regular expression to exclude spaces in the string

Similar Questions

• Regular expression to remove space after the tag < a... >?

  Hello

  I'm relatively new to using regular Expressions, but am in need who will help me find all the tags < a... > with space immediately after this tag and replace it with the exact same tag, but not space after the tag < a... >.

  Thus, for example, a regular expression that there is:

  < a href = "somelink.html" > Somelink < /a >

  as well as:

  "< a href =" # "MM_swapImgRestore" onmouseover = "MM_swapImage (' sub-nav_button_professional_portal',", ' images/sub-nav_button_ professional_portal_f2.png', 1 ') "> Somelink < /a >

  And just remove the space in the hyperlink.  Does anyone know how to do this?

  Find-

  \ (.*)\

  Replace-

  $2

  Post edited by: Murray * CPA *.

• Insert a space in the string

  Hello

  I'm trying to put spaces in the string after each 4 characters (2 bytes).

  Any help will be greatly appreciated.

  Thank you

  hiNi.

  Several ways to do so.  My favorite is the regular expressions...

  

• Can I do regular expressions or Boolean logic in the search?

  Can I do regular expressions or Boolean logic in the search? (Windows + F) Suppose I want to search pdf files or text files. I can go * .txt | * .pdf?

  I searched for about an hour for a simple answer to this and this is the closest, I came, but it still not answering the question.

  Can I use expressions simple boolian in instant search and if yes what are.

  I think that AND and WOULDN'T cover 90% of what I want.

  I want to search for emails for things such as [Minutes AND project x]

  Currently, this property returns all the messeges with minutes and all messages with project x.
  Since I have minutes of many projects, and many emails with project x not the minutes that returns are many.   I would use a simple AND to get the intersection.

  If and expression exist, I have found no reference to it.

  According to me, the back had these expressions in the search function.

  Thank you

• Count the spaces in the string...

  Hi, I use "PL/SQL Developer" program.
  I need to count the number of spaces in the string...
  Can someone help me with this...?
  
  String, for example: 'One AsdA qwe' should return number = 2 - (2 spaces in this string)

  Raivis wrote:
  Hi, I use "PL/SQL Developer" program.
  I need to count the number of spaces in the string...
  Can someone help me with this...?

  String, for example: 'One AsdA qwe' should return number = 2 - (2 spaces in this string)

  This should do it.

  LENGTH() - LENGTH(REPLACE(, ' '))
  

• Remove the spaces from the string

  Hello
  Does anyone have a simple routine/function to remove all but 1 space between "words" (row charactives no spaces).
  
  For example, I would convert the string (I used '.' as a 'space'):
  
  'This... too... a lot of people of spaces' to
  'This.has.too.many.spaces '.
  
  I need this to compare strings of registration where a user can hit the space bar one extra time or 2.
  
  Thanks in advance.

  hefterr wrote:
  > Thank you Dan. This seems good.
  > Can I use...
  > Is not sure the quotes for 2
  > spaces

  You need quotes replacement in case of nesting them like this... Make your choice.

  
  OR
  

• Trim white spaces in the string.

  Hi all
  
  I want to know how to cut the white space in the whole of the line when reading the file. For that, the threshold is,
  
  October 7, 2010, DRI DEV, 07A0002EE, BIOS 0.75FAWARM 2.003, DRI, 215461 D, 2.0.5-O,, manson, COMPUTING equipment, 07A0002EE '
  I use the regexp_replace function. but I'm not able to use the appropriately. See the following query:
  
  Select trim (regexp_replace (October 7, 2010, DRI DEV, 07A0002EE, BIOS 0.75FAWARM 2.003, DRI, 215461 D, 2.0.5-O,, manson, COMPUTING equipment, 07A0002EE ',' [[: space :]]')) text]]))
  Double;
  
  o/p: 10/7/2010,,DRIDEV,,07A0002EE,BIOS0.75FAWARM2.003,DRI,,215461D,2.0.5-O,,,manson,GearIT,07A0002EE. Here, the space also replacement including white spaces. I don't want to replace only the remaining spaces space need as expected o/p.
  
  EXPECTED O/p: 07/10/2010, DRI DEV, 07A0002EE, BIOS0.75FAWARM2.003, DRI, 215461 D, 2.0.5-O,, manson, Gear IT, 07A0002EE
  
  Oracle version:
  
  Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
  PL/SQL Release 10.2.0.4.0 - Production
  CORE 10.2.0.4.0 Production
  AMT for IBM/AIX RISC System/6000: release 10.2.0.4.0 - production
  NLSRTL Version 10.2.0.4.0 - Production
  
  Please help me.
  
  
  
  Thanks in advance!
  
  Kind regards
  Florian...

  select
  regexp_replace(
    regexp_replace('10/7/2010,,DRI DEV,,07A0002EE                           ,BIOS 0.75FAWARM 2.003     ,DRI,,215461 D ,2.0.5-O     ,,,,manson,Gear IT,07A0002EE'
    ,'^\s+|\s+$')
  ,'\s*,\s*'
  ,',')
   from dual
  

  Kind regards
  Sayan M.

• How can we remove the present space in the string in BlackBerry?

  Hello

  Suppose I have string like "New York". How can I remove the space between and do like "New York". As BlackBerry does not support the Java replaceAll() method, I am not able to do.

  Thank you...

  StringUtilities.removeChars (...)

• #... expression # leaves extra space for the first #.

  Hello

  I run CF 9.01 and have a nasty white space problem:

  Fill out a form like this field:

  ---

  ... value = "#application.utils.FormatTS (now ()) ' # '..."

  ---

  This leaves behind in the generated source code the following:

  ----

  " value= ' 2011-01-22 03:34:03"

  ---

  Later, the empty space causes a problem. I saw this bug throughout all versions of CF. One can avoid this problem in all cases where we can concatenate strings as in "some test" & application.utils.FormatTS ()... This is not geneate empty space.

  IsDate (CF) returns if the space is there... LSIsDate () returns YES... in my humble OPINION inconsistent behavior.

  BTW: The FormatTS () function generates NO space left. Is that the way that CF generates code

  BTW: I know that I can get rid of the space with trim()... However, it's a mess to be forced to do so in all THE processes of these constructions...

  No known cure for this? Or any reason WHY CF like this? How can I seduce do not generate the code of juice?

  Thank you

  Martin

  BTW: The FormatTS () function does NOT generate the main

  space. Is that the way that CF generates code

  Are you positive? It's certainly the most likely cause. Check the cffunction and make sure you have set output = "false".

• Validation of regular expression delimiter alphanumeric and hose

  I have a header in the file as below

  EMP_ID | EMP_NAME | DEPARTMENT | SALARY | ACTIVE1

  past to a string

  String test = ' EMP_ID | EMP_NAME | DEPARTMENT | SALARY | ACTIVE1;

  I need to check if the header is only of alphanumeric characters and pipedelimiter is allowed.

  other than these, I need to trigger an error.

  Any suggestions.

  had the regular expression pattern

  It's the string pattern = "^ ([a-zA-Z0-9 |]) '. +)$";

• 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

• 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})$ /))

  {

  ...

• Helps the understanding of regular Expressions

  Hello people,
  
  I need help to understand Regular Expressions.
  

  -- This returns the Expected string from the Source String. ", Redwood Shores,"
  SELECT
    REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                  ',[^,]+,', 1, 1) "REGEXPR_SUBSTR"
    FROM DUAL;
  
  REGEXPR_SUBSTR
  -------------------------------
  , Redwood Shores,
  
  However, when the query is changed to find the Second Occurrence of the Pattern, it does not match any. IMV, it should return ", CA,"
  
  SELECT
    REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                  ',[^,]+,', 1, *2*) "REGEXPR_SUBSTR"
    FROM DUAL;
  
  REGEXPR_SUBSTR
  -------------------------------
  NULL

  Can someone help me understand why second query not returning ", CA, '?
  
  I did research on this forum and found the link on the thread "https://forums.oracle.com/forums/thread.jspa?threadID=2400143" for the basic tutorials.
  
  
  
  Kind regards
  P.

  The reason is that the comma between 'Redwood Shores' and 'CA' already represents the first occurrence.
  So it can not match the second occurrence at the same time.

  You can replace to (remove the trailing ',' in the regex):

  SELECT  REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',',[^,]+', 1, 1) REGEXPR_SUBSTR  FROM DUAL;
  
  SELECT  REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',',[^,]+', 1, 2) REGEXPR_SUBSTR  FROM DUAL;
  

  Published by: hm on 14.06.2012 00:52

  When you remove also the leading comma you get to the BlueShadows solution.

• Print every word in the new line for a space in a string

  Hi all

  I would print every word in the new line, if there is a space in the string.

  Is there a better way using regexp?

  declare

  number of v_count;

  v_text varchar2 (1000): = 'Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - Production | " ';

  v_single_text varchar2 (1000);

  number of v_space_pos;

  Start

  Select regexp_count (v_text,' ') in double v_count;

  While v_count > 0

  loop

  v_space_pos: = instr (v_text,' ');

  v_single_text: = substr(v_text,1,v_space_pos-1);

  dbms_output.put_line (v_single_text);

  v_text: = substr (v_text, v_space_pos + 1);

  v_count: = v_count-1;

  end loop;

  end;

  Thank you

  Rambeau

  HI - try as

  with t as (select 'Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - Production' |) "" double str)

  Select regexp_substr (str,'[^] +', 1, level)

  t

  connect by level<=regexp_count(str,'[^>

• Another question about regular expressions with String.matches

  don't match String.matches () method expressions when a substring of the string matches, or must match the whole string? So if I have the string '123ABC', and I ask match "1 or more letters" will be it fail because there are other that the letters in the string, but then spend if I add "1 or more letters AND numbers 1 or more? Thus, in the second case each character in the string is recorded in the research, as opposed to the first. Is that correct, or are there ways to JUST matching a substring in the string instead of all this? I'll do some examples too... but that makes sense?

  It must match the entire string. Use Matcher.find () to match on just a sub-string)