Chips with 3 delimiter characters using regular expressions

Similar Questions

• Changeparticular the characters of a string using regular expressions...

  Hi all
  I'm writing a function using the function of expression regular oracles REGEXP_REPLACE but I could not succeed until now.
  
  My problem is as follows, I have a text in a column say "Scott Sdfdfs Sdfd" I want to replace all the s and S with X characters and make the text looks like "XdfXdf XdfdfX Xdfd".
  
  It is possible by using regular expressions in oracle?
  Can you give me some clue?
  
  Thank you

  select
    regexp_replace('sdfsdf Sdfdfs Sdfd',
                   's|S', 'X') Replaced
  from dual;
  

  REPLACED
  ------------------
  XdfXdf XdfdfX Xdfd 
  

• Find the words (wild cards) using regular expressions

  I'm testing to see if the words are present for revision 1 of a drawing of the cartridge.

  The script search the digit 1 followed by a date, a title, and 4 sets of initials.

  The number 1 is static, (date, title and original are the cards that they are different for each design).

  I use regular expressions to match the words.

  The regular expression highlighted in blue is the number 1 and the date.

  Him remains highlighted in orange does not match the title and initials.

  If anyone can help with the regular expression that is most appreciated.

  Once I got that work will add the form fields for the initials, noting only the console at this point for the tests.

  numWords = this.getPageNumWords (0);

  number of words on the page

  loop through the words on the page

  for (var j = 0; j < numWords-1; j ++)

  {/ / get the pair of words to test}

  ckWords = this.getPageNthWord (0, j) + ' ' + this.getPageNthWord (0, j + 1); test words

  Check if 1 26.05.16 THE STRENGTHENING REVISED MM SB AE GM word string is present

  If (ckWords.match(/ ^ 1\s [0-9] {1,2}.)) [0-9] {1,2}. [0-9] {2} \s\w+(\s+\w+){1,7}/))

  {

  Console.println (ckWords);

  }

  }

  You can use something like this:

  ckWords = this.getPageNthWord (0, j) + ' ' + this.getPageNthWord (j, 0, + 1) + ' ' + this.getPageNthWord (0, + 2 j) + ' ' + this.getPageNthWord (0, j + 3) + ' ' + this.getPageNthWord (0, j + 4) + ' ' + this.getPageNthWord (0, j + 5) + ' ' + this.getPageNthWord (0, j + 6);

  If (! ckWords.match (/ ^ 1\s\d {1,2} \.\d {1,2} \.\d {2} \s\w+ (?:-s + \w +) {1.8} \s([A-Z]{2})\s([A-Z]{2})-s ([A - Z] {2}) \s([A-Z]{2})$ /)) ckWords + ckWords + "" + this.getPageNthWord (0, j + 7);

  If (! ckWords.match (/ ^ 1\s\d {1,2} \.\d {1,2} \.\d {2} \s\w+ (?:-s + \w +) {1.8} \s([A-Z]{2})\s([A-Z]{2})-s ([A - Z] {2}) \s([A-Z]{2})$ /)) ckWords + ckWords + "" + this.getPageNthWord (0, j + 8);

  If (ckWords.match (/ ^ 1\s\d {1,2} \.\d {1,2} {2} \s\w+ \.\d (?:-s + \w +) 1.8 ([A - Z] {2}) \s([A-Z]{2})\s([A-Z]{2})\s {} \s([A-Z]{2})$ /))

  {

  ...

• A special character validation using regular expressions in ADF

  Hi guys,.

  I want to put the validation of a special as character (,.') ((en) &, -) using regular expressions.

  I asked the posting as [a-zA-Z0-9'(.),--/ &] but it does not work properly.

  Special characters should be like:

  Comma,

  Hyfan-

  Dot.

  Open and close braces (and).

  Ampercent &

  Apastrophy '

  Space ""

  Please help if anyone has idea.

  And I also tried to put under expression as...

  [a-zA-Z] + (\\s* [0-9] * [a-zA-Z] *-* & * \\(*\\) *, *'*. *) * [a-zA-Z0-9] + but we need the validation if we put special characters between the charater as "ab," chain "& (bc).

  his does not work if I put a special character at the beginning and the end of the string in the ADF

  Thanks Timo...

  its working fine...

• ADF Email of Validation using regular expressions

  Hello
  
  Wanted to add search Email Validation VO.
  
  It works if I put

  <af:validateRegExp pattern="[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}" 
                                           messageDetailNoMatch="The value {1} is not a valid email address:"/>

  However, this requires identification of email to be entered in capital letters.
  
  I tried with below the option does not work.
  

                 <af:inputText value="#{bindings.xxEmail.inputValue}" label="Email"
                                        required="#{bindings.xxEmail.hints.mandatory}"
                                        columns="#{bindings.xxEmail.hints.displayWidth}"
                                        maximumLength="#{bindings.xxEmail.hints.precision}"
                                        shortDesc="#{bindings.xxEmail.hints.tooltip}" id="it5">
                              <f:validator binding="#{bindings.xxEmail.validator}"/>
                              <f:validateLength minimum="6"/>              
                              <af:validateRegExp pattern="[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}" 
                                           messageDetailNoMatch="The value {1} is not a valid email address:"/>
                          </af:inputText>

  I got over information of
  ADF Email of Validation using regular expressions
  The user enter email id without @.
  
  
  
  Please suggest this model to reach.
  
  
  Thank you
  JIT
  
  Published by: appsjit on January 25, 2013 19:08

  Hello.
  My English is not very good.

  I use below format and it works

  
  "^ [_A-Za-z0 - 9-] + (\.). [_A-Za-z0-9-]+)*@[A-Za-z0-9][A-Za-z0-9-]+ (-.)] ([A Za-z0-9] +) * (-.) [A-Za - z] {2,}) $»
  messageDetailNoMatch = "the value {1} is not a valid email address" / >

  Habib

  Published by: Habib Eslami on January 26, 2013 01:22

• validate cfinput using regular expressions

  Hello

  can someone help me valditing a field of cfinput using regular expressions?

  First digit must be a number or a 'R '.

  Figure 2-15 can be anything without special characters. Figure 2-15 can also be empty.

  I try this, but it does not work (sorry I'm a beginner using regex).

  < cfinput type = "text" name = "field1" required = "yes" validate = 'regular_expression' pattern = "[0 - 9Rr] [0 - 9a - zA - Z]" * "maxlength ="15">"

  Thank you for advice!

  Claudia

  ^ [0 - 9Rr]([0-9a-zA-Z]{1,14})? $

• Using regular expressions to solve sys_refcursor of a record

  Regarding my Question about sys_refcursor with record type of thread, I thought it can be solved differently. It is:
  
  I have a string like ' 8:1706, 1194, 1817 ~ 1:1217, 1613, 1215, 1250'
  
  I need to do a few things using regular expressions and get something like
  

  select * from <table> where
  c1 in (8,1)
  and c2 in (1706,1194,1817,1217,1613,1215,1250);

  Is it possible by using regular expressions in a single select statement?

  Hello

  Game 6' - 8 "wrote:
  Your understanding is quite correct. But unfortunately it doesn't Frank.

  SQL> SELECT COUNT (*)
  2    FROM (SELECT sp.*
  3            FROM spml sp, spml_assignment spag
  4           WHERE sp.spml_id = spag.spml_id
  5             AND spag.class_of_svc_id = 8
  6             AND spag.service_type_id IN (1706, 1194, 1817)
  7             AND spag.carrier_id = 4445
  8             AND NVL (spag.haulage_type_id, -1) = NVL (NULL, -1)
  9             AND spag.effdate = TO_DATE ('01/01/2000', 'mm/dd/yyyy')
  10             AND spag.unit_id = 5
  11             AND sales_org_id = 1
  12          UNION ALL
  13          SELECT sp.*
  14            FROM spml sp, spml_assignment spag
  15           WHERE sp.spml_id = spag.spml_id
  16             AND spag.class_of_svc_id = 1
  17             AND spag.service_type_id IN (1217, 1613, 1215, 1250)
  18             AND spag.carrier_id = 4445
  19             AND NVL (spag.haulage_type_id, -1) = NVL (NULL, -1)
  20             AND spag.effdate = TO_DATE ('01/01/2000', 'mm/dd/yyyy')
  21             AND spag.unit_id = 5
  22             AND sales_org_id = 1);
  
  COUNT(*)
  ----------
  88
  
  SQL> SELECT COUNT (*)
  2    FROM spml sp, spml_assignment spag
  3   WHERE sp.spml_id = spag.spml_id
  4     AND spag.carrier_id = 4445
  5     AND NVL (spag.haulage_type_id, -1) = NVL (NULL, -1)
  6     AND spag.effdate = TO_DATE ('01/01/2000', 'mm/dd/yyyy')
  7     AND spag.unit_id = 5
  8     AND sales_org_id = 1
  9     AND REGEXP_LIKE ('8:1706,1194,1817~1:1217,1613,1215,1250',
  10                      '(^|~)' || spag.class_of_svc_id || ':'
  11                     )
  12     AND REGEXP_LIKE ('8:1706,1194,1817~1:1217,1613,1215,1250',
  13                      '(:|,)' || spag.service_type_id || '(,|$)'
  14                     );
  
  COUNT(*)
  ----------
  140
  
  SQL> 
  

  Published by: release 6' - 8 "August 11, 2009 20:04

  Serving what you ordered!

  Originally, you said that you are looking for something that produces the same result as

  where   c1 in (8, 1)
  and      c2 in (1706, 1194, 1817, 1217, 1613, 1215, 1250)
  

  in other words, the c1s could be coupled with any of the c2s.
  Now, it seems that what you want is

  where     (     c1 = 8
       and     c2 IN (1706, 1194, 1817)
       )
  or     (     c1 = 1
       and     c2 IN (1217, 1613, 1215, 1250)
       )
  

  in other words, c1 = 8 and c2 = 1250 is not good; is not c1 = 1 and c2 = 1706.

  In this case, try

  WHERE     REGEXP_LIKE ( s
                , '(^|~)' || c1
                       || ':([0-9]+,)*'
                       || c2
                       || '(,|~|$)'
                )
  

• How to extract the value of a tag XML using regular Expressions

  We get a response XML from a WEB SERVICE.

  I convert it to VARCHAR2.

  Now, I want to get the real answer that is inside the tag of the response.

  I tried this:

  DECLARE

  V_1 VARCHAR2 (30000): = ' < soap: Body > < ns:ProcessArgusFeedsResponse xmlns:ns = "urn: PegaRULES:SOAP:ArgusToPegaFeeds:Services" > ' |

  '< response > good < / answer >< / ns:ProcessArgusFeedsResponse > < / soap: Body > ';

  v_response VARCHAR2 (100);

  BEGIN

  DBMS_OUTPUT. PUT_LINE (V_1);

  v_response: = REGEXP_SUBSTR (v_1, ' / < >(.+?) < \/Response > answer /');

  dbms_output.put_line (v_response);

  END;

  It does not work.

  Any help would be greatly appreicated.

  Hello

  user12240205 wrote:

  We get a response XML from a WEB SERVICE.

  I convert it to VARCHAR2.

  Why?  XML has its own native ways of analysis; Why not use them?

  If you use regular expressions, then REGEXP_REPLACE, as shown above, is a good option in Oracle 10, but starting in Oracle 11.1, you can use REGEXP_SUBSTR like this:

  REGEXP_SUBSTR (v_1

  , '(.+?)'

  1

  1

  NULL

  1

  )

  The 6th argument is like a backreference; "He tells REGEXP_SUBSTR did not return to the entire organization, but only the part inside the 1st left '(' et correspondant à sa droite).

  The '?' to make it non-greedy is necessary only if v_1 can contain more than one response.

  Now, I want to get the real answer that is inside the tag of the response.

  I tried this:

  DECLARE

  V_1 VARCHAR2 (30000): = "" |

  'Good';

  v_response VARCHAR2 (100);

  BEGIN

  DBMS_OUTPUT. PUT_LINE (V_1);

  v_response: = REGEXP_SUBSTR (v_1, ' /(. +?)) <\ esponse="">/');

  dbms_output.put_line (v_response);

  END;

  It does not work.

  That's because it's looking for a slash ("/") before the '' tag and another after the ' tag.

  The backslash ("\") is not necessary here, but it is not nothing wrong.

• String format using regular expressions

  Input string output format...
  
  SELECT q'<select ab_c "ABC", efg "EFG" from dual>' str FROM DUAL
  
  Output:
  
  STR                                  
  -------------------------------------
  select ab_c "ABC", efg "EFG" from dual
  
  
  Required output format using regular expression...
  
  
  STR                                  
  -------------------------------------
  select 'ab_c' "ABC", 'efg' "EFG" from dual

  Regular expressions have many limitations as tools of analysis, and you specify the rules you want. This expression puts quotation marks around a non-empty string before a quoted string:

  SELECT regexp_replace(q'',
                        '([^" ]+)( +"[^ ]*")' , '''\1''\2' ) str FROM DUAL;
  
  STR
  ------------------------------------------
  select 'ab_c' "ABC", 'efg' "EFG" from dual
  {code}
  
  It is not robust - a missing " will confuse it, and you should be using bind variables anyway.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                
  		    
              
             
   
  		     
  One for the era: how to get this output using REGULAR EXPRESSIONS?
   
  			 How to get the bottom of output using REGULAR EXPRESSIONS?
  SQL> ed
  Wrote file afiedt.buf
  
    1* CREATE TABLE cus___addresses    (full_address                   VARCHAR2(200 BYTE))
  SQL> /
  
  Table created.
  
  SQL> PROMPT Address Format is: House #/Housename,  street,  City, Zip Code, COUNTRY
  House #/Housename,  street,  City, Zip Code, COUNTRY
  SQL> INSERT INTO cus___addresses VALUES('1, 3rd street, Lansing, MI 49001, USA');
  
  1 row created.
  
  SQL> INSERT INTO cus___addresses VALUES('3B, fifth street, Clinton, OK 74103, USA');
  
  1 row created.
  
  SQL> INSERT INTO cus___addresses VALUES('Rose Villa, Stanton Grove, Murray, TN 37183, USA');
  
  1 row created.
  
  SQL> SELECT * FROM cus___addresses;
  
  FULL_ADDRESS
  ----------------------------------------------------------------------------------------------------
  1, 3rd street, Lansing, MI 49001, USA
  3B, fifth street, Clinton, OK 74103, USA
  Rose Villa, Stanton Grove, Murray, TN 37183, USA
  
  SQL> The REG EXP query shouLd output the ZIP codes: i.e. 49001, 74103, 37183 in 3 rows.
  Published by: user12240205 on June 18, 2012 03:19			 
  /* Formatted on 2012/06/18 17:25 (Formatter Plus v4.8.8) */
  SELECT REGEXP_SUBSTR ((REGEXP_SUBSTR (full_address, '[^,]+', 1, 4)), '[[:digit:]]+') RESULT
    FROM cus___addresses
  		   
   
              
             
   
  		     
  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
  		   
   
              
             
   
  		     
  How to capture multiple line String using regular expressions?
   
  			 
  Hello
  I have a simple program like this:
  
  What I want to accomplish is to capture everything between > start and > to end with a single regular expression matching node. It seems that the definition of multiples? true or False does not help.
  I'm using LabVIEW 2012.
  If it is impossible to capture using a single node, that's fine. But I want to assure you that I can make full use of this node without combining several others.
  Thank you!
  			 
  > start([\w\s]*) > end
  A point matches any character except line break characters.  You have two of them.
  		   
   
              
             
   
  		     
  Validation using regular expressions or oracle fn
   
  			 
  Hello
  I want to apply validation to a column. I'm trying to see if I can use the regexp_replace function to find a boss and see if I can replace it to a better format. Can someone help me with these rules?
  1. only the characters a - z, A - Z, 0-9, ', - is allowed. If there are no other characters, then these characters must be removed.
  2. name should not begin with a symbol.
  3. If there is any symbol permitted in the name, (for example the hyphen) then the hyphen should be preceded and followed by a space.
  I would like to know if the above can be obtained by regular expressions or any other function will be useful to achieve the same.
  
  Thank you very much
  Cherkaoui
  			 
  HI, Charan,
  Whenever you have a problem, please post a small example data (CREATE TABLE and only relevant columns, INSERT statements) of all the tables involved, so that people who want to help you can recreate the problem and test their ideas.
  Also post the exact results you want from this data, as well as an explanation of how you get these results from these data, with specific examples.  Post in this case, maybe 5 or 10 strings, and what you would like to see (a string corrected, or a 'yes' / 'No' flag saying if this string follows rules.)
  Always say what version of Oracle you are using (for example, 11.2.0.2.0).
  See the FAQ forum: Re: 2. How can I ask a question on the forums?
  2784427 wrote:
  Hello
  I want to apply validation to a column. I'm trying to see if I can use the regexp_replace function to find a boss and see if I can replace it to a better format. Can someone help me with these rules?
  1. only the characters a - z, A - Z, 0-9, ', - is allowed. If there are no other characters, then these characters must be removed.
  2. name should not begin with a symbol.
  3. If there is any symbol permitted in the name, (for example the hyphen) then the hyphen should be preceded and followed by a space.
  I would like to know if the above can be obtained by regular expressions or any other function will be useful to achieve the same.
  Thank you very much
  Cherkaoui
  Sorry, we don't know what you want to do.
  1 seems to mean you want to change the given string so that it follows the rules.  (What exactly are special characters?  Are there only 2 special characters: single quotation mark and hyphen?)
  2 seems to tell us that you want to check if it meets the rules
  3. can be a
  I'll assume that you want to modify the string so that the returned value corresponds to rules.  This means that if the given string violates, 2 symbols will be removed from the beginning of the string, and if it violates 3, then spaces will be added.
  I don't think that there is a way to do this with a regular expression.
  You can make each with a separate REGEXP_REPLACE corrections, and make all in the same query by one nesting inside each other, like this:
  SELECT REGEXP_REPLACE (REGEXP_REPLACE (REGEXP_REPLACE (: str)))
  , "[^ a-zA-Z0-9 cm-]"
  )
  , '^[''-]+'
  )
  , ' *([''-]) *'
  , ' \1 '
  ) AS new_str
  OF the double
  ;
  of course, I can't test it, until post you some sample data and the exact results you want from these data.
  Here, the inner REGEXP_REPLACE rule 1 applies.  It deletes all characters except a to z small letters, capital letters A to Z, numbers, spaces, single quotes and hyphens.
  Rule 2 applies the REGEXP_REPLACE middle.  It removes the single quotes and hyphens that appear at the beginning of the string.
  
  The external REGEXP_REPLACE rule 3.  It guarantees that each single quotation mark or a hyphen is preceded and followed by a space.  If there is already a space (or several spaces) before and/or after the special symbol, they are replaced by a single space.  If this isn't what you want, it can be fixed, but it will make the expression even messier.
  You might consider to write a user-defined function to normalize strings.  In a function procedure, you can apply the rules one after the other. you need not nesting complicated like that.  Of course, a user-defined function will be slower, but maybe it's not a problem.
  Still, the above query converts the string so that the production rules.
  If you just want to check whether a string conforms to the rules, you can compare the original string to the output of the above expression.  If they are identical, then the string is in accordance with the rules.
  		   
   
              
             
   
  		     
  Use regular expressions to extract .llb file from the path name
   
  			 
  I'm trying to be smart (always a dangerous thing) and use a regular expression to extract the name of a library to a filepath converted to a string.   While I appreciate there are other ways to do this, regex seems to be a very powerful way of neat, I would be able to operate.
  IE, if I have a string of type, C:\applications\versions\library.llb\toplevel.vi, I want to be able to extract the library.llb of the string, because it will be of variable length, may include the numbers & spaces and may be located in a file of variable depth hierarchy.   In other words, I want to extract the part of the string between the last-which ends with .llb
  The best I managed so far is \\+.*llb who all returned less the drive letter and the toplevel.vi
  Can someone help me to achieve this goal, or I'm better by using an alternative method (for example filepath string in array, search for .llb)
  Thank you
  Matt
  			 
  Hi Matt,
  attached you will find two other options.
  Mike
  		   
   
              
             
   
  		     
  Validation of attribute text WITHOUT using regular expressions
   
  			 
  Hello world
  I'm working on a few validation rules for a text attribute and one of the conditions is that the text string can contain only alphanumeric characters.
  Because of the requirements that I can't use a regular expression here (the output should be a Boolean value confirming if the text is valid or not), so I need to find a way to write this in the rules...
  Yes, is there an easier way of writing this rule that the use of the ' Contains (< text >, < substring >)' function? Like this:
  [Etc. for all the non alpha numeric characters]
  If not, is it possible to write a rule that States that the text contains ONLY the following characters
  [Etc for all alphanumberic characters]
  Or is there some other function I could use it here?
  Appreciate any thoughts or input about how this could be solved, thank you in advance for your help.
  			 
  Unfortunately, regular expressions are the ideal solution for exactly the scenario you are exhibitor. The only alternative other than a broad 'or' statements is regular expressions, or write a custom function.
  		   
   
             
            
           
  		
   
            
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   Validation using regular expressions or oracle fn 
   
         
   Use regular expressions to extract .llb file from the path name 
   
         
   Validation of attribute text WITHOUT using regular expressions 
   
        
   
       
   
      
   
      
     
     
   
     
   
      
  
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