Regular expression matches the string starts by &

Similar Questions

• 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.

• Analyze the Mac address with the regular expression matching

  Hello world

  I have a problem with the function of regular expression matching,

  I try to analyse the response both a query arp - a 192.168.0.15 to retrieve the MAC address of the remote IP address, I used the following regular expression: ^ ([0-9a-fA-F]{2}[:-]){5}([0-9a-fA-F]{2})$

  I wonder why should I do a subset of the first string to extract only the part of the MAC address. The regular Expression function is not able to recognize the regular expression directly in the middle of a string?

  I only works when I extracted the subset of tring right as in the picture below.

  Thanks for your replies.

  Get rid of the "^" at the beginning of your regular expression. You are ordering him to find the model at the beginning of the string.

• 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 ^.

• 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="">

• 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 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'

• 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.

• Multiline - Regular Expression Match string

  I'm trying to understand the format of a regular expression to pull select off multi-line string lines and fill in these lines as the individual elements of an array of strings using regular expressions to Match. The total length of the multiline string may vary as well as the text in the string. The string can contain letters, numbers and special characters. I've attached an example VI. In the example of VI, I want to only return lines starting by "device #" in the table. The number of lines starting by "device #" can vary, but I want to capture them all.

  Or is there a better functioning to be used instead of the corresponding regular Expression that will give me the desired result?

  aaronb wrote:

  I'm trying to understand the format of a regular expression to pull select off multi-line string lines and fill in these lines as the individual elements of an array of strings using regular expressions to Match. The total length of the multiline string may vary as well as the text in the string. The string can contain letters, numbers and special characters. I've attached an example VI. In the example of VI, I want to only return lines starting by "device #" in the table. The number of lines starting by "device #" can vary, but I want to capture them all.

  Or is there a better functioning to be used instead of the corresponding regular Expression that will give me the desired result?

  Corresponding regular expression works well for this.

  

  Ben64

• 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

• 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.

• 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

• 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

• 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

• pattern by using regular expressions match

  I'm playing (and wrong) with regular expressions

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + > ') twice;

  give < PSN > - I understand that

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + >, 1.2 ') twice;

  give < 231 > - I understand that

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' < [[: alnum:]] + >, 1.3 ') twice;

  give < ABc > which confused me until I realized that < 3 25 > has not been matched, because it has a space inside

  so I changed it to

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' <. * + > ', 1.3) twice;

  who gave a syntax error, so I changed it again to

  Select regexp_substr (1 < PSN > / # < 231 > # < 3 25 3 > / < ABc > ',' <. * + > ') twice;

  that works, but gives < PSN > / # < 231 > # < 3 25 3 > / < ABc >

  I guess because the. * corresponds to anyting that all but the last closure >

  The question I have is how can I retrieve the text of mounting bracket included (even if it has multiple spaces)?

  Hello

  9c5dfde3-EAAE-45A7-80a1-bba8a71c826c wrote:

  Thanks for that - is there a way to return all 4 surrounded by extracts of <> without resorting to PL/SQL?

  Of course;

  REGEXP_REPLACE (str

  , '(^|>)[^<>

  , '\1'

  )

  Returns a copy of str with all outside rafters removed, for example

  <231><3 25="">

  .  It doesn't matter how many pairs of sharp hooks - there is.