Extract values with regular Expression

Similar Questions

• Mask a number with regular expressions

  Hi @ll!
  
  Is it possible to hide a given number
  
  of "12345678" in "XXXX5678".
  or '987458' to 'XX7458 '.
  
  with Regular Expressions and without substr()? To display only the last four digits and the 'X' value for the rest. The size of the number is not always the same.

  Alex, I think the OP wanted the first 4 characters to hide ;)

  Something like that?

  select translate(substr('12345678',1,length('12345678')-4),'1234567890','XXXXXXXXXX')
                  || substr('12345678',length('12345678')-3)
            from dual
  /
  TRANSLAT
  --------
  XXXX5678
  

  More

  with test_data as
  (
  select '12345678' card_no from dual
  union all select '123456' from dual
  union all select '5678900' from dual
  ) -- End of test data
  select translate(substr(card_no,1,length(card_no)-4),'1234567890','XXXXXXXXXX') ||
            substr(card_no,length(card_no)-3)
      from test_data
  /
  
  TRANSLATE(SUBSTR
  ----------------
  XXXX5678
  XX3456
  XXX8900
  

  Arun-

• HELP WITH REGULAR EXPRESSIONS

  Hello
  I'm just experimenting with regular expressions
  
  My information in table are as shown below
  

  TABLE NAME: -  EMP_TEST
  
  STRUCTURE WITH DATA
  
    X_NAME
  ---------------------
  12456com
  ab1245com
  AXM4554.com

  Now, I pulled the following query to find the rows that have 1
  

  select *
  from emp_test
  where regexp_like (x_name,'[1]')
  
  RESULT AS EXPECTED : -
  
  
    X_NAME
  ---------------------
  12456com
  ab1245com

  Now, I want to do a query that will give me tose lines that do not have 1
  
  Initially I have run the following query
  

  select *
  from emp_test
  where regexp_like (x_name,'[^1]')
  
  RESULT NOT AS EXPECTED : -
  
    X_NAME
  ---------------------
  12456com
  ab1245com
  AXM4554.com

  After a bit of searching on Google I found an application that works as expectd
  

  select *
  from emp_test
  where regexp_like (x_name,'^[^1]*$')
  
  RESULT AS EXPECTED : -
  
    X_NAME
  ---------------------
  AXM4554.com

  Where my basic question is that all seeking a 1 in a row we do not put the beginning ^ end $ and so on and the
  Works of query.
  So why do [^ 1] does not work as shown, while excluding an amd why is ^ [^ 1] * $ necessary while excluding a

  Hello

  Elessar wrote:
  So why do [^ 1] does not work as described

  Because regexp_like(str,'[^1]') means: search str who have at least one character that is not 1.
  When regexp_like(str,'^[^1]*$') means: search str done characters that are not 1 start (^) at the end ($)

• Need help with regular expressions

  Hi all
  
  I need your help, because I have no ideas more...
  
  I have the following problem: in the column of the database table, I have the string with the names of files already uploaded to the database. For example: + File1_V01.txt +, + File1_v02.txt +, + File1_01_v01.txt +, etc. The string _vxx * + or _Vxx * + (non-case sensitive) represents the version of the file.
  
  Now the problem: I'll upload the file with the name + File1_v02.txt + (already exists in the table).
  If the file name already exists in the table the pl/sql function should get the name of the file with the following version number. In my case it takes + File1_v03.txt +.
  
  Is it possible to do this using SELECT with regular expressions?
  
  Best regards and thanks!

  with t as (
    select 'File_V05.txt' fn from dual union all
    select 'File_V04.txt' fn from dual union all
    select 'File2_v03.doc' fn from dual union all
    select 'File2_v115.doc' fn from dual union all
    select 'File2_v15.doc' fn from dual union all
    select 'File1_v03.doc' fn from dual union all
    select 'File1_v115.doc' fn from dual union all
    select 'File1_v999.doc' fn from dual union all
    select 'File2.doc' fn from dual union all
    select 'File2_v05.doc' fn from dual union all
    select 'File1_v01.txt' fn from dual union all
    select 'File1_v02.txt' fn from dual union all
    select 'File1_v1.txt' fn from dual union all
    select 'File1_v1.doc' fn from dual union all
    select 'File1_v2.txt' fn from dual union all
    select 'File2_v01.doc' fn from dual union all
    select 'File2_v02.doc' fn from dual union all
    select 'File1_ABC_v01_DEF.docx' fn from dual union all
    select 'File1_ABC_V02_ABC.docx' fn from dual union all
    select 'File1_ABC_v01_12_04_17.docx' fn from dual union all
    select 'ABC_V1_QWERT.pdf' fn from dual
  )
  
  select fn,
  case when fn!=fn_new then
   last_value(fn_new)
   over(partition by regexp_replace(upper(fn),'V[[:digit:]]+','') --(.*?V0*)([1-9]+)(\..*?)$
   order by nv
   rows between unbounded preceding and unbounded following
   )
  else fn
  end fn_new
  from (
      select
      case when v-1 <= 0 then fn
      else
             regexp_replace (fn,
                      '(_v|_V)(\d*)',
                      case
                      when length(substr(fn,v+1,p-v-1)+1) > (p-v-1)
                      then '\1'||to_char(substr(fn,v+1,p-v-1)+1)
                      else '\1'||lpad(substr(fn,v+1,p-v-1)+1,p- v-1,0)
                      end
             )
      end fn_new
      ,fn
      ,case when v-1 <= 0 then -1
       else
          substr(fn,v+1,p- v-1)+1
       end nv
      from (
          select fn, regexp_instr(upper(fn),'_V[[:digit:]]+',1,1,1) p, instr(upper(fn),'_V')+1 v from t
      )
  )
  order by fn
  
  FN     FN_NEW
  ABC_V1_QWERT.pdf     ABC_V2_QWERT.pdf
  File_V04.txt     File_V06.txt
  File_V05.txt     File_V06.txt
  File1_ABC_v01_DEF.docx     File1_ABC_v02_DEF.docx
  File1_ABC_v01_12_04_17.docx     File1_ABC_v02_12_04_17.docx
  File1_ABC_V02_ABC.docx     File1_ABC_V03_ABC.docx
  File1_v01.txt     File1_v3.txt
  File1_v02.txt     File1_v3.txt
  File1_v03.doc     File1_v1000.doc
  File1_v1.doc     File1_v1000.doc
  File1_v1.txt     File1_v3.txt
  File1_v115.doc     File1_v1000.doc
  File1_v2.txt     File1_v3.txt
  File1_v999.doc     File1_v1000.doc
  File2.doc     File2.doc
  File2_v01.doc     File2_v116.doc
  File2_v02.doc     File2_v116.doc
  File2_v03.doc     File2_v116.doc
  File2_v05.doc     File2_v116.doc
  File2_v115.doc     File2_v116.doc
  File2_v15.doc     File2_v116.doc
  

• Problem with regular Expression

  Hello!!
  I have a problem with the regular expression. I want to validate only one word, and second are the same. To do this, I wrote a regex
  
  Model p=Pattern.compile("([a-z][a-zA-Z]*)\\s\1");
  Matcher m = p.matcher ("nikhil nikhil");
  Boolean t = m.matches ();
  If (t)
  System.out.println ("it's a game");
  on the other
  System.out.println ("is no match);
  
  The result I get is always 'there no match. "
  
  Your timely help will be very appreciated.
  
  Concerning

  Hello.

  You are missing a slash in the regex

  Pattern p = Pattern.compile("([a-z][a-zA-Z]*)\\s\\1");
  Matcher m = p.matcher("nikhil nikhil");
  boolean t = m.matches();
  if (t) {
      System.out.println("There is a match");
  } else {
      System.out.println("There is no match");
  }
  

• Need help with regular expression

  I have a string like: separation by a comma as G, H, L
  
  the table has values like this
  
  Col1 Col2
  ROW1 G
  ROW2 F
  ROW3 L
  
  What is trying to achieve is to find out if G or H or L in the string of separated by commas are in col2. is it possible by using regular expressions or not we do split and the loops?
  
  Thanks in advance.

  Hello

  ora1001 wrote:
  I have a string like: separation by a comma as G, H, L

  the table has values like this

  Col1 Col2
  ROW1 G
  ROW2 F
  ROW3 L

  What is trying to achieve is to find out if G or H or L in the string of separated by commas are in col2. is it possible by using regular expressions or not we do split and the loops?

  Thanks in advance.

  You don't even need regular expressions

  INSTR ( ',' || col1 || ','
        , ',' || col2 || ','
        )
  

  will be greater than 0 if (and only if) co12 is one of the elements of col1.

  If you're curious, a way of using regular expressions is

  REGEXP_LIKE ( col1
           , '(^|,)' || col_2 || '($|,)'
           )
  

  but it will be less effective than Instr.

  Published by: Frank Kulash, November 5, 2010 11:38

• Validation with Regular Expressions

  I am validating data using regular expressions.
  Here is a copy of the code.
  
  < cfinput type = "text" name = "FQID_ #ATTRIBUTES. "FQID #_officer" width = "36" validate = 'regular_expression' pattern = "^ \bna\b$ |" ^ \d * [0-9](|. \d*[0-9]|,\d*[0-9])?$» required="#ATTRIBUTES.required#» activé = «Oui» message = «0,0 à 100 ou na, requis» value = «#ATTRIBUTES.) "FQID_Response.Officer #" style = "textAlign:right"; onchange = "removeDash (FQID_ #ATTRIBUTES. FQID #_officer); "/ >
  
  I don't get any errors, but I'm still able to I want to enter in the text box. I try to limit to a number between 0 and 100 and excludes the decimal amount. It must also accept na if not applicable. As it is now, I can enter anything.

  Thanks, that explains it.

• Need help with regular Expression (RegEx)

  Try to wrap your head around a regular Expression for the following format example: 0022-C-4452 OR 0022-C-4452-C

  
  

  * The 4 digits are always numbers
  

  * The last 5 digits are alpha numeric

  * Last (if used) digit is always 'C' (in reference to the second structure)

  Hold the dashes for "auto fill" if possible? This would be in the Custom Format? Sequence of keys? Or Validation? I appreciate any help!

  I still think that you did not correctly describe your problem.

  1 figures of 4 characters.

  Optional separator.

  Then 6 alphanumeric characters and ' - '.

  OR

  1 figures of 4 characters.

  Optional separator.

  Then 6 alphanumeric characters and ' - '.

  Optional separator.

  Character 'C '.

  Note that the "-" is not a number and not an alphabetic character. It is a white space character.

  Try:

  function {MyRe (cString)
  var cFormatted = "";
  var RE_MyCode0 = /^(\d{4})[-.]) {0,1} ([A-Za-a0 - 9-] {6}) $/;
  var RE_MyCode1 = /^(\d{4})[-.]) {0,1} ([A-Za-a0 - 9-] {6}) [-.] {0,1} ([C]) $/;
  If (RE_MyCode0.test (CString) == true) {}
  cFormatted = RegExp. $1 + '-' + RegExp. $2;
  }
  If (RE_MyCode1.test (CString) == true) {}
  cFormatted = RegExp. $1 + '-' + RegExp. $2 + '-' + RegExp. $3;
  }
  Return cFormatted;
  } / / end of MyRe function;

  some tests;
  var MyString = "0022-C-4452; good channel;
  Console.println ("Input:" + MyString);
  Console.println ("result:" + MyRe (MyString));

  var MyString = "0022-C-4452-C; string of Goo;
  Console.println ("Input:" + MyString);
  Console.println ("result:" + MyRe (MyString));

  var MyString = "A022-C-4452" / / bad string;
  Console.println ("Input:" + MyString);
  Console.println ("result:" + MyRe (MyString));

  var MyString = '0022-4452-CZ' / / bad string;
  Console.println ("Input:" + MyString);
  Console.println ("result:" + MyRe (MyString));

• Grouping and backreferences with regular expressions on the window to replace the text

  I'm really appreciate the inclusion of regular Expressions in the search and replace functionality. One thing miss me that East of backreferences in the replacement expression. For example, in unix tools vi or sed, I could do something like this:
  

   s/\(firstPart\) \(secondPart\) \(oldThirdPart\)/\2 \1 newThirdPart/g 

  that allow me to switch the places of first and secondPart and substitute totally thirdPart. If grouping and backreferences are already present in the window replace text, how do you properly call them?
  
  Published by: Justin.Warwick on August 23, 2011 08:26

  You can vote on the request for this to the exchange of SQL Developer, to add weight to the implementation as soon as possible: https://apex.oracle.com/pls/apex/f?p=43135:7:3693861354483465:NO:RP, 7:P7_ID:16761

  Kind regards
  K.

• Mask creditcard number with regular expressions

  Hi @ll!
  
  Is there a way to hide a number of credit card of
  
  "1234-1324-1234-1234" to "1234-1324-1234-XXXX.
  
  using regular expressions without using substr?
  
  The following works on a .net Application
  
  regexp_replace (ccnum, "[0-9] (?)") =. {} 4})', 'X')
  
  but doesn't seem to work in a sql statement.
  
  Best regards.
  
  Published by: m8r-qbkka9, November 16, 2009 05:11

  Alex Nuijten wrote:

  regexp_replace (str, '(-[[:digit:]]{4})', '-XXXX') rstr
  

  http://forums.Oracle.com/forums/Ann.jspa?annID=719

  Oops, I meant:

  regexp_replace (str, '([[:digit:]]{4}-)', 'XXXX-') rstr
  

• Please help with regular expression

  Hello
  
  With the help of my previous answers of the detachment, Re: procedure to extract several substring of a string , I've updated the query. But I don't get the answer you want in all cases. Could you please help me? My query is based on the previous announcement. Any other way to do this?
  I'd really appreciate it.

  select 
         ltrim ( regexp_substr(txt, '\[(\w+)', 1, level), '[')      as id, /* id is number */
         ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '\w+-*\d*', 1, 1), ':')  as qid, /* Qid could be char/number/space any combination except ':' */
         ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '\w+', 1, 2), ':')      as num,
        to_date( ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '[^:]+', 1, 3), ':'),'MM/DD/YY')   as effdate
  from  (
                          select  '[10946:M100:N:][10947:Q1222:N:][38198:PPP-2:N:][13935:PPP-6:N:][38244:QQQ-4:Y:01/01/10]'     as txt
                          from     dual
           )
          connect by level <= length(regexp_replace(txt, '[^[]'));

  I should get:

  ID             QID          NUM         EFFDATE
  10946     M100     N     
  10947     Q1222     N     
  38198     PPP-2     N     
  13935     PPP-6     N     
  38244     QQQ-4     Y     01-JAN-10

  But, be
  

  ID             QID          NUM          EFFDATE
  10946     M100     N     
  10947     Q1222     N     
  38198     PPP-2     2     
  13935     PPP-6     6     
  38244     QQQ-4     4     01-JAN-10

  Thank you

  Hello

  If the column number is wrong, isn't it?
  Describe what should be the num column. For example "num is the 3rd part of :-delimited list placed in square brackets.

  If this is what you want, and then change the definition of number of

  ...                     ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '\w+', 1, 2), ':')      as num,
  

  TO

  ...                      REGEXP_SUBSTR  ( REGEXP_SUBSTR ( txt
                                        , '[^]]+'
                                              , 1
                                     , LEVEL
                                     )
                           , '[^:]+'
                           , 1
                           , 3
                           )       AS num,
  

• Kind of cool thing I noticed with regular expressions

  SELECT REGEXP_SUBSTR (' first pitch, second, third field, ',' [^,] *,', 1.2 ")

  DOUBLE;

  Looks like it's supposed to return:, third field.

  but I'm null

  try to run to another position later than 1

  SELECT REGEXP_SUBSTR (' first pitch, second, third field, "," [^,] *,',14, 1)

  DOUBLE;

  output:, third field.

  Now try adding 2 comma after "second field" while retaining the original code

  SELECT REGEXP_SUBSTR ("first pitch, second, third field, ',' [^,] *,',1, 2")

  DOUBLE;

  Released:third field (as opposed to null)

  I think here that oracle treats , the second field,

  as first occurrence and leaves it there, so in fact there is no second occurrence because we have no comma to start the second occurrence.

  The example comes from a manual I think the author meant that there is to be occurrence 2. I just thought it was sloppy. Maybe I'm wrong on my statement

  "I think here that oracle treats the second field .

  as first occurrence and leaves to whom, so actually there is no second occurrence because we have no comma to start the second occurrence. »

  ... but it seems OK for me.
  

  other thoughts?

  Hello

  2776946 wrote:

  ...

  I think here at Oracle treats, second field.

  as first occurrence and leaves it there, so in fact there is no second occurrence because we have no comma to start the second occurrence.

  ...

  Exactly!  Occurrences do not overlap.  Looking for a model that starts and ends with a comma 2 comma by disaster.  You would need 4 commas in the chain to get a 2nd accident.  If the comma after "second field" is part of the 1st appearance, then it can also be part of the 2nd event.

• Doubt with regular Expression

  Hi all,
  
  How can I get the result using regexp_replace and regexp_substr?
  

  with tab as
  (
  Select 'TEST( XX_XXXXXX, 12 ) ; AAAA'  txt from dual union all
  Select 'TEST( AAAAAAAAA , 67 ); 1234'  txt from dual union all
  Select 'TEST( 92233 ,   47   ); 5234'  txt from dual union all
  Select 'TEST( AAAAAAAAA , AA ); 897'   txt from dual union all
  Select 'TEST( CCCCC 25 );'             txt from dual union all
  Select 'TEST CCCCC, 45 );'             txt from dual union all
  Select 'TEST( EDCCCC, 45 ;)'           txt from dual union all
  Select 'TEST(BBBBBBBBB,12);'           txt from dual
  )
  --
  Select regexp_substr(txt, '[^.*,. *][[:digit:]][^\).*$]') from tab
  /
  
  --
  REGEXP_SUBSTR(TXT,'[^.*,.*][[:
  ------------------------------
  12
  67
  922
  897
  25
  45
  45
  
  8 rows selected

  Mandatory Criteria
  
  "TEST" And "(" And "," And ")" And ";"

  expected result
  12
  67
  47
  Null
  Null
  Null
  Null
  12

  Kind regards

  Something like:

  with tab as
  (
  Select 'TEST( XX_XXXXXX, 12 ) ; AAAA'  txt from dual union all
  Select 'TEST( AAAAAAAAA , 67 ); 1234'  txt from dual union all
  Select 'TEST( 92233 ,   47   ); 5234'  txt from dual union all
  Select 'TEST( AAAAAAAAA , AA ); 897'   txt from dual union all
  Select 'TEST( CCCCC 25 );'             txt from dual union all
  Select 'TEST CCCCC, 45 );'             txt from dual union all
  Select 'TEST( EDCCCC, 45 ;)'           txt from dual union all
  Select 'TEST(BBBBBBBBB,12);'           txt from dual
  )
  select  txt,
          case
            when regexp_like(txt,'^TEST\(.+, *\d+ *\) *;.*$') then regexp_replace(txt,'(^TEST\(.+, *)(\d+)( *\) *;.*$)','\2')
          end expected_result
    from  tab
  /
  
  TXT                          EXPECTED_RESULT
  ---------------------------- ---------------
  TEST( XX_XXXXXX, 12 ) ; AAAA 12
  TEST( AAAAAAAAA , 67 ); 1234 67
  TEST( 92233 ,   47   ); 5234 47
  TEST( AAAAAAAAA , AA ); 897
  TEST( CCCCC 25 );
  TEST CCCCC, 45 );
  TEST( EDCCCC, 45 ;)
  TEST(BBBBBBBBB,12);          12
  
  8 rows selected.
  
  SQL> 
  

  SY.

• Number of shaped with preg_replace Regular Expression and PHP

  Hello

  I would like to add a 'dash' after every 3 digits in a given number (10 digits). For example, 9785678941 became 978-567-894-1. How could I achieve this with regular expression using PHP preg_replace?

  Thank you.

  The next solution is based on the example of "The use of backreferences followed literals digital" published on the php.net site.

  In accordance with the $string, $pattern, $replacement nomenclature which is the php.net example use, here´s my modification:

  <>

  $string = '9785678941';

  $pattern = ' / (\\d{3})(\\d{3})(\\d{3})(\\d{1)} /';

  $replacement = ' ${1}-{2}-${3}-${4}';

  echo preg_replace ($pattern, $replacement, $string);

  ?>

• Allow specific characters - Regular Expression

  Hello everyone

  I am new to the regular expression and I have a very simple question. I use the function "read from the text file" to load a file delimited by tabs with 3 columns in my VI. Then, the string is converted to table and I use the values.

  Nevertheless, I would like to develop a "filter" that allows only digits (0-9), colon, comma , and point to strings.

  Using the function "matches regular expression", I tried a regular expression like this:

  [^ 0-9] | [^\]. [|^:]| [^,]

  But it does not work.

  Could someone help me with this problem?

  Thank you

  Dan07

  Use search and replace with regular Expression String selected.