regular expression help

Similar Questions

• Regular expression help please. (extraction of a subset of the string between two markers)

  I haven't used regular expressions before, and I can't find a regular expression to extract a subset of the string between two markers.

  The chain;

  Stuff of header I want
  Stuff of header I want
  Stuff of header I want

  Stuff of header I want
  Stuff of header I want
  Stuff of header I want

  6 ERRORS
                       
  Info I want to line 1
  Info I want line 2
  Info I want line 3
  Info I want to line 4
  Info I want to line 5
  Info I want line 6
  END_ERRORS

  From the string above (it is read from a text file), I try to extract the subset of string between ERRORS 6 and END_ERRORS. The number of errors (6 in this case) can be any number from 1 to 32, and the number of lines I want to extract will correspond with this number. I can provide this number of a caller VI if necessary.

  My current solution, which works, but is not very elegant;

  (1) using Match Regular Expression for the return of the string after you have synchronized the 6 ERRORS

  (2) uses the Regular Expression matches to return all characters before game END_ERRORS of the string returned by (1)

  Is there a way this can be accomplished using 1 Regular Expression Match? If so someone could suggest how, as well as an explanation of the work of the given regular expression.

  Thank you very much

  Alan
  

  

  I used a character class to catch any word or whitespace characters.  This put inside parentheses a substring matching the criteria that you can get by developing the node for regular expression matching.  The \d matches the numbers and the two * s repetition of the previous term.  So, \d* will find the '6', as well as "123456".

• Regular expression help numbers

  Hello
  
  I want an exact match of 9 digits or 12 digits, my query should give "No Match found" because the input value is actually 10 digits
  
  Select case when regexp_like (regexp_replace (' 123 4567 890 ',' '), ' ^ ([0-9] {9}) |) () [0-9] {12}) $')
  then Match "found."
  another "No Match Found"
  end test
  Double;
  
  Need help, I have to do something about something very basic, bad.
  
  Kind regards
  Ash

  Remove the 2 brackets:

  SQL> select case when regexp_like(regexp_replace( ' 123 4567 890', ' ' ), '^([0-9]{9}|[0-9]{12})$')
    2  then 'Match Found'
    3  else 'No Match Found'
    4  end as test
    5  from dual;
  
  TEST
  --------------
  No Match Found
  
  SQL> select case when regexp_like(regexp_replace( ' 123 4567 89', ' ' ), '^([0-9]{9}|[0-9]{12})$')
    2  then 'Match Found'
    3  else 'No Match Found'
    4  end as test
    5  from dual;
  
  TEST
  --------------
  Match Found
  SQL> 
  

• Need help with a regular expression

  I have a Zebra printer string that I want to analyze some information, but I can't get a single regex to do.

  The string is like the following:

  ↑ FT342, 695 ^ A0N, 83, 81 ^ FH\ ^ FDS/N: [[WIDGET]] ^ FS

  ^ FT793, 1170 ^ A0N, 67, 67 ^ FH\ ^ ea FD1. Widget #00123 ^ FS

  ^ FT793, 1170 ^ A0N, 67, 80 ^ FH\ ^ ea FD2. #00456 Widget Deluxe ^ FS

  ^ FT793, 1170 ^ A0N, 67, 90 ^ FH\ ^ FD #0789 ^ FS

  I want what is in red. It will vary from one label to the label. I will use the results to show to the user what label it is printing. Who is blue only varies, if that helps anything.

  Thanks for any help!

  Here would be a way using regular expressions. Again, you have to ignore the first entry and this version will add a blank entry in the last position of the table. Wil you have more entries if your label contains more than what you have posted...

  

• 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;
  }

• Regular Expression search and replace question - please help!

  I was wondering if someone could help me I have a lot of paper with notes like this: [1], but I need them to look like in this ^[1].

  I don't know what to put in the section find [[\d]*] , but I need help with what to put in the field replace to make < sup > numbers]

  In fact, your regular expression is false. What you need is the following:

  (\[\d+\])

  The field replace must contain this:

  ^$1

• 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));

• I need help you rename a file using regular expressions in Bridge.

  Hello

  I work at a University, and we are working through files for our theses and Dissertations. We were renamed to make them more coherent. I wonder if there is a regular expression that could help in this process?

  Examples come from current file names;

  • THESIS H343G 1981
  • Thesis of 1981 g996e
  • THESIS-1981-A543G

  I just need to change the actual file names. how they are formatted.

  Where appropriate on the thesis.

  Hyphens (-) in all white space.

  First letter, last letter is lower case on appeal no (H343g)

  If the list above should look like;

  • Thesis-1981-H343g
  • Thesis-1981-G996e
  • Thesis-1981-A543g

  I've seen people do some pretty cool things with regular expressions! Any help would be greatly appreciated. Thank you!

  You would be better to use a script to do this as an example because I don't think it would be possible in the new name of bridge.

  Using ExtendScript Toolkit or a text editor to copy the code in any event and save it to sub Filename.jsx

  This must be recorded in the appropriate folder. It is located by going to preferences in Bridge, select Startup Scripts, this will open the folder where the script should be saved.

  Once this is done close and restart Bridge.

  Usage: Goto the Tools Menu and select Rename PDF files

  Be sure to only test the code with some files copied to a separate first folder to make sure it's what you want.

  The script will make all PDF files in the selected folder.

  #target bridge
  if( BridgeTalk.appName == "bridge" ) {
  renamePDFs = MenuElement.create("command", "Rename PDFs", "at the end of Tools");
  }
  renamePDFs.onSelect = function () {
  app.document.deselectAll();
  var thumbs = app.document.getSelection("pdf");
  for( var z in thumbs){
  var Name = decodeURI(thumbs[z].spec.name);
  var parts = Name.toLowerCase().replace(/\s/g,'-').match(/(.*)(-)(.*)(-)(.*)(\.pdf)/);
  var NewName = parts[1].replace(/^[a-z]/, function(s){ return s.toUpperCase() });
  NewName += parts[2]+parts[3]+parts[4]+parts[5].toUpperCase().replace(/[A-Z]$/, function(s){ return s.toLowerCase() });
  NewName += parts[6];
  thumbs[z].spec.rename(NewName);
      }
  };
  

• Helps the understanding of regular Expressions

  Hello people,
  
  I need help to understand Regular Expressions.
  

  -- This returns the Expected string from the Source String. ", Redwood Shores,"
  SELECT
    REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                  ',[^,]+,', 1, 1) "REGEXPR_SUBSTR"
    FROM DUAL;
  
  REGEXPR_SUBSTR
  -------------------------------
  , Redwood Shores,
  
  However, when the query is changed to find the Second Occurrence of the Pattern, it does not match any. IMV, it should return ", CA,"
  
  SELECT
    REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                  ',[^,]+,', 1, *2*) "REGEXPR_SUBSTR"
    FROM DUAL;
  
  REGEXPR_SUBSTR
  -------------------------------
  NULL

  Can someone help me understand why second query not returning ", CA, '?
  
  I did research on this forum and found the link on the thread "https://forums.oracle.com/forums/thread.jspa?threadID=2400143" for the basic tutorials.
  
  
  
  Kind regards
  P.

  The reason is that the comma between 'Redwood Shores' and 'CA' already represents the first occurrence.
  So it can not match the second occurrence at the same time.

  You can replace to (remove the trailing ',' in the regex):

  SELECT  REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',',[^,]+', 1, 1) REGEXPR_SUBSTR  FROM DUAL;
  
  SELECT  REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',',[^,]+', 1, 2) REGEXPR_SUBSTR  FROM DUAL;
  

  Published by: hm on 14.06.2012 00:52

  When you remove also the leading comma you get to the BlueShadows solution.

• 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
  

• Help... Regular expression for characters [special] /: ^ &amp; * () @# $

  Regular expression to parse the string in square brackets:
  I am trying to parse a string in square brackets, but as [] are special characters used in regular expressions to start a character class, I want to remove its special meaning. This regular expression **_user=[a-zA-Z]*@[a-zA-Z]{2}+.abc.com_** works for the analysis of [email protected]* but I want to analyze _user = [[email protected]] _ *, I use the term regular *user=\[[a-zA-Z]*@[a-zA-Z]{2}+.abc.com\]*_* but is in error because it makes special sense of character class. I tried to use \ backslash before *------[]] * but it not working giving error * "INVALID ESCAPE SEQUENCE."
  
  [JAVA CODE: user = "user=\[[a-zA-Z]*@[a-zA-Z]{2}+.abc.com\"];
  Pat pattern is Pattern.compile (user);.
  Carpet to match = pat.matcher ("2011-03-11 02:08:44, 653: User = [[email protected]], Doing report: account.server.regprocmemunix.daily");
  If (mat.find ()) {}
  User1 = mat.group ();
  }
  This room code gives error _ _ "SEQUENCE EXHAUST" INVALID and throw a PatternSyntaxException*.
  
  Please help me parse the string within large brackets [].

  You realize that, although ' [' is a special character in a regular expression, ' \' is a special character in Java, right?] You also need to escape to the ' \'.

  So, if you want to use ' [' in your regular expression, you must use something in the sense of]

  \\[
  

  Published by: almightywiz on April 25, 2011 14:05

  DOH... too slow...

• Form validation helps with the regular Expression [a-zA-Z]

  I'm trying to use the regular expression [a-zA-Z] to allow only upper or lowercase WITHOUT SPACES. With the help of [a-zA-Z] allows space and numbers.

  Could someone give me a point in the right direction?

  Thank you!

  RGNelson wrote:

  I'm trying to use the regular expression [a-zA-Z] to allow only upper or lowercase WITHOUT SPACES. With the help of [a-zA-Z] allows space and numbers.

  Could someone give me a point in the right direction?

  Please try with the following regular expression, which should work for text entry fields 'a line' well standard.

  ^ [A-Za-z] + $

  See you soon,.

  Günter

• Find and replace - regular expression to help?

  How can I find ' <!-* PAGE footer AREA *-> "and replace it and evething after with my new coding of footer on all my pages.

  The problem is:

  The code in my existing page footer area is not the same on all pages.

  I want to replace it with a new one on all pages

  I watch the operators of regular expressions, not found patern that works...

  Thanks in advence for your help

  The following regular expression is the and everything up, but not including the closing tag:

  [\s\S]+(?=<\/body>)
  

  Use it in the search field and put the comment and the code to the footer in the field replace. Select use regular expression.

  Always make a backup before using a regular expression on many pages.

• The clob data analysis - helps in the regular Expression

  Hello

  Need to analyze a clob and and get its value.

  with the temp as

  (select '-dn=cn=9245fe4a-d402-451c-b9ed-9c1a04247482,ou=people,dc=idauth-userstore,dc=standardlife,dc=com' as double val)

  Select * Temp;

  "Analysis to include only the characters after ' dn = cn =" and ending with a comma, so I need to value the O/P as feadcbedca

  
  

  How to achieve this using regular expressions.

  
  

  Also is it a useful site to learn Oracle regular expressions for beginners like me.

  If yes give me the link, it will be useful.

  
  

  Thanks in advance
  

  SQL> with temp
    2  as
    3  (
    4  select '- dn=cn=9245fe4a-d402-451c-b9ed-9c1a04247482,ou=people,dc=idauth-userstore,dc=standardlife,dc=com' as val from dual
    5  )
    6  select regexp_replace(ltrim(regexp_substr(val, 'dn=cn=[^,]+'), 'dn=cn='), '[^[:alpha:]]') val
    7    from temp;
  
  VAL
  ----------
  feadcbedca
  
  SQL>
  

• Need help with a simple regular expression replacement

  Hello everyone,
  
  He comes to the table that I have to work with.
  

  CREATE TABLE "TBL_ACCOMMODATION" ("ACCOMMODATION_ID" NUMBER, "HOTEL_NAME" VARCHAR2(100), "ADDRESS" VARCHAR2(200), "LOCATION" VARCHAR2(100), "PHONE" VARCHAR2(50), "EMAIL_ADDRESS" VARCHAR2(60), "CONTACT_PERSON" VARCHAR2(60), "STATUS" CHAR(1), "CREATED_BY" VARCHAR2(10), "CREATED_DATE" DATE, "MODIFIED_BY" VARCHAR2(10), "MODIFIED_DATE" DATE, "MOBILE" VARCHAR2(50))
  REM INSERTING into TBL_ACCOMMODATION
  Insert into TBL_ACCOMMODATION (ACCOMMODATION_ID,HOTEL_NAME,ADDRESS,LOCATION,PHONE,EMAIL_ADDRESS,CONTACT_PERSON,STATUS,CREATED_BY,CREATED_DATE,MODIFIED_BY,MODIFIED_DATE,MOBILE) values (147,'Testt','Auckalnd','Henderson','565756776','[email protected]','Jasmine','A',null,null,'JEEJJ',to_date('23/10/12','DD/MM/RR'),null);
  Insert into TBL_ACCOMMODATION (ACCOMMODATION_ID,HOTEL_NAME,ADDRESS,LOCATION,PHONE,EMAIL_ADDRESS,CONTACT_PERSON,STATUS,CREATED_BY,CREATED_DATE,MODIFIED_BY,MODIFIED_DATE,MOBILE) values (129,'Kirby Hotel','25A Aitken Street
  Wellington','Wellington','04 918 8513','[email protected]','Deahdoow Maharg','A',null,null,'LEAN',to_date('14/02/13','DD/MM/RR'),'027 356 4333');
  Insert into TBL_ACCOMMODATION (ACCOMMODATION_ID,HOTEL_NAME,ADDRESS,LOCATION,PHONE,EMAIL_ADDRESS,CONTACT_PERSON,STATUS,CREATED_BY,CREATED_DATE,MODIFIED_BY,MODIFIED_DATE,MOBILE) values (167,'Avenue ee','10 Wellington Street
  Wellington','Wellington','4444444','[email protected]','James','A',null,null,'LEAN',to_date('21/02/13','DD/MM/RR'),null);
  Insert into TBL_ACCOMMODATION (ACCOMMODATION_ID,HOTEL_NAME,ADDRESS,LOCATION,PHONE,EMAIL_ADDRESS,CONTACT_PERSON,STATUS,CREATED_BY,CREATED_DATE,MODIFIED_BY,MODIFIED_DATE,MOBILE) values (185,'Quadrant Hotel','10 Waterloo Quadrant,
   Auckland','Auckland','9555555','[email protected]','Quentin QQ','A',null,null,'LEAN',to_date('04/03/13','DD/MM/RR'),null);

  My select statement is

  SELECT acc.hotel_name || '(' || replace(replace(acc.address,chr(13),', '),',,',',') || ')'
  FROM TBL_ACCOMMODATION acc
    inner join ijs_seminar s ON acc.accommodation_id = s.accommodation_id 
  where s.seminar_id = :P27_SEMINAR_ID

  I have to use the function replace twice. One is to replace the Chr (13) a comma is second band a comma where there are two commas.
  
  I don't know much about regular expressions.
  If someone can show me a better way to handle this using the regular expression rather than a heavy means above.
  
  Thanks in advance
  Ann

  Hi, Ann.

  Ann586341 wrote:
  Hello everyone,

  He comes to the table that I have to work with.

  CREATE TABLE "TBL_ACCOMMODATION" ("ACCOMMODATION_ID" NUMBER, "HOTEL_NAME" VARCHAR2(100), "ADDRESS" VARCHAR2(200), "LOCATION" VARCHAR2(100), "PHONE" VARCHAR2(50), "EMAIL_ADDRESS" VARCHAR2(60), "CONTACT_PERSON" VARCHAR2(60), "STATUS" CHAR(1), "CREATED_BY" VARCHAR2(10), "CREATED_DATE" DATE, "MODIFIED_BY" VARCHAR2(10), "MODIFIED_DATE" DATE, "MOBILE" VARCHAR2(50))
  REM INSERTING into TBL_ACCOMMODATION
  Insert into TBL_ACCOMMODATION (ACCOMMODATION_ID,HOTEL_NAME,ADDRESS,LOCATION,PHONE,EMAIL_ADDRESS,CONTACT_PERSON,STATUS,CREATED_BY,CREATED_DATE,MODIFIED_BY,MODIFIED_DATE,MOBILE) values (147,'Testt','Auckalnd','Henderson','565756776','[email protected]','Jasmine','A',null,null,'JEEJJ',to_date('23/10/12','DD/MM/RR'),null);
  ...
  

  Thanks for posting the sample data. You post the results, but the validation of the existing query which I suppose, produced good results. I can't run this query, because it requires the ijs_seminar table and a connection variable that also has no post, but I can just comment on the join and the WHERE clause.

  Is this really the best set of sample data for this question? This problem involves Chr (13) and repeated commas, but I don't see any s CHR (13) or repeated commas in the sample data. In addition, it seems that there are a lot of columns that play no role in this issue and just to make things difficult to read.

  My select statement is

  SELECT acc.hotel_name || '(' || replace(replace(acc.address,chr(13),', '),',,',',') || ')'
  FROM TBL_ACCOMMODATION acc
  inner join ijs_seminar s ON acc.accommodation_id = s.accommodation_id
  where s.seminar_id = :P27_SEMINAR_ID
  

  I have to use the function replace twice. We need to replace the Chr (13) by a comma,.

  As posted, inside REPLACE replaces Chr (13) with a comma and a space which could be important if you then pick up consecutive commas.

  second is the band a comma where there are two commas.

  I don't know much about regular expressions.
  If someone can show me a better way to handle this using the regular expression rather than a heavy means above.

  Assuming you want to replace Chr (13) with just a comma, then an equivalent would be:

  SELECT    acc.hotel_name || '('
                    || REGEXP_REPLACE ( acc.address
                                        , '[,' || CHR (13) || ']{1,2}'
                              , ','
                              )
                  || ')'            AS h
  FROM        tbl_accommodation acc
  INNER JOIN  ijs_seminar       s   ON   acc.accommodation_id = s.accommodation_id
  WHERE s.seminar_id = :P27_SEMINAR_ID
  ;
  

  In the argument to REGEXP_REPLACE 2nd

  [xy]{1,2}
  

  medium 1 to 2 characters of set of x and y. This could be
  x or
  there or it could be 2 characters
  XY or the other way
  YX or it could be the same characters 2
  XX or
  YY

  REGEXP_REPLACE is slower that REPLACE. Even if your original expression is longer, it may be more effective. (Performance may be not a problem in this case.)