Regular expressions - problem

Similar Questions

• the simple regular expression problem

  Hello
  
  I need assistance with regular expressions. I have a situation when I need to get data from one table to the other and I think that my problem can be solved using REG EXP, but I don't know how to use them properly.
  
  
  I need to separate fileld varchar2 which is basically number/number in 2 separate number fields
  

   
  
  
  CREATE TABLE tst (CODE VARCHAR2(10)); 
  
  INSERT INTO tst VALUES('10/15'); 
  INSERT INTO tst VALUES('13/12'); 
  INSERT INTO tst VALUES('30'); 
  INSERT INTO tst VALUES('15'); 
  
  CREATE TABLE tst2 (po NUMBER, co NUMBER); 

  I need to get the code in the co and po columns. I think the result should look something like this, but:
  

   
  
  INSERT INTO tst2 
  SELECT regexp_substr(CODE 'something here to get the number before /') AS po, 
         regexpr_substr(CODE 'something here to get number after') AS co 
  FROM tst;   

  Any help appreciated

  Agree with the above,

  However, if you really want to know how to do with regular expressions you can do it like this...

  SQL> select regexp_substr('10/15','[^/]+',1,1) from dual;
  
  RE
  --
  10
  
  SQL> select regexp_substr('10/15','[^/]+',1,2) from dual;
  
  RE
  --
  15
  

• The regular expression problem

  Dear friends,

  In my script I have some sections that test the contents of an edit field before it is processed further.

  Perfectly things like the following:

  var re_Def = /#[A-Za-z][A-Za-z0-9_]+/;          // valid variable name ?
  items = ["#correct", "notcorrect", "#This_is4", "#thisIs", "@something", "#ALLOK", "", ];
  // search    0            -1          -1!!        -1!!          -1        -1!!     -1      <--- incorrect method
  // test    true         false         true        true        false       true    false    <--- correct method
  for (var j = 0; j < items.length; j++) {
    var item = items[j];
    alert ("'" + item + "' ==> " +  item.search(re_Def) + "\n" + re_Def.test(item));
  }
  var re_Def = /(\[ROW +\d+\]|\[COL +\d+\]|\[CELL +\d+, +\d+\]|Left *\(\d*\)|Right *\(\d*\)|Above *\(\d*\)|Below *\(\d*\))/;
  items = ["[ROW 17]", "[Row n]", "[ROW n]", "[CELL 3, 9]", "[CELL 3 9]", "Abbove()", "Right(3)"];
  // result  true        false      false         true         false        false         true    
  for (var j = 0; j < items.length; j++) {
    alert ("'" + items[j] + "' ==> " +  re_Def.test(items[j]));
  }
  

  But what follows always returns false, independly of the content of the string element:

  var re_Def = /{[EFJ]\d*}|{I}/;    // valid format def?
  var item = "{E27}";
  var result = re_Def.test(item);
  alert (result);                   // false !!
  

  RegEx buddy told me, that
  -l' REGULAR expression is correct
  -the result must be true, not false

  -The verbose definition of the RegEx is:
  Match is the following regular expression (attempting the next alternative only if this one fails) "{\d* [EYF]}."
  Match the character "{" literally "{}".
  Match a single character present in the list "J" "[EYF]."
  Match a single digit 0. 9 paper"\d*»
  Between zero and unlimited times, as many times as possible, giving as needed (greedy) «*»
  Match the character "}" literally "}".
  Or match number 2 below (the entire match attempt fails if it cannot match) regular expression "{i}".
  Match the characters "{i}" literally "{i}".

  Typo unrecognized? Test the faulty method?

  Results are fake, as soon as I use the list of characters []] - but look at the first block of code: there are also lists of character they are treated properly.

  The braces in the regular expression must be escaped to be taken literally:

  var re_Def = /\{[EFJ]\d*\}/;
  

  Kind regards

  JoH

• Regular expression, problem when literals have content "I_".

  Hello
  
  I guess it seems to be a bug/defect of Oracle. If I do this wrong, please let me know.
  I need to remove the "_VERSION' coming to the latest in a string column. For example, if the string is WINDOWS_XP_VERSION, then the output should be WINDOWS_XP.
  
  But if you run the query below
  
  SELECT
  REGEXP_SUBSTR (['AA_TEST_3_XI_VERSION ',' \w+[^_VERSION]') t1.
  REGEXP_SUBSTR (['AA_TEST_3_Xi_VERSION ',' \w+[^_VERSION]') t2.
  REGEXP_SUBSTR (['AA_TEST_3_XIA_VERSION ',' \w+[^_VERSION]') t3
  FROM DUAL;
  
  
  T1 T2 T3
  AA_TEST_3_X AA_TEST_3_Xi AA_TEST_3_XIA
  
  1 selected lines
  
  It seems that the problem exists only when I use "I_" and works well with "i_". I checked with a few other characters and everything worked fine.
  Any ideas? I use 11g.
  
  Thank you
  Sharmin

  any string that does not "_VERSION", it would return NULL

  You would need a small correction:

  SQL> with t as (
   select 'AA_TEST_3_XV_VERSION' str from dual union all
   select 'AA_TEST_3_XE_VERSION' from dual union all
   select 'AA_TEST_3_XIR_VERSION' from dual union all
   select 'AA_TEST_3_XIR' from dual
  )
  
  select regexp_substr (str, '(\w+?)(_VERSION)?$',1,1,null,1) str
    from t
  /
  STR
  -------------------------------
  AA_TEST_3_XV
  AA_TEST_3_XE
  AA_TEST_3_XIR
  AA_TEST_3_XIR                  
  
  4 rows selected.
  

• Chain problem: not being able to escape a bracket for a possible regular expression

  So, I have barely any idea of how to use regular expressions except a little knowledge of it in PHP.
  
  What I'm trying to do is to separate the first part of a string of qualifier, so I need a regular expression since ideally I wish I could. The problem I have is that my attempt of separation does not work. I tried to do what I think the model or class Matcher, I tried to use String.split, and I tried using the regular String.replace method. What I'm trying to do is illustrated (I hope) in the following NBS:
  

  public class SSCCE
  {
      public static void main(final String args[])
      {
          final String testString = "I want to keep this sentence. (I wish to "+
              "discard this sentence and the space in between the sentences.)";
          final String testExtract = testString.replace(" \(.+\)","");
          System.out.println(testExtract);
      }
  }

  From my understanding, the parentheses are supposed to be a wild card of sorts, so they would need to be escaped in order to be taken as a literal, which means that line 7 is supposed to correspond to any of the characters inside the brackets where the opening is preceded by a space and then delete them. The problem I have is that escaping brackets generates an error and does not escape their product the original string. So I don't know how to tell Java I use actually literal parentheses.
  
  I tried to configure my preferences of error to ignore everything, but it does not work.
  
  I don't know if it's important, but I use Java 1.6 and Eclipse for PHP 1.3.2.20110218 - 0812.

  rp0428 wrote:
  Shouldn't be using replaceAll if you use a regular expression?

  http://docs.Oracle.com/javase/6/docs/API/Java/lang/string.html

  and, on the assumption that you actually use a replaceAll(), you need to escape to the '-' characters as they are themselves i.e. Java escape characters

  " \\(.+\\)"
  

  In addition, the use of ".". +"can lead lead to important feedback. In this simple example you probably won't notice it, but in general, you'd better avoid the return back by using the reluctant modifier or by replacing the '.' by a character set that does not include the character terminal.

• 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");
  }
  

• Regular Expression Validator problem

  Hello
  I have a text box that I want to use a regular expression validator to check that one or more 9 char alphanumeric serial numbers have been entered.
  < mx:RegExpValidator
  source = "{entryField}.
  property = "text".
  Flags = "IM".
  expression="{/[a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9] ({|, | \\n|\\r|\\t)/} ')
  valid = "handleResult (Event)" "
  Invalid = "handleResult (Event)" "
  trigger = "{entryField}.
  triggerEvent = "textInput".
  / >
  
  for the test
  private void handleResult(eventObj:ValidationResultEvent):void {}
  If (eventObj.type is ValidationResultEvent.VALID)
  {
  entryField.setStyle ("borderColor", "green");
  entryField.setStyle ("largeurBordure", "2");
  }
  on the other
  {
  entryField.setStyle ("borderColor", "red");
  entryField.setStyle ("largeurBordure", "2");
  }
  }
  
  The problem is the handler function always comes back valid, even when it must be valid, as: 123456789 a12345678 lk231jkop
  
  Can anyone tell where I go wrong? Thank you!

  I finally got it works like this:
  triggerEvent = "keyUp".
  expression = "^ ([A-Za-z0-9] {9} \ (|, |))" \n|\r)*)+$ »

  The big takeaway from this is that the textInput event does NOT work as advertised. Specifically, it does not occur during a deletion. In addition, it forces me to add a space at the end of a serial number using the above expression so that the validation is successful. Fortunately, 'keyUp' works very well :-)

• Problem on regular expressions

  Hi all

  I'm working on a project that require me to separate the following examples:

  + 0, + 0.0000 E1, - 4.33 + E1, - 4.222 + E1, - 6.33 + E2, - 6.55 + E2

  What I need is the final four results separately:

  -4.33 + E1.

  -4.222 + E1,

  -6.33_E2,

  -6.55 + E2.

  I'm totally cool regular expression. Any help is appreciated!

  Thank you

  + Kunsheng

  Hi, Kunsheng

  Good evening exercise to learn regular expressions...

  For example, quick and dirty:

  

  I have the strong feeling that something's wrong here (I guess in ([0-9] + [1-9] +)), but in any case the above code is just starting point for you.

  Andrey.

• problem of regular expression with brackets

  Hello

  I have a Zebra printer string that contains tokens that I want to search.

  ^ BY2, 3, 55 ^ FT50, 219 ^ BCN, Y, N... ^ FD >: [[number 1]] ^ FS
  ^ BY2, 3, 55 ^ FT525, 221 ^ BCN, Y, N... ^ FD >: [[serial number 2]] ^ FS
  ^ FT170, 192 ^ XG000. GRF, 1, 1 ^ FS
  ^ FT220, 430 ^ A0N, 27, 27 ^ FH\ ^ ea FD2. #00931 drum kit / ^ FS
  ↑ FT350, 460 ^ A0N, 27, 27 ^ FH\ ^ FD #00602 ^ FS
  ^ FO10, 32 ^ GB780, 480, 8 B, 1 ^ FS
  ^ FT720, 500 ^ A0N, 10, 10 ^ FH\ ^ FD [[Date Code]] ^ FS

  I want to find any instance of double hooks regardless of the bracketed text. I find [[SerialNumber 1]] for example, but I also want to analyze the entire string and ensure that there no case [[text]] left in the string. I was not able to build a regular expression that will find them.

  Hi elrathia,.

  Please find below screenshot

  

  Kind regards

  SrikrishnaNF

• problem of the regular expression?

  Hi I have a file that contains the following text
  
  < cfset objNews.strNewsTickerLink = "index.cfm? pageid = 83" >
  
  Now what I want to do is to scrape the value of this variable for example "index.cfm? pageid = 83. Now, this variable can be different... so basically regardless of the value of the variable objNews.strNewsTickerLink is on... I want to get it out. Now, I'm not that great with regular expressions... can someone help me?
  
  THX

  Thank you... but I found a even better who does the work for me

  http://CFLib.org/UDF/GetContainer

• problem with the validation type: regular expression

  Hi all
  I have a text field element call p40_opt
  in the past, this element could accept only the values 1-9
  I made a posting type: regular expression
  1 validation expression:
  p40_opt
  
  2 validation expression:
  [1-9]
  
  works very well.
  
  Today, I wanted to add an option that p40_opt can also contains the value: 10
  so I changed the range [1-10].
  
  and now, it's not good because if I choose another option plus 1 and 10 to Ant
  
  he do the validation.
  
  what I'm missing here?
  Thanks in advance
  
  Under the direction of: naama on August 6, 2009 01:31

  The forum software has mutilated regular expressions in your message. You should edit, wrapping up the regular expressions in.

   tags so they are properly displayed.
  

  ^ [1-9] {1} $. ^ $ 10

  would be one regular expression that performs the required validation, however a regular expression doesn't appear to be the most straightforward way to validate such a range of values. A SQL or PL/SQL validation would be clearer, using the expression:
  

  : p40_opt between 1 and 10

• regular expression for alarm url

  Hello

  I have a problem with the regular expression. I want to filter only the name of a dsc alarm url: \\EW-MONITOR\monitor\000_t-ist_STB-M1-AS. Alarms.HI must be 000_t-ist_STB-M1-AS. Someone has a tip for me?

  Best regards

  --
  Joachim

  If as jcarmody asked, the name will always be between 'monitor\' and '. '. Alarms"use the output matching sub regular expression corresponds to

   

  

  It takes the name from alam of the entire game.

• Build a regular expression for data series of Arduino.

  All,

  I have a few Arduinos Council integrated into a test configuration that I can't save the data of.  I now need to be able to see my data in real-time as it comes on the serial port.  I found a VI that seems like it should work, my problem is that I can't get a regular expression to work.

  The VI is not mine, but if I can get this working, I can easily put it in my VI.

  Here's my Arduino code; This is the timestamp, followed four data points, with delimiters tab. It prints on the serial port as

  190876 762314 814437 1108235 1091719

    Serial.print(sTime);
    //Serial.print(", ");
    Serial.print("\t");   
  
    data = getdata(dataRead);
    data = data>>4;
    Serial.print(data);
    //Serial.print(", ");
    Serial.print("\t");  
  
    data = getdata(dataRead1);
    data = data>>4;
    Serial.print(data);
    //Serial.print(", ");
    Serial.print("\t");  
  
    data = getdata(dataRead2);
    data = data>>4;
    Serial.print(data);
    //Serial.print(", ");
    Serial.print("\t");  
  
    data = getdata(dataRead3);
    data = data>>4;
    Serial.println(data);
  

  I think it is especially a problem with regular expression.  Any advice or pointers would be great.

  I wish there was a place where you just have to copy and paste my thong in and get a regular expression.

  Do not use bytes to the port!

  Take advantage of the termination character you have.  Activate it on the series set up and set it to the newline character.

  Now the VISA Read will wait until she has the entire line (or timeout if it does not).  Then you can analyze your data out there.

  I think the problem with your code, it's that you're looking for a backslash n.   Not a line break.  Turn it on the style of display of this constant and you will see it is fixed for normal display and backslash is no bar display.

• Regular expression matching is not what matches Pattern

  I read a lot of posts on how match model does not match what match regular expressions will be due some characters does not.

  However, I found a problem with the other way. A simple Reg - Ex who works in the match pattern but not regular Expression match.

  What I have here is just an example. I want to use regular Expression Match then I can specify some matches under.

  The reg - ex's: one or more non-numeric characters, a space, one or more numeric characters. At the beginning of the string.

  How can I get this to work in regular expression matching? I work in LabVIEW 2010f2 32 bit. Here is the code snippet and the results:

  

  Rob

  One of the subtle differences is the operator of negation for the character classes.  In match pattern is ~, but for the Match RegEx is ^.

• 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.