Using regular expressions to store the first 3 letters of a string

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• 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})$ /))

  {

  ...

• 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

• 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

• 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.

• 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
  

• 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 
  

• How can I use a regular expression to validate the email address

  How can I use a regular expression to validate the e-mail address.

  var emailExp:RegExp = /([a-z0-9._-]+?) @([a-z0-9.-] +). ([a - z] {2,4});

• Regular expression to remove the space in the HTML tag

  Hi all

  My HTML string is as below.

  Select ' < CityName > RICHMOND < / Nom_ville >

  < StateCd > ABCD CDE

  < StateCd / >

  < CtryCd > CAN < / CtryCd >

  < CtrySubDivCd > BC < / CtrySubDivCd > ' double Str

  Output desired is

  < CityName > RICHMOND < / Nom_ville > < StateCd > ABCD CDE

  < StateCd / > < CtryCd > CAN < / CtryCd > < CtrySubDivCd > BC < / CtrySubDivCd >

  I want to remove these spaces of the tag value box with only spaces otherwise leave it as what. Please help to implement the same using regular expressions.

  Hello

  We don't know what you want.  This site seems to be formatting your message in a weird way.

  As the statement

  SELECT «...» "THE DOUBLE;

  without formatting, to show your entry and after the exact output desired friom as, with as little in shape as possible.  It might be useful if you use some character like ~ instead of spaces (just for viewing; we will find a solution that works for spaces).

  To remove the text which consists of spaces and nothing else between the tags, you can say

  REGEXP_REPLACE (str

  , '> +<>

  , '><>

  )

  How is this string generated?  Maybe there is an easier and more effective way to keep the bad wrtings sub off the chain in the first place.

• regular expression to remove the zeros on the right

  I need a regular expression to remove the zeros after the decimal point. I tried (?.)<=\.\d+?)0+(?=\D|$) but="" i="" get="" a="" error="" about="" look="" behind="" not="" a="" fixed="" length="" or="" something="" like="" that.="" i="" am="" not="" a="" regex="" expert="" and="" i="" was="" wondering="" just="" how="" to="" do="" this="" with="" regular="" expression="" or="" some="" other="">

  Z.K. wrote:

  [...] or some other way.

  

  I tried and I tried but I couldn't crack with a regular expression, so I took the easy way.  The first match found pattern the comma and the other removes the zeros to the right of the rest.  It is not discriminate between numbers and all the rest, though.

• Chips with 3 delimiter characters using regular expressions

  Hello world

  I have a function that is able to mark the input in a collection string using regular expressions.

  In case the input string is a character such as the comma or semicolon delimiter,

  We can just get the result we want like the example below.

  SQL> select * from v$version;
  
  
  BANNER
  --------------------------------------------------------------------------------
  Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
  PL/SQL Release 11.2.0.1.0 - Production
  CORE    11.2.0.1.0      Production
  TNS for 64-bit Windows: Version 11.2.0.1.0 - Production
  NLSRTL Version 11.2.0.1.0 - Production
  

  SQL> with tab1 as (
    2      select 'abc,dfg,h,,1234' as col1 from dual
    3  )
    4  select regexp_substr(col1, '([^,]*)(,|$)', 1, level, 'i', 1) as result
    5  from tab1
    6  connect by level <= regexp_count(col1, ',')+1;
  
  
  RESULT
  ---------------
  abc
  dfg
  h
  
  1234
  

  But in the case where the channel of entry has 2 types of delimiter and each delimiter consists of 3 characters as below

  I wonder if it is possible to get the result as below.

  The input string: 01| ^ | ABCD| ^ | 111| * | 02| ^ | efgh| ^ | 222

  Separators: | * | is divided into lines, | ^ | is divided into columns

  Expected result:

  col1 col2 col3

  Row1 - > 01 abcd 111

  row2-> efgh 02 222

  Simply put, take a next

  The input string: 01| ^ | ABCD |^| 111 |*| 02 |^| efgh |^| 222

  Separator: | * |

  Result:

  01. ^ | ABCD | ^ | 111

  02. ^ | efgh | ^ | 222

  How can I achieve this using regular expressions?

  Kind regards

  Euntaek

  You need to know the number of the column from the outset:

  with tab1 as)

  Select ' 01 | ^ | ABCD | ^ | 111. * | 02. ^ | efgh | ^ | 222' as double col1

  )

  Select rownum,

  regexp_substr (regexp_substr (col1 '(.*?) ((\|\*\|)| $) ', 1, level, null, 1),'(.*?) ((\|\^\|)| $) ', 1, 1, null, 1) col1,.

  regexp_substr (regexp_substr (col1 '(.*?) ((\|\*\|)| $) ', 1, level, null, 1),'(.*?) ((\|\^\|)| $) ', 1, 2, null, 1) col2.

  regexp_substr (regexp_substr (col1 '(.*?) ((\|\*\|)| $) ', 1, level, null, 1),'(.*?) ((\|\^\|)| $) ', 1, 3, null, 1) col3

  of tab1

  connect by level<= regexp_count(col1,'\|\*\|')="" +="">

  /

  ROWNUM COL1 COL2 COL3

  ---------- ----- ----- -----

  1 01 abcd 111

  2 efgh 02 222

  SQL >

  SY.

• 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...

• 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.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                
  		    
              
             
   
  		     
  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
  		   
   
              
             
   
  		     
  Regular expression TO validate the name of column
   
  			 Hi all
  
  
  I want to write a regular expression to validate the name of column names.i.e enterted must follow the convention of valid column name
  contain only A - Z, a-z, 0-9, _, $, and #.
  
  
  Please someone help me to do this.
  Padma			 
  Hey padma.
  Regular expression TO validate the name of column
  to set jaspx for example in the Validation number:
  **
  If your use na na... chain mention you model as etc. from A to z...
  otherwise if the two method nt answer using java Scrpit na... it's better ground for you.
  If however hepatitis is any querys na msg me... If possible I try best,.
  Thnks,
  Suresh V
  Published by: V Suresh on April 2, 2012 16:55
  		   
   
             
            
           
  		
   
            
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