's why you have to mention your db version 4-digit during the validation of the questions:

  Select

  LTRIM (regexp_substr ('ORA-20102: cw_upload_queue_item_meta | ec3_unique_file_ref:434234234235435345634345 |)) ORA-00001: unique constraint (IBIS. ((CW_UPLOAD_QUEUE_ITEM_META_PK) violated ',': [^ |] +', 1, 2),': ')

  f

  of the double

• REGEXP_SUBSTR I need any word val, including space, or null

  Hi all

  I need to separate the last value in the field, but my query spaces not considered between the delimiters, please help thanks...

  SELECT RTRIM (REGEXP_SUBSTR ('C | 00013010101 |)) C00009 | 009402 | 00000100.000 | 1. AZUSDINE 500 MG TABLET | AZULFIDINE | 00016010101 | 1. 19950118 | 20150330 | 00000000 | 00000000 | 0 | F | 9. 1. 1. 0 | 1. 1. AB | 00000000 | 1. 0 | 000 | 0 | 0000001 | 00,000 | 00,000 | 00,000 | 0000012 | 0000012 | AB | TAB | 00000001.000 | 19900930 | 19900930 | 00000000 | 1. 19950701 | 2. 0 | BOTTLE | 0 | 20080925 | 20110928 | 0 | 0 | 0 | AB | 00000100.000 | 0 | 00 | AB | 2. 2. 9. 3. 1. I have | AZULFIDINE 500 MG TABLET '

  (' [a-zA-Z0-9] +--------|?', 1, 72),' | ') Double val

  output

  VAL

  -----

  AZULFIDINE

  Mandatory

  PUT out, VAL: AZULFIDINE 500 MG TABLET

  Thank you

  Olivier

  If the number of fields is fixed, if you always want (in this case) the 69th field, you could do that (I replaced your channel hardcoded with a column called "input_string" for clarity):

  SELECT RTRIM (REGEXP_SUBSTR (input_string |'|)) ',' [^ \ |] *\|',1,69),'|') Val from your_table

  If you want just the entire last section, you can do it without using regular expressions at all:

  SELECT SUBSTR (input_string, INSTR(input_string,'|',-1) + 1) val from your_table

• REGEXP_SUBSTR simple query

  Hi all

  Select RTRIM (REGEXP_SUBSTR('675809532|11787148|0','[a-zA-Z0-9]+\|'),'| ') double cardid

  Output

  --------

  675809532

  form above query I had the first value 675809532 how can I second valueor third value only pls help for these

  I need to pass the value to the variable individally

  Thank you...

  By adding the occurece to the RegExp as:

  Select RTRIM (REGEXP_SUBSTR ("675809532 |")) 11787148 | 0','[a-zA-Z0-9] + \| ?', 1,2),'| ') the double cardid

  and the regular expression would be easier that way:

  Select REGEXP_SUBSTR ('675809532 | 11787148 | 0', ' [^ \ |] +' 1, 3) double cardid

  HTH

• using regexp_substr

  Hello

  I need to extract the first five characters digital of the specified column. I'm able to reach the requirement using substr and definition of functions.

  I learned that the use of regular expressions we can achieve in a simple way. Please find below the test data.

  with t as (select '123456suri' double empid

  Union all select 'suri436789' of the double

  Union all select '521521suri' of the double

  Union select all 'suri785643abc' of the double)

  select * from t ;

  Desired output:

  EmpID

  ---------

  12345

  52152

  Please help me how to extract the results required using REGEXP_SUBSTR.

  I tried with query below, but not able to restrict the result to only five digits.

  Select double regexp_substr('3456789suri123','[0-9]+',1,1)

  Thank you

  Suri ;-)

  Post edited by: Suri - added query

  Good first try

  \d is a shortcut for [0-9]

  What happens when you try to \d\d?

  How about \d\d\d?

  You should be able to guess the rest of there

  You can also use {n} means "repeat exactly n times.

• REGEXP_SUBSTR - how to use case-sensitive

  I use REGEXP_SUBSTR with case-insensitive option

  while I use as below

  Select REGEXP_SUBSTR ("uk123 research", "UK", "i") double column;

  It displays error: ORA-01722: invalid number

  Help, please...

  Oracle 11.2.0.1.0

  You provide all the required parameters. See the http://docs.oracle.com/cd/B28359_01/server.111/b28286/functions138.htm#SQLRF06303 documentation

  Here's the correct query:

  select REGEXP_SUBSTR('search for uk123', 'UK',1,1,'i' )  col_name from dual;