HELP WITH REGULAR EXPRESSIONS

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• 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
  

• 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));

• 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

• 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,
  

• 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-

• 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");
  }
  

• have dell dimension 4700 xp need help with outlook express I have error message and can not send email

  I need help to figure out why my outlook express does not work when I try to send a link, or email. I have a dell dimension 4700 xp restored. The error message cannot find the server with an error code number. I am a newbee so one who responds to it will have to break it down for me. So far, I had lots of help from proceeding to audio and graphics, and I am totally grateful for the help I get.

  Thank you, Gina Davis

  I have

  He has already worked with Outlook Express? Using OE? Hotmail is a Webmail service.

  How to add your Hotmail e-mail account to Microsoft Outlook Express
  http://windowslivehelp.com/solution.aspx?SolutionID=99d4b13d-13dB-40D8-9cdf-172002d4194c

  If you don't want to use OE to access Hotmail and want to just use IE to access Hotmail, then just copy the link and paste it into a new message from Hotmail.

• Help in regular Expression for the beaches of limitation

  Hi, I'm working on the provision of a text field is limited to dates, it's just a part of the code. I already have the validation of the dates, but I am now limiting what the user enters using a regular expression. This code works a little however, it does not limit me for example I can enter more than 2 digits, but then he limits based on the total amount allowable so for example 8 digits are allowed if I just type. I need to stop after 2 digits then have a - then 2 other numbers then one - and then followed by 4 digits. I tried to limit each section and grouping as well. Any help would be greatly appreciated. Thank you.

  It is in the format code and I am the appellant in the key sequence.

  function DateKS () {}

  var value = AFMergeChange (event);

  If (! event.willCommit) {}

  Allow only characters that match the regular expression

  Event.RC = /^([0]{0,1}[1-9]{0,1}|[_1]{0,1}[012]{0,1}) ([-] {0,1}) ([0] {0,1} [1-9] {0,1} |) [12] {0,1} [0-9] {0,1} | ([3] {0,1} [01] {0,1}) ([-] {0,1}) ([0-9] {0,4}) $/ .test (value);

  }

  }

  I decided that control for 100 and 400 was not necessary because this event does occur that all 400 years. But I'm working on it further and changed even more. Here is my code to work.

  function isLeapYear (year) {}
  year return % 4 = 0;
  }

  function checkDaysInMonth (day, month, year) {}
  daysInMonth var = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];

  If (month = 2) {}
  If (isLeapYear (year)) {}
  daysInMonth [1] += 1;
  }
  }

  return daysInMonth [month - 1] > = day;
  }

  function checkDateFormat (dateStr) {}
  var errorMsg = ",
  maxYear = (new Date()) .getFullYear (),
  minYear = maxYear - 1,.
  match = dateStr.match(/^(\d{2})-(\d{2})-(\d{4})$/),
  months,
  day,
  year;

  If {(matches)
  month = parseInt (matches [1], 10);
  day = parseInt (matches [2], 10);
  year = parseInt (matches [3], 10);

  If (month < 1="" ||="" month=""> 12) {}
  errorMsg = "invalid value for the month: ' + matches [1];"
  } ElseIf (day = 0) {}
  errorMsg = "invalid value for the day:" + match [2];
  } else if (! checkDaysInMonth (day, month, year)) {}
  errorMsg = "number of days for invalid month: ' + match [2];"
  } ElseIf (year < minyear="" ||="" year=""> maxYear) {}
  errorMsg = "invalid value for the year:" + match [3] + "-must be between" + minYear + "and" + maxYear;
  }
  } else {}
  errorMsg = "invalid date format: ' + dateStr + ' \r\nPlease use format: dd-mm-yyyy ';"
  }

  return errorMsg;
  }

  function checkReceivedDate() {}
  var value = AFMergeChange (event),
  errorMsg = ";
     
  ignore control if the value is blank, because this field is not mandatory
  If (! value) {}
  return;
  }

  If {(event.willCommit)
  errorMsg = checkDateFormat (value);

  If (errorMsg) {}
  App.Alert (errorMsg, 0, 0, "error");
  Event.value = ";

  Returns false;
  }
  } else {}
  Allow only characters that match the regular expression
  Event.RC = /^(?:0) [1-9]? 1 [012]?) ? -? ( ? : 0 [1-9] ? | [12] [0-9]? 3 [01]?) ? - ? 2? 0? [0-9] {0,2} $/ .test (value);
  Event.RC = / ^ \d{0,2}-?\d{0,2}-?\d{0,4}$/.test(value);
  }

  Returns true;
  }

• Extract values with regular Expression

  How to extract the values in [] using regular expressions.

  With data As
  (
  Select 'AAAAAA[10] AAA: 19C' Txt From Dual Union all
  Select 'XX[450]-10A' Txt From Dual Union all
  Select '[5]AVC19C' Txt From Dual Union all
  Select 'FVD[120]D2AC' Txt From Dual
  )

  I hope that this return

   10
  450
  5
  120

  Thanks in advance

  Or another:

  SQL> ed
  Wrote file afiedt.buf
  
    1  With data As
    2  (
    3  Select 'AAAAAA[10] AAA: 19C' Txt From Dual Union all
    4  Select 'XX[450]-10A' Txt From Dual Union all
    5  Select '[5]AVC19C' Txt From Dual Union all
    6  Select 'FVD[120]D2AC' Txt From Dual
    7  )
    8  --
    9  select regexp_replace(txt, '^.*\[(.*)].*$','\1')
   10* from data
  SQL> /
  
  REGEXP_REPLACE(TXT,'^.*\[(.*)].*$','\1')
  ---------------------------------------------------------
  10
  450
  5
  120
  
  SQL>
  

  Edit: even though I think it is misleading not escape the right hook, so I prefer:

  SQL> ed
  Wrote file afiedt.buf
  
    1  With data As
    2  (
    3  Select 'AAAAAA[10] AAA: 19C' Txt From Dual Union all
    4  Select 'XX[450]-10A' Txt From Dual Union all
    5  Select '[5]AVC19C' Txt From Dual Union all
    6  Select 'FVD[120]D2AC' Txt From Dual
    7  )
    8  --
    9  select regexp_replace(txt, '^.*\[(.*)\].*$','\1')
   10* from data
  SQL> /
  
  REGEXP_REPLACE(TXT,'^.*\[(.*)\].*$','\1')
  ---------------------------------------------------------
  10
  450
  5
  120
  
  SQL>
  

  just for cleanliness.

  Published by: BluShadow on June 29, 2009 13:35

• 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.

• 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.

• 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
  

• Need help with GREP expressions

  I know - KNOW - there is a way to do it, but keep my GREP expressions.

  I want to add an em dash before any sentence that is in a paragraph style. So I obviously want the em dash to be in the 'start of the paragraph' location. Lack of keep all my GREP searches. Can anyone help?

  Adobe of the 'start of the subsection' implementation is just a little off the coast of standard GREP (which, in itself, is less a set of hard and fast rules but rather as "guidelines"). Yes - you can Search for beginning of paragraph using plain ^, but you cannot use it with a replace operation.

  The trick is to give something to look for in InDesign. Search

  ^.
  

  (it's just - a wildcard just after the start of a paragraph). You don't want to lose the character it finds, so replace it with

  ~_$0
  

  where the first 2 codes is your dash and the last 2 is 'the text together. This is your original found generic character!

  Don't forget to put your paragraph style in the box of formatting to find.

  Of workaround works around the failure of implementation, because indeed, you want to add something to an existing text. Unfortunately, another pretty standard GREP to search for empty paragraphs-^ $-does not work, because then you have no room for a wildcard... (and Yes, you pouvez use a hard return instead of code \r is first or last, but then it will not work at the beginning or at the end of the story). It's one of those things that I wish to see corrected with CS5.

  [Edit] HA! Peter beat me to it, but he has made a typo. In addition, my story is longer.

• Help with Airport Express and Configuration capabilities

  I have a new Airport Express which is set to expand the network wireless (created by Time capsule).  Is it possible to also provide a wired connection to another device in this mode?  If so, how do I need to change the settings? The other device is a hub of lighting (not wireless) which must have an Internet connection to allow remote control of lights.  It is currently connected to the CT but its location does not cover a part of the House, the new AE is the perfect place.

  Is it possible to also provide a wired connection to another device in this mode?

  Yes

  If so, how do I need to change the settings?

  Some settings need to be changed. By default, both the "O" WAN port and port LAN <>- both act as LAN ports. Choose one or both to an active Ethernet connection.

  The other device is a hub of lighting (not wireless) which must have an Internet connection to allow remote control of lights.

  Turning off the hub of lighting, make the Ethernet connection to the AirPort Express and the hub of power upward. You should be in business.