w3hello.com logo
Home PHP C# C++ Android Java Javascript Python IOS SQL HTML videos Categories
Regular Expression that matches text between a specific string and a character
$var = 'Variable1 = Value1, Variable2 = Value2, Variable3 = Value3, Variable4 = Value4,'; $new = preg_split( '/,/', $var); print_r($new); That will break it up into an indexed array Array ( [0] => Variable1 = Value1 [1] => Variable2 = Value2 [2] => Variable3 = Value3 [3] => Variable4 = Value4 [4] => )

Categories : Regex

Regular expression match any character at the beginning end with specific string
Try this as a regex: /S+-line(?![-w])/ The word anchor is not suitable here since dashes are not considered part of a word, so /S+-line/ would mistakenly match text-with-line-not-to-be-replaced. Hence the lookahead construct. Of course, according to your use case, S may seem a little coarse. If your words really only consist of letters then dashes etc, then you can use the normal* (special normal*)* pattern: /[a-z]+(-[a-z]+)*-line(?![-w])/i (normal: [a-z], special: -) (edit: changed the lookahead construct, thanks to @thg435)

Categories : Javascript

How to exclude a specific string/pattern from a regular expression in oracle?
try to replace the string you want to exclude with a string that would not be found by your regexp. e.g with str as ( select '1013;' as s from dual union select '1013' from dual union select '1013:' from dual union select '10133;' from dual ) select s , regexp_substr(s, '1([^34][:;]|[0-9][^:;].*)') --> regexp , regexp_replace(s,'^1013($|:|;)','x') --> replaced string , regexp_substr(regexp_replace(s,'^1013($|:|;)','x') , '1([^34][:;]|[0-9][^:;].*)') --> regexp with replaced string from str ;

Categories : Regex

How to identify using a regular expression a specific string written over two lines within a text?
Simple S+ ? S+ S+ : match non whitespace characters one or more times ? : match newlines in windows/linux s* : match whitespaces zero or more times S+ : match non whitespace characters one or more times Online demo

Categories : Regex

Need a specific regular expression
Try this: [^[]*[[^]]*] ^----^ anything but [ ^^ - [ ^----^ - anything but ] ^^ - ] Optionally you can wrap your whole regex in ^ and $ to match whole string only. Example: > /^[^[]*[[^]]*]$/.test("test[1,2]") true > /^[^[]*[[^]]*]$/.test("test[1,test2]") true > /^[^[]*[[^]]*]$/.test("test_wrong") false > /^[^[]*[[^]]*]$/.test("test[wrong") false

Categories : Javascript

Constructing a specific regular expression
Does this regular expression work for what you need to do? String pattern = "([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+)"; Alternatively, you could split the String using spaces. line.split(" "); Do note that in each of these cases, the value you are getting is a String. You'll need to convert them to integers or doubles using Integer.parseInt() or Double.parseDouble(). EDIT To ignore the first section - control.avgo String pattern = "[^ ]+ ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+) ([^ ]+)"; Matcher.group(0) specifically matches the entire pattern. So simply don't use it.

Categories : Java

Regular Expression for a specific set of words in a sentence
Description This expression will match the your phrases, and ensure they are not embedded inside another larger word. ^.*?(?:s|^)(ofstheshouse|time|thississhow|home)(?=W|$).* PHP Code Example: You didn't specify a language so I'm just providing this php example to simply show how this works. Sample Text 1) "I was coming out of the house" 2) "I remember the time when I used to be a baby" 3) "Well, I am not sure what you did, but this is how I fix my problems" 4) "When are you coming home?" 5) "This is howard Timey said of the houseboat" 6) "The last word in this line is home Code <?php $sourcestring="your source string"; preg_match_all('/^.*?(?:s|^)(ofstheshouse|time|thississhow|home)(?=W|$).*/imx',$sourcestring,$matches); echo "<pre>".print_r($matches,true); ?> Mat

Categories : Regex

regular expression for strings with specific values
You need to anchor correctly, use the correct grouping and repetition syntax and allow for presence of other characters. See this: import re re.search(r'^M.*(CD|EF)+.*F$', "M 123ABNCDandEF78949XZ F").group() 'M 123ABNCDandEF78949XZ F'

Categories : Python

Specific decimal number regular expression
You could prohibit the number from containing all zeros with a negative lookbehind. That would also invalidate all entries that consist of zeros and dots, but that's probably a good thing: it is very likely that since you do not want a standalone zero, you do not want 00, 000, or 0.000 as well: ^(d{1,4}(.d{1,3})?)(?<!^[0.]+)$ // ^^^^^^^^^^^^ // ||| // The lookbehind part The part that I added to your expression checks that the string does not consist entirely of zeros and dots. Here is a demo on ideone.

Categories : C#

Regular expression to extract a specific value from HTML anchors
Please don't use regex to parse HTML. Use something like BeautifulSoup. It's so much easier and better :p from bs4 import BeautifulSoup as BS html = """<a href='http://xyz.com/1' class='page larger'>2</a> <a href='http://xyz.com/2' class='page larger'>3</a> <a href='http://xyz.com/3' class='page larger'>4</a> <a href='http://xyz.com/4' class='page larger'>5</a> <a href='http://xyz.com/5' class="next">»</a>""" soup = BS(html) for atag in soup.find_all('a', {'class':'next'}): print atag['href'] With your example, this prints: http://xyz.com/5 Also, your regular expression works fine.

Categories : Python

Java regular expression to match everything between same specific character sequences
Just change your second group into a lookahead foo(.*?)(?=foo|$) See it on Regexr The problem is you are matching the "foo" that you want to use as next start point. You can avoid this by using the lookahead assertion. This way the following "foo" is not matched and therefor used as start of the next match.

Categories : Java

Regular expression to split by a specific occurrence of a token while ignoring other occurrences
You can use a negative lookbehind: (?<!\), As a Java string, that would be "(?<!\\),".

Categories : Java

Python Regular Expression - Regular Expression that finds one pattern per line?
There's no point loading the whole file into memory to do an re.findall and by doing so you lose the ability to short circuit upon finding the first match import re with open('data.txt') as f: print sum(1 if re.search(r"Line", line) else 0 for line in f) 5

Categories : Python

Regular expression returns empty array in php even though the regular expression is correct
I assume you were trying to do a noncapture group for <price... but you missed the :. Or you should take out the question mark. If the price group is optional, try like the regex below. You should use the following website to help you with regex. I find it extremely helpful. <td>( |s)*?(<span( |s|.)*?</strong>( |s)*?$(<price>)*([0-9.]*).*?)$(.*?)( |s)*?< Edit live on Debuggex In the above example, your first match would have the following captures: 0: "<td><span class="offer"><strong>xscre:<br></strong>$299 xxxxx&x;xx<span class="fineprint_number">2</span></span><br>de&ea;s $399<" 1: "" 2: "<span class="offer"><strong>xscre:<br></strong>$299 xxxxx&x;xx<span clas

Categories : PHP

Compare the string with other string using regular expression in oracle
Select 'Y' from dual where regexp_like('test|xyz|test1|xyz1','xyz','i'); It will return result only for matched values. If you want 'n' for unmatched values then try regular expression instring. Or you can do it simply with lower() function on both side and checking for instring. select case when regexp_instr('test|xyz|test1|xyz1','XYz',1,1,1,'i') > 0 then 'y' else 'n' end col1 from dual

Categories : SQL

Get string with PHP regular Expression
Quick & Dirty: #w+ #b+ d+(?:.?d+)? (.*) Example: <?php $string = "France Gros Frère et Sur Hte-Cote de Nuit Blc 2008 #wwww #bbbbb 8.5 Nice yellow fruit nose, some vanilla notes, good crispness"; $regex = "/#w+ #b+ d+(?:.?d+)? (.*)/"; preg_match ($regex, $string, $output); echo $output[1]; ?> But if there can be a string WITHOUT any number after #bbbbb, you better use this: #w+ #b+s*(?:d+(?:.d+)?)?s*(.*) So you don't have to put any number after #bbbbb and you can use as many spaces as you like between #bbbbb, the number (if there's any) and the string you want to extract. Most of it is optional, so your string could look like this: blabla #w #bb Hello World Or like this blabla #wwwwwwwwwwwwwwww #bbb 1337 Hello World Or like this: #w #bHello World

Categories : PHP

EDU String to Regular Expression
I would implement two validations: check for correct email format validates :email, :format => { :with => /A([^@s]+)@((?:[-a-z0-9]+.)+[a-z]{2,})/i, :on => :create } check for .edu domains validates :email, :format => { :with => /.edu/i, :on => :create }

Categories : Ruby On Rails

Infopath Regular Expression End of String
You can use a negative look-ahead to mimic $: [0-9]{6}(/[0-9]{6})*(?!.) Using [0-9] instead of d because the latter could have unicode digit chars. If needed, you can use a negative look-behind to mimic ^ as well: (?<!.)[0-9]{6}(/[0-9]{6})*(?!.)

Categories : Regex

Extract from string using regular expression
## Preparation s <- "test.test AS field1, ablh.blah AS field2, faslk.lsdf AS field3" pat <- "(\w+)(?:,|$)" ## Note the doubly-escaped \w ## Use the powerful gregexpr/regmatches one-two punch m <- gregexpr(pat, s) paste(regmatches(s, m)[[1]], collapse=" ") # [1] "field1, field2, field3"

Categories : Regex

Match a string in a csv using regular expression
The '+' symbol in a regular expression is a quantifier, it stands for "one or more" so you are using it wrong you don't need to concatenate if that is what you are trying. You also need to use the method matches.

Categories : Regex

Can't change regular expression string
You can create a regex simply by creating a new RegExp class. public function RegExp(re:String, flags:String) function tagCheck(tag:String, rez:String) : String { var tagRegex:String = tag.replace(new RegExp('[.\\+*?\[\^\]$(){}=!<>|:\' + ('/' || '') + '-]', 'g'), '\$&'); var regExp:RegExp = new RegExp('\[' + tagRegex + '\](.*?)\[\/' + tagRegex + '\]', 'g'); var matches:Object = regExp.exec(rez); return matches[1]; } Should work (though untested). The first line is to make sure that a tag called my.tag gets changed into my.tag.

Categories : Actionscript

Alphanumeric Regular Expression String
I think you will have to escape the +, and the metacharacter after [BC] needs to be * for "zero or more" instead of + for one or more. You also don't need the ? at the end since there's no need to make this a non-greedy match; you're matching till the end of the string anyway: /^+?d+[BC]*$/ Otherwise, what you have is not really a valid regex. There is nothing to repeat for the + at the beginning, and the ^ is just an anchor for the beginning of the string.

Categories : Regex

Regular expression to replace a string
The @ makes it necessary to escape all the " with a second ", so "". Without it to escape the " you would have used ", but I consider it better to always use @ in regexes, because the is used quite often, and it's boring and unreadable to always have to escape it to \. Let's see what the regex really is: Console.WriteLine(@"""[^""~]+""([^~]|$)"); is "[^"~]+"([^~]|$) So now we can look at the "real" regex. It looks for a " followed by one or more non-" and non-~ followed by another " followed by a non-~ or the end of the string. Note that the match could start after the start of the string and it could end before the end of the string (with a non-~) For example in car"hello"help it would match "hello"h

Categories : C#

String matching with regular expression
Tested: $subject = "I want to match occurence of function gcb_process."; $pattern = '/function gcb_process/'; preg_match($pattern, $subject, $matches); print_r($matches); Output: Array ( [0] => function gcb_process ) Read PHP Doc more details

Categories : PHP

Why this string matches the regular expression?
In your regular expression, there's trailing |: # ^More word lists and tips at http://wwwmajortests.com/word-lists$| # ^ Empty pattern matches any string: >>> import re >>> re.match('abc|', 'abc') <_sre.SRE_Match object at 0x7fc63f3ff3d8> >>> re.match('abc|', 'bbbb') <_sre.SRE_Match object at 0x7fc63f3ff440> So, Remove the trailing |. BTW, you don't need ^ becasue re.match checks for a match only at the beginning of the string. And, I recommend you to use raw strings(r'....') to correctly escape backslahes. ADDITIONAL NOTE d matches only a single digit. Use d+ if you also want to match multiple digits.

Categories : Python

Split string using regular expression in php
you can split them if they are always like this vith substr() function reference: http://php.net/manual/en/function.substr.php but if they are dynamic in lenght. you need to get a ; or any other sign that is not likely to be used there before 2nd "http://" and then use explode function reference: http://php.net/manual/en/function.explode.php $string = "http://something.com/;http://something2.com"; $a = explode(";",$string);

Categories : PHP

regular expression for finding string between [[ ]] and {{}}
You must double escape the opening square brackets (you can do the same with the closing) since they are used to define character classes: Pattern p = Pattern.compile("\[\[(.*?)]]"); (exactly the same with curly braces, that are used for quantifiers) You can read this fabulous post about the incredible world of square brackets.

Categories : Java

Regular Expression Replace String
Your expression is close. Are there any conditions where this won't work? (d*.d*)s\w{2}s(d*.d*) with the replace pattern being (for JS) [$1, -$2, $2] or for emacs (according to http://www.emacswiki.org/emacs/RegularExpression) [1, -2, 2]

Categories : Regex

Regular expression string followed by numbers
This is very easy, the regexp you are looking for is /#Question[0-9]+#/ If you need to extract the number you can just wrap the [0-9]+ part in parenthesis /#Question([0-9]+)#/ making it a group. How you use a captured group depends on the specific regexp implementation (e.g. python, perl, javascript ...). For example in python you can replace all those questions with corresponding answers from a list with answers = ["Andrea", "Griffini"] text = "My first name is #Question1# and my last name is #Question2#" print re.sub("#Question([0-9]+)#", lambda x:answers[int(x.group(1)) - 1], text)

Categories : Regex

Extract xml from string using regular expression
why aren't you matching from <HotelML to </HotelML? something like: <HotelML .*</HotelML> Or, just go through the file line by line, and whenever you find a line matching ^.* PM >>Response:.*$ read the following lines as xml until the next matching line...

Categories : C#

Simple string regular expression
if you go to http://gskinner.com/RegExr/ you can write this expression: ^(C0[0-9]*[0-9]{4})[^0-9] And in the content you put: C012345 - legal C047851 - legal C*1*54821 - not legal (does not starts with 'C0') C0412 - not legal (mismatch length) C0*a*4587 - not legal And you will see that it only matches what you want.

Categories : C#

Regular expression to take 2 parts of a string in c#
This pattern: (^[^(]*).*[(.*)project.csproj]$ Captures these groups: dir1dir2dir3file.aspx.cs C:dirdirdirdir amespace.namespace.namespace.namespace If the name project.csproj file could change, you can use this pattern instead: (^[^(]*).*[(.*\)[^\]*]$ This will match everything up to the last path fragment within the brackets.

Categories : C#

JavaScript: How to pull out a string from a URL using a regular expression
Use a lookahead and a lookbehind, (?<=http://localhost:8080/bladdey/shop/).+?(?=/hired3) Check here for more information. Also, there is no need to escape the : or / characters.

Categories : Javascript

How to extract parts of a particular string with a regular expression?
You can use regex d matches a single digit + is a quantifier which matches preceding pattern 1 to many times So d+ would match 1 to many digits Your code would be public int addAllInts(String s) { int temp=0; Matcher m=Pattern.compile("\d+").matcher(); while(m.find()) { temp+=Integer.parseInt(m.group()); } return temp; }

Categories : Java

Creating Regular Expression to search a string
Remove the anchor ^ from the regex and allow variables with a length of more than one character. /^{(w):(w)}/ becomes: /{(w+):(w+)}/

Categories : PHP

Regular Expression to select first five CSVs from a string
Following negative look-behind based regex will work: Pattern pattern = Pattern.compile("(?:.*?(?<!(?:(?<!\\)\\)),){0,5}"); However for full fledged CSV parsing better use a dedicated CSV parser like JavaCSV.

Categories : Java

Regular expression - also empty string allowed
I believe that the following would work: /^$|^+?[0-9]+(([0-9]+))?[0-9-]*[0-9]$/i I've inserted ^$| at the beginning of your regex. The pipe (|) is called an "alternation" indicator and says that either the expression before it or after it must match. ^$ will simply match a string with no characters between the start and end of input. The result: an empty string or the original expression.

Categories : Javascript

Regular Expression to get a string between parentheses in Javascript
You need to create a set of escaped (with ) parentheses (that match the parentheses) and a group of regular parentheses that create your capturing group: var regExp = /(([^)]+))/; var matches = regExp.exec("I expect five hundred dollars ($500)."); //matches[1] contains the value between the parentheses console.log(matches[1]); Breakdown: ( : match an opening parentheses ( : begin capturing group [^)]+: match one or more non ) characters ) : end capturing group ) : match closing parentheses Here is a visual explanation: tinyurl.com/mog7lr3.

Categories : Javascript

Can a string be padded to match a regular expression?
Have you considered splitting the regex on path separators and only applying the portion of the regex that applies at the current depth of your search? This seems like a much more efficient way of proceeding.

Categories : Regex

Regular Expression to remove underscore and string
Try using str_replace; it's much more efficient than regex for something like this. However, if you want to use regular expressions, you need a delimiter: preg_replace('|_normal|','', $url);

Categories : PHP



© Copyright 2017 w3hello.com Publishing Limited. All rights reserved.