Analyze the Mac address with the regular expression matching

Similar Questions

• Is the function of the regular Expression

  Hi guys, using this model, I got this error:

  -4600 error has occurred during the regular expression match.

  I have attached the VI.

  can you help me? Thank you

  inuyasha84 wrote:

  Hi well I want to save (create) a file and do a check to see if the new file that I want to create already exists or not. If the idea was to see if the path of the new file is equal to the old path

  Why not just use 'check if the file or folder exists? (E/s files-> Adv file functions)

  Cameron

• 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

• Sense of the regular expression in detail

  Hi all
  
  I don't know if this question is to type correct forum but my question is
  
  What is retail mean more regex? Please help me understand this character by character.
  
  Expression is ^ [(\\w\\-]([\\.\\w])+[\\w]+@([\\w\\-]+\\.) + [A - Z] {2,4} $]
  
  Thank you

  sabre150 wrote:
  In general terms, that the regular expression is a poor attempt to check email address syntax. Very bad because it does not cover the complete specification.

  Yes. And after that: If your application looks at the e-mail address (valid) of someone and told them that this is not an email address valid, they will think you are a fool. And if your application is used for commercial purposes, which would be a black mark against your company. You're better not to validate e-mail addresses to all that the implementation of a solution with false negatives.

• The Regular Expressions and GUID.

  Hello gurus, I hope you can help me!
  I need to select GUID from a table and to do this, I need the regular Expression. My
  Perl is not good and not good Regular Expression. My database is Oralce 11.2.0.2.0 and
  Linux (Oracle Version 6) is the operating system of the Machine. If you need further information,
  I'll look closely. Thank you. Jehangir.

  >

  Hi Jehangir and welcome to the forums.

  I need to select the GUID of a table and to do this,

  Well, the first thing we do is read the forum FAQ and also the post by BluShadow
  at the top of the messages on the home page of the forum. You should have provided code (DDL
  and DML) showing your particular problem, but since it's your first time, I'll be gentle ;)

  We have it done - clients have sometimes GUID as PKs, and we need to send data to
  their systems, but it is not as simple as it may first appear.

  GUID may arise in three formats.

  The Oracle one - SELECT Sys_GUID() from DUAL which is just a string of 32 hexadecimal characters.
  Then the chain with the hyphen, then the string with dashes and {} at the beginning and end (see
  examples of data).

  with datax as
  (
    SELECT '79A864CCD8E44CD8B0A2765DF9EF337B' as guid  FROM DUAL  UNION ALL
    SELECT '79A864CFD8E44CD7B0A2765DF9EF337B' FROM DUAL UNION ALL
    SELECT '8gdfsgsgfdg' FROM DUAL UNION ALL  -- dummy for testing
    SELECT '21EC2020-3AEA-1069-A2DD-08002B30309D' FROM DUAL UNION ALL
    SELECT '21EC5550-3AEA-1069-A2FF-08002B30309D' FROM DUAL UNION All
    SELECT '{21CC2020-3AFA-1A69-A2DD-08002B30309D}' FROM DUAL
  )
  
  -- first one is the Oracle format
  select * from datax where regexp_like(guid, '[0-9a-fA-F]{32}'); -- Oracle select sys_guid();
  
  -- second one is with hyphens
  
  select * from datax where regexp_like(guid, '[0-9a-fA-F]{8}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{12}');
  
  -- third one is with hyphens and curly brackets.
  
  select * from datax where regexp_like(guid, '^\{[0-9a-fA-F]{8}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{12}\}$');
  
  -- This converts both of the last two formats back into Oracle format, which is what
  -- we use. Notice, that I haven't used regualar expressions to do this. Regexes are
  -- computationally expensive, and you should use Oracle's string furnctions if possible
  
  SELECT REPLACE(REPLACE(REPLACE(GUID, '{', ''), '}', ''), '-', '')  FROM Datax;
  

  HTH,

  Paul...

  Jehangir.

• The regular expression Dilimit

  How can I delimit the regular expression with the number sign (#) and then use an apostrophe in the expression.
  Can someone give some explamples.
  
  As regular expressions are placed in quotation marks, and I have an apostrophe in a string also how would I be able to specify the regular expression with a sign by dilimiting #.
  
  
  Thank you
  
  Published by: LostWorld Sep 15, 2010 05:40

  Hello

  Not sure I understand the question you are faced with, but I think using q citing might help:

  SQL> with t as (
    2  select 'abcd ''1234''' str from dual)
    3  -- end of sample data
    4  select str, regexp_substr(str, q'#'1234'#') str from t;
  
  STR         STR
  ----------- -----------
  abcd '1234' '1234'
  

  The symbol # now includes the string in the example above.

  You can view the documentation for more information in the link below:
  http://download.Oracle.com/docs/CD/E11882_01/server.112/e17118/sql_elements003.htm

• I know by logic box to get the output using the regular expression?

  Hello
  
  I am now only study the notion of Regexp. I had seen the best of Mr. Blus. Can I know how it works. I need this logic of the functionality of wildcards in sting matching Regexp (wildcard string Matching).
  

  SQL> ed
  Wrote file afiedt.buf
   
    1  WITH test_data AS (
    2  SELECT 'c:\temp\folderA\fileA.txt' t FROM DUAL UNION ALL
    3  SELECT 'c:\temp\fileA.txt' t FROM DUAL UNION ALL
    4  SELECT '\\mymachine\A\fileB.txt' t FROM DUAL UNION ALL
    5  SELECT '\\mymachine\A\B\fileB.txt' t FROM DUAL UNION ALL
    6  SELECT '\\mymachine\A\B\C\image.jpg' t FROM DUAL UNION ALL
    7  SELECT '\\mymachine\A\B\C\D\music.mpg' t FROM DUAL UNION ALL
    8  SELECT 'c:\myfolder\folderD\folderE\4969-A.txt' t FROM DUAL
    9  )
   10  select regexp_replace(t, '^.*[\]([^\]*)[\][^\]*$','\1')
   11* from test_data
  SQL> / 
   
  REGEXP_REPLACE(T,'^.*[\]([^\]*)[\][^\]*$','\1')
  ------------------------------------------------------------------
  folderA
  temp
  A
  B
  C
  D
  folderE
   
  7 rows selected.
   
  SQL>

  Please help me in this matter.
  
  Iqbal

  Sabrina wrote:
  One last question what is the average of

  The final "\1" in

   select regexp_replace(t, '^.*[\]([^\]*)[\][^\]*$','\1') 
  

  Iqbal

  It is a reference.

  See here:
  http://download.Oracle.com/docs/CD/B19306_01/AppDev.102/b14251/adfns_regexp.htm#CHDHCIGH

  and in the middle of the table here:
  http://download.Oracle.com/docs/CD/B19306_01/AppDev.102/b14251/adfns_regexp.htm#CHDIEGEI

  Matches the nth previous subexpression, in other words, either grouped in parentheses, where n is an integer between 1 and 9. The parentheses cause > an expression be remembered; a backreference refers to him. A backreference account subexpressions from left to right, starting with the opening > bracket of each subexpression preceding. The expression is not valid if the source string contains less than n subexpressions preceding the \n.

  Oracle supports the expression of backreference in the regular expression pattern and the replacement of the REGEXP_REPLACE function string.
  The expression (abc: def) matches the strings abcxyabc and defxydef xy\1, but does not abcxydef or abcxy.

  A backreference allows you to search for a string repeated without knowing the actual string advance. For example, the expression ^(.*) \1$ > matches a line consisting of two adjacent instances of the same string.

  As explained in table 4-2, backreferences store sub-expressions matched in a temporary buffer, which allows to reposition the characters. You access the pads with the notation \n, where \n is a number between 1 and 9. Each subexpression brackets and is numbered from left to right.

• 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

• How can I refer to a variable in the regular expression

  Hello friends,
  
  I have this Regexp, extract the County code: (971)
  

  Select regexp_replace (regexp_replace ('05-000971 7910-324324', '\D'),'^ 0 * (971)? 0?') of double;

  It is very good and the need...
  
  But, thinking about the future, someone may need to remove the country code (961), so it is better if I put the code in a variable, but
  
  How can I list the County code via a variable since the Regexp:
  

  declare 
  a varchar2 (15);
  code number := 971;
  begin
  select regexp_replace(regexp_replace('000971 05 7910 - 324324','\D'),'^0*(code)?0?') into a from dual;
  dbms_output.put_line ( a);
  end;

  but it does not work?
  
  Best regards
  Fateh

  You must link the value of the variable code in the regular expression pattern

  
  select regexp_replace(regexp_replace('000971 05 7910 - 324324','\D'),'^0*('||code||')?0?') into a from dual 
  

• Regular expression matching receives only two digit in brackets

  Hello

  I use the regular expression of the correspondence with the following expression. It is only able to get all the numbers that are not mere numbers. I want to retrieve all the values which lie between >< in="" the="" string="" and="" create="" an="">

  Then

  182
  2

  would be output

  182

  2

  Any help would be appreciated. I've attached what I have so far. Right now I still have the >< and="" it="" can="" only="" grab="" numbers="" that="" are="" not="" single="">

  Use (>.) [ ^(<>)]*)<    *="" instead="" of="">

• 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 ^.

• How to change the mac address with a mac myself?

  Hello

  I need help to change the MAC address of one of my VM in a configuration.

  I need it because I need to do some testing on a program and the program's tie to the MAC address.

  I see how to reset the MAC, but not how set one I want... could someone help me with this.

  Lafa91

  BTW I have french speaking first of all sorry in advance for my bad English.

  Lafa im not very familiar with labmana. but the alternative is that you change your MAC address in the guest NETWORK adapter settings OS

• Help with the Regular Expressions and regexp_replace

  Oh great guru Oracle can I can receive assistance
  
  I need to clean the phone numbers that have been entered in the table per_phones of Oracle e-Business. Some of the phone numbers have hyphens, some have spaces and some have tank. I just want to get out all the figures and then re - format the number.
  
  E.g.
  914-123-1234... out (914) 123-1234
  9141231234... new (914) 123-1234
  914 123 1234... (914) 123-1234
  MyPhone... just null
  (914)-123-1234... (914) 123-1234
  
  I really tried to understand the instructions of regular expressions, but for some reason, I can't understand it.

  For example:

  SQL> with sample_data as (
    2   select '914-123-1234' phone_number from dual union all
    3   select '9141231234'                from dual union all
    4   select '914 123 1234'              from dual union all
    5   select '(914)-123-1234'            from dual
    6  )
    7  select regexp_replace(
    8           regexp_replace(phone_number, '\D')
    9         , '(...)(...)(....)'
   10         , '(\1) \2-\3'
   11         ) as formatted_num
   12  from sample_data
   13  ;
  
  FORMATTED_NUM
  --------------------------------------------------------------------------------
  (914) 123-1234
  (914) 123-1234
  (914) 123-1234
  (914) 123-1234
   
  

• Reset the virtual machine MAC address with API VCD

  Is it possible to get the same functionality that the VCD UI where for the properties of a virtual computer, you can go to drop-down list MAC address and select "Reset". After the virtual computer is updated, a new MAC address. I was not able to achieve this goal of the API by using the following link to change.

  API/v1.0/vApp/VM-2008703626/networkConnectionSection

  and by the way < MACAddress > Reset < / MACAddress >

  What makes the user interface behind the scenes? The same is possible via the API?

  Hello

  Try this element sends do not at all. This would reset the Mac address.

  Reset

  Kind regards

  Rajesh Kamal.

• Evaluation of the regular expression with logical operator

  Hi all
  
  I am little confused with expression with logical operator evaluation. I'm trying to understand the in expression.
  
  -----
  eXa.getTrue () & & eXa.getFalse () | eXa.getFalse () & & eXa.getTrue () | eXa.getFalse () is false and True Count: 1 False Count: 3
  
  According to the agreement, it should be true with True Count: 1 False Count: 3
  It must run the 1 getTrue() then 1stGetFalse() and getfalse() then 2nd jump 2nd getTrue() and must run 3rd fetFalse()
  
  -----
  eXa.getTrue () & & eXa.getTrue () | eXa.getFalse () & & eXa.getTrue () | eXa.getFalse () is true and true count: 2 False Count: 0
  
  According to the agreement, it should be true with True Count: 3 False Count: 0
  He must run getTrue() 2 1 1 getFalse() jump and run 3rd getTrue() and jump the last getFalse().
  
  -----
  eXa.getTrue () | eXa.getFalse () & & eXa.getFalse () | eXa.getTrue () & & eXa.getFalse () presents itself as true and true count: 1 False Count: 0
  
  According to the agreement, it should be true with True Count: 2 False Count: 2
  It must run 1 getTrue() and jump getFalse() 1st run 2nd getFalse() and then run getTrue() 2nd and 3rd then getFalse()
  
  Please, help me understand above expressions.
  
  
  Here is the definition of methods:
  
  Private boolean getTrue() {}
  trueCount ++;
  Boolean retrunValue = 4 > 3;
  Return retrunValue;
  }
  
  Private boolean getFalse() {}
  falseCount ++;
  Boolean retrunValue = 3 > 4;
  Return retrunValue;
  }
  
  
  Thanks for your help

  >

  added parentheses to make order of the most obvious BSP. adding a "?" to show calls not executed.

  (eXa.getTrue() && eXa.getFalse()) || (eXa.getFalse() && eXa.getTrue()) || eXa.getFalse()  comes as false and True Count: 1 False Count: 3
  
  (T && F) = F
  (F && ?) = F
  (F) = F
  
  F || F || F = F
  
  (eXa.getTrue() && eXa.getTrue()) || (eXa.getFalse() && eXa.getTrue()) || eXa.getFalse()  comes as true and True Count: 2 False Count: 0 
  
  (T && T) = T
  (? && ?) = ?
  (?) = ?
  
  T || ? || ? = T
  
  eXa.getTrue() || (eXa.getFalse() && eXa.getFalse()) || (eXa.getTrue() && eXa.getFalse())  comes as true and True Count: 1 False Count: 0
  
  (T) = T
  (? && ?) = ?
  (? && ?) = ?
  
  T || ? || ? = T