Regular expressions in the APEX

Similar Questions

• 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.

• Regular expression for the format: 000-000 - 000000 000000

  Hi guys,.

  I need to validate the columns in a regular expression for the format of 000 000 - 000000 000000.

  For example - if the column contains a value such as "500 110 - 500044 000100" then it should return 'true' otherwise 'false. '

  Your timely help is well appreciated.

  Thanks in advance.

  Hello

  inDiscover wrote:

  Hi guys,.

  I need to validate the columns in a regular expression for the format of 000 000 - 000000 000000.

  For example - if the column contains a value such as "500 110 - 500044 000100" then it should return 'true' otherwise 'false. '

  Your timely help is well appreciated.

  Thanks in advance.

  If you want a regular expression

  REGEXP_LIKE (str

  {{' ^ {\d{3} \d{3}-\d{6} \d{6}$'

  )

  According to your needs.

  You don't need regular expressions for this.  It will be more efficient to use

  TRANSLATE (str

  '012345678'

  '999999999'

  ) = 999 999 - 999999 999999'

• regular expression for the xml tags

  Dear smart people of the labview world.

  I have a question about how to match the names of xml text elements.

  The image that I have some xml, for example:

  Peter

  13

  and I want to match all of the names of elements, that is to say: no, son, grandson, age, regardless of any attribute have these items. There is a regular expression, I can loop, that can do this? (Something like "\<.+\> ". "") It is no good because it matches the entire xml string.) I'd really only two different expressions, one for the match start elements, e.g. and one for the correspondence of the elements, for example.

  Thanks for your help in advance!

  Paul.

  The site Of regular Expressions will be very convenient.

  They have some good tutorials on regexp with a demo of the XML tags:

  Here is a small excerpt:

  The regular expression <\i\c*\s*>matches an opening of the XML without the attributes tag corresponds to a closing tag. <\i\c*(\s+\i\c*\s*=\s*("[^"]*"|'[^']*'))*\s*>corresponds to an opening with a number any attributes. Put all together, <(\i\c*(\s+\i\c*\s*=\s*("[^"]*"|'[^']*'))*| i\c*)\s*="">corresponds to an opening with attributes or a closing tag.  (source)

  If you want advanced XML analysis I suggest JKI XML toolkit.

  Tone

• Oracle regular expressions - splits the string into words for

  Hello

  Nice day!

  My requirement is to split the string into words.

  So I need to identify the new line character and the semicolon (;), comma and space like terminator for string entry.

  Please note that I am currently embedded blank and the comma as separator, as shown below.

  Select regexp_substr('test)
  TO
  string in words, "([^, [: blanc:]] +) (', 1, 1) double;"

  How to integrate the semicolons and line break characters in regular expression Oracle?

  Please notify.

  Thanks and greetings

  Sree

  This has nothing to do with REGEXP. Is SQL * more parser does not not a semicolon at the end of the line:

  SQL > select ' testto, mm\;
  ERROR:
  ORA-01756: city not properly finished chain

  SQL >

  Just break the chain:

  SQL > select regexp_substr ('testto, mm\;' |) '
  2 string into words
  3 \w+',1,level ',') of double
  4. connect by level<= regexp_count('testto,mm\;'="" ||="">
  5 string in words
  6      ','\w+')
  7.

  REGEXP_SUBSTR ('TESTTO, MM\;' |') STRINGIN
  --------------------------------------
  Testto
  mm
  string
  in
  Words

  SQL >

  Or modify SQL * more the character of endpoints:

  SQL > set sqlterm.
  SQL > select regexp_substr ('testto, mm\;)
  2 string into words
  3 \w+',1,level ',') of double
  4. connect by level<=>
  5 string in words
  6      ','\w+')
  7.

  REGEXP_SUBSTR ('TESTTO, MM\;) STRINGINTOWO
  --------------------------------------
  Testto
  mm
  string
  in
  Words

  SQL >

  SY.

• regexp_substr: a regular expression for the separate comma operator of witn of string literals IN

  The following regular expression separates simple values separated by commas (SELECT regexp_substr (: pCsv,'[^,] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,] +', 1, level) IS NOT NULL); Exampple: 300100033146068, 300100033146071, 300100033146079 returns 300100033146068 300100033146071 300100033146079

  This works very well if we use the regex with SQL IN operator select * from mytable where t.mycolumn IN (SELECT regexp_substr (: pCsv,'[^,] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,] +', 1, level) IS NOT NULL);

  But this query does not work if the comma-separated value is a single literal string 'one', 'two', 'three '.

  Thanks for your reply. my request was mainly on regexp_substr.  Need to request was simple: any table with a column of varchar type can be used. Next time I'll give you an example.

  All ways working answer for my question is is SELECT regexp_substr (: pCsv,'[^, "] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,"] +', 1, level) IS NOT NULL

• 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;
  }

• Using Regular Expressions in the Code of edge

  Hello.

  I am quite new to the Edge Code but find it quite interesting to use.

  The find/replace feature is pretty but I got a little confused on how to use regular expression matching.

  I tried to clean up the coordinates of a file Adobe Edge animate full of 120.17px (heavy and less accurate)

  So basically you're looking \d+\.\d+px

  First of all, she says "use /re/ for regex" even though I know the Code is made for developers who * courses * know that it took me a little time to understand I should just put my regex between slashes.

  /\d+\.\d+PX/

  then the time comes to replace them. ALT-cmd-F and it says "replace".

  Puzzled again.

  Will I type my full sentence there? Maybe a «...» "written up somewhere around would avoid this question because, there, of course, a second field to come.

  And there, I got stuck.

  I can not find how to get the replacement of the model.

  tried to \1 $1 \1/ $1 / nothing seems to work... any hint of welcome.

  Franck,

  You are right that it does not work. I open a topic. Here is the link if you want to follow: https://github.com/adobe/brackets/issues/1861

  FYI, I know exactly how you want to replace the search results, but here are a few tips:

  You must set the text that you want to retrieve by using parentheses. Thus, for example, if you want the integer part of the number of the result, then your regexp would be: / (\d+)\.\d+px/

  Then you must specify the first result using $1, so (when it's fixed), you can use something like: $1px

  Thank you

  Randy

• Regular expression to the substring

  Hi people;
  
  I need to extract substrings dynamically an attribute A.
  A varchar2 attribute is set like this: "LXXXXX/111111 (+), (-), LXXXXXX/111111... LXXXXX/111111. Always the same series.
  
  I need to store all the "111111 (+)" "111111 (-)" "111111" the same file in a new attribute named B.
  
  In my view, regular expressions that could help me, but I'm not a very good...
  
  Thanks for your help. ^^

  select  regexp_replace('LXXXXX/111111(+),LXXXXX/111111(-),LXXXXXX/111111,etc...',',*LX+/') from dual
  /
  
  REGEXP_REPLACE('LXXXXX/111111(+
  -------------------------------
  111111(+)111111(-)111111,etc...
  
  SQL> 
  

  SY.

• Regular expressions in the field of Validation does not.

  I really hope someone can help me.  I'm quite familiar with regular expressions, but for some reason, I can't it to work.

  Environment: Windows XP

  Adobe Acrobat version 8.12

  Scenario:

  In a text field of the form, the user enters a full path.

  This is the code that runs when the user leaves the field:

  / * Test alternatives

  var validChars = / ^ \w [:\\]$/
  var validChars = / ^ \w :\\$/

  var validChars = / ^ [A - Z] [a - z] [0-9] [:\\_]$/

  */

  
  var validChars = / ^ [A-Za-z0-9 :\\_]$/

  If (event.value! = "")
  {
  If ((event.value) validChars.test is false)
  {
  App.Alert ("Invalid character.  Only alphanumeric, colon, underscore, backslask and spaces are valid characters. ») ;
  Event.RC = false;
  }
  }

  The characters allowed in the path are: alphanumeric, underscore, colon, backslash (\), and space.

  I tried all the expressions above with c:\aoa_apps and they all fail: alert the invalid character appears.

  I've added alerts to print the event.value and that seems to be ok.

  I tried all of them with c:aoa_apps and they still fail.  Finally, I tried caoa_apps and they still aren't.

  I checked each of the expressions in a small validator of expression that I have and they seem to be ok.

  I'm cross-eyed looking for a reason why it's a failure.   So, apparently, that I do something stupid, again, I'm new to scripting Adobe Acrobat.  But for the life of me, I can't.

  Can anyone spot the problem?

  Any help is greatly appreciated.

  caerial.

  How you set up your regexp, only the first character is tested. Try

  place a * before the $.

• regular expression - get the numbers from a string

  Hello world
  
  I'm trying to use regular expressions to get all the numbers in a string. The only problem is that the chain can vary.
  
  For example:
  
  It's my rope 3 and 8 I want 2 get out of those 7 numbers
  Random text 9 with 5 for everyone weekend 8
  
  How can I do?
  
  Thanks in advance :)
  

  Quote:
  Posted by: Scott Stroz
  

  #numbersOny #.

  Who will be only to crush all the numbers in a large number.

  The attached code will return a nice list of numbers.
  Though the numbers may contain commas or decimal points, the code can easily be adjusted

• 10g Express to the Apex 3.1.2 update problem

  I have Oracle Database 10 g Express Edition Release 10.2.0.1.0 installed on windows XP. I want to develop an application using oracle express (apex), so I thought it would be a good idea to upgrade the apex (of the 2.0 default apex supplied with 10g xe) at the latest, free 3.1,2 before you start. I tried to update twice & it failed. Without going into a ton of details, from errors in the paper that I saw every time, it seems that the installation script, apexins.sql, expects 1 parameter to be the default tablespace instead of the password. Any ideas on this subject?
  My other questions are:
  1) before moving on to rel 3.1.2 can I first create the user/passwd apex? The doc did not initially create the account. (so I did not). It is said apexins are the followiing, but errors in the newspaper that I seem to disagree:
  @apexins < password > < tablespace_htmldb > < tablespace_files > < tablespace_temp >
  < images > < connect >
  
  (2) I'm following the instructions for apex rel 3? From what I read, I think I can go directly from apex rel 2.0 to 3.1.2. I found 2 different install docs for module 3 and they were different. The two documents supposed to run change passwd script (apxchpwd) after installation, but a doc editing had 2 additional scripts (which I did not work). Can someone tell me please for the right installation instructions?
  
  (3) after research I found an old 2.2 installaction apex tutorial that says you have to install HTTP before the APEX. Apparently I misinterpreted the doc 3.1 install it says http was already included in oracle 9.2 or later. I guess I don't have no need to install HTTP since oracle xe is rel 10. Now, I think I have to install HTTP from the accompanying CD first before moving on to APEX 3.1.2. Is this correct?
  
  (3) it is said that it takes in the SYSTEM tablespace for apex 3 85 M. the size of the system tablespace is 76800 (default).
  Should I increase it before attempting to upgrade again? (I've waited the defect of being large enough since the apex is delivered with the db)
  
  (4) suggestions/advice for starting more are welcome! Again, I've seen 1 > doc on the way to a bad installation of cleaning. Both said to drop the flows_0300 user, but I told exec wwv_flow_upgrade.switch_schemas, I did until I tried the 2nd install. Am I supposed to run the switch_schemas or not to bad cleaning install? In addition, in the troubleshooting doc I don't see an option to call switch_schemas for my apex release (2.1) so I just followed the examples and by the following 'FLOWS_020100 '. for the parameter "FLOWS_020200". Was it a good or bad idea?
  
  (5) is there any other software (i.e. application, server?) that must be installed before or after the apex 3.1.2? Anything else I missed? The documentation is unclear and has omissions.
  
  Thank you.

  Hello

  I guess you do this on a Windows-based computer and Apex 3.1.2 has already been downloaded from the Web Application Express page and will use the http port 8080.

  NOTE: Make sure your Oracle XE database does not automatically start when Windows starts. If this is the case, then remove gently Oracle Database XE before proceeding with any of these steps.

  Step 1: Unzip Apex in an appropriate place (preferable "c:\oraclexe\apex").

  Step 2: Start a session SQL more than c:\oraclexe\apex and connect as a 'sysdba '.
  sqlplus / as sysdba

  Step 3: Install the Apex.
  SYSAUX SYSAUX TEMP/i +@apexins / +.

  Step 4: Reconnect the server Oracle XE.

  Step 5: Assign/Set PATH to APEX of the images in the APEX_HOME folder where we decompressed APEX that is c:\oraclexe.
  + @apxldimg. SQL APEX_HOME +.

  Step 6: Assign / define credentials administrator to the APEX.
  + @apxxepwd. SQL votre_mot_de_passe +.

  Step 7: Restart the Oracle XE database.

  Step 8: Top configuration completed. The availability will be as follows:
  Availability of apex: http://localhost: 8080/apex
  Administrator of the apex: http://localhost: 8080/apex/apex_admin

  Follow these steps and let me know if it works for you or not.

  Kind regards
  Naveed.

• Matches regular expression in the collection

  Hello
  
  I need to do the following:
  I have a long string with repeated similar data. I would like, by using a regular expression, to excerpts from all matches in a collection. Is there a way to perform this task?
  
  I look through the owa_pattern package, but as far as I saw, I can extract only a single match. Here's an exact quote:
  "" If the regular expression can match several strings that overlap, this function is the longest game."- http://download.oracle.com/docs/cd/B28359_01/appdev.111/b28419/w_patt.htm
  
  So what can I do if I want to get all the games?
  
  Thank you in advance. Any help would be appreciated.
  
  Best regards
  beroetz

  Maybe this will help:

  SELECT REGEXP_SUBSTR ('{A1:5} {BB:10}{CC:5}', ' (<.*?>) |) () {. *?})', 1, LEVEL) token
  OF THE DOUBLE
  CONNECT REGEXP_INSTR ('{A1:5} {BB:10}{CC:5}', ' (<.*?>) |) () {. *?})', 1, LEVEL) > 0
  ORDER BY CSA LEVEL;

  If you want to eliminate the unwanted characters <> {} then use REGEXP_REPLACE in addition:

  SELECT REGEXP_REPLACE (REGEXP_SUBSTR ('{A1:5} {BB:10}{CC:5}', ' (<.*?>) |)) () {. *?})', 1, LEVEL), "[<>{}]') token.
  OF THE DOUBLE
  CONNECT REGEXP_INSTR ('{A1:5} {BB:10}{CC:5}', ' (<.*?>) |) () {. *?})', 1, LEVEL) > 0
  ORDER BY CSA LEVEL;

  See you soon

  Edited by: Nuerni the 17.10.2008 at 11:02

• Description Resource path location Type InvalidRegex: the value of model "([a-zA-Z_]{1,63}[\s-a-zA-Z_0-9.] 0.63 {} [---]?) *' is not a valid regular expression. The error reported is "-' is an invalid character range. Write '--'. ', at the level of the co

  Hello world

  I want to debug my code now, but I have this error on BlackBerry_App - Descriptor.xml (see title).

  Here is the text of my xml gall:

  
   
   
   
     
       
     
   
   
     
   
   
     
   
   
     
     
   
   
  

  I read it could be an issue with jre, or material of the installation, but I have reinstalled, I use the blackberry jre 6.0, but the error is still there...

  What is the case?

  And how can I solve it?

  Everyone answered cordially

  Thanks for your response!

  No accents in my properties, but it wasn't for me a problem of jre. I used the jre7.0

  I also note that blackberry offers default download of the jrs 6.27

  My problem is solved

• How to use a regular expression using the model concept and match for this scenario?

  Hi guys,.

  I have a string "we have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle.

  I need to replace the numbers based on the condition below.

  If more then 5, replace with many

  If less than 5 then replace by a few

  If it is 1, replace by "only one".

  Here is my code, I'm missing the part equates to replace the numbers could one of please help me solve you this.

  private static String REGEX = "(\\d+)"; "

  private public static String INPUT = "we have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle."

  String pattern = "(.*) (\\d+)(.*)";

  private static String REPLACE = "replace with many";

  Public Shared Sub main (String [] args) {}

  Create a model object

  Model r = Pattern.compile ("REGEX");

  Now create object match.

  Matcher m = r.matcher (INPUT);

  Change the value to 7 by the replacement string

  How to assimilate (\\d+) greater than a number and use the code below.

  ENTRY = m.replaceAll (REPLACE);

  Print the final result;

  System.out.println (Input);

  Thank you and best regards,

  Hello

  Try the following makes use of 'appendReplacement"instead with the methods 'start' and 'end' to locate and check the search string"regExp"before dynamically set the string"replace ":

  String regExp = "\\d+";
  String input = "We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle";
  String replace;
  Pattern p = Pattern.compile(regExp);
  // get a matcher object
  Matcher m = p.matcher(input);
  StringBuffer sb = new StringBuffer();
  while (m.find()) {
     Integer x = Integer.valueOf(input.substring(m.start(), m.end()));
     replace = (x >= 5) ? "many" : (x == 1) ? "only one" : "few";
     m.appendReplacement(sb, replace);
  }
  m.appendTail(sb);
  System.out.println(sb.toString());
  

  HTH.

  Kind regards
  Rajen

  PS: Please mark as answer/useful if it solves your problem to the benefit of all members of the community.