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regex pattern to match specific number pattern , skip if there different pattern
You could try this: 57d{7}(?:-d)? Here's what it looks like: In Java, that would be Pattern.compile("\b57\d{7}(?:-\d)?\b").

Categories : Java

Regex to match paths that don't match a specific pattern: Express Router
The following regex will match any path except those starting with /foo/ app.get(/^/([^f][^o][^o]|.{1,2}|.{4,})/.*$/, routes.index); I assume that this is a standard javascript regex.

Categories : Regex

Regex wildcard to only match strings which don't have an underscore
Your pattern won't do what you think it does. It will match a single A, n, y or 1 character, followed by a literal _from, followed by a single A, n, y or 2 character. The start (^) and end ($) anchors around your string will also ensure the entire string must match the pattern, and not just a substring. Perhaps you want a pattern like this: ^(.*)_from([^_]*) This will match zero or more of any character, captured in group 1, followed by a literal _from, followed by zero or more of any character other than underscores, captured in group 2. It will also allow any other characters to follow the matched substring. Or possibly this: ^([^_]*)_from([^_]*) This will match zero or more of any character other than underscores, captured in group 1, followed by a literal _from, followed by zer

Categories : Dotnet

how to match everything except a particular pattern using regex
^/user/(?!login)(.*) I altered your regex by requiring there to be a word boundary after login for the lookahead to match, e.g. login followed by a non alphanumerical character (space or / for example). It seems to work.

Categories : Regex

Regex pattern to match three different things
I tried to get a pattern for it, but I can't exclude the "by" word: Use positive lookbehing instead of positive lookahead: (?<=by )[A-Za-z ]+(?=,) Working DEMO

Categories : Regex

How to match alphanumeric pattern in Regex?
Say you have the following string: T071108452T 42D896D5O 3587O Note that I added an extra O --^ If you want to match until the last O, you may use the following pattern: [ws]+(?=O) This means: [ws]+ match words and whitespaces one or more times, greedy. (?=O) Zerowidth lookahead assertion to match until O found Now if you want to match until the first occurence of O then you may use the following pattern: [ws]+?(?=O). Note the added question mark, it's to match ungreedy. Note: w will also match an underscore, you may replace [ws] by [^WS_] to prevent that. Note the negation and the uppercased letters.

Categories : Regex

Match pattern in regex expression
Try using this pattern: ^(?< pol>w+)_(?< fac>[^_]+)_(?< end>w+)_(?< op>w+) The [^_] bit a character class which means 'match anything except an underscore'. If the delimiter is |, you'll have to use | in your pattern, because | has special meaning in regular expressions (although you won't need to escape it inside a character class). Like this: ^(?< pol>w+)|(?< fac>[^|]+)|(?< end>w+)|(?< op>w+) On a side note, I find it's much easier to use verbatim strings when specifying a regular expression pattern because you won't have to type in so many escape sequences: new Regex(@"^(?< pol>w+)|(?< fac>[^|]+)|(?< end>w+)|(?< op>w+)"); However, in this case, you might be better off just using Split: var result = input

Categories : C#

How can I match this complex pattern with regex?
If I understood you correctly you can match it like this: var campuses = ["Barrie", "Orangeville"]; var terms = ["Winter 2014", "Summer 2014"]; var your_regex = new RegExp("(" + terms.join("|") + ")\s-\s(" + campuses.join("|") + ")"); Then test if your record matches: if (your_regex.test(your_record)) { // Record matches }

Categories : Javascript

How to extract which combination of regex match from my pattern?
You can use matcher.group() method once you get Matcher out of that Pattern Matcher m = ANIMALS.matcher("The boy is good"); while(m.find()) { System.out.println(m.group()); }

Categories : Java

multiline regex pattern that match 3 digits
So the entire file must look like this? Then try new Regex(@"A(?:d{3} ? )*z") Explanation: A # Start of string (?: # Match the following (non-capturing) group: d{3} # - three digits ? # - one CRLF or LF (linebreak) )* # any number of times (0 or more) z # until the very end of the string If the file might not end with a newline (not sure from your description), you can use new Regex(@"A(?:d{3} ?$ ?)*z", RegexOptions.Multiline) This initially makes newlines optional ( ? ?) but ensures that there is a line ending after every three-character bit by placing the end-of-line anchor $ between CR and LF, which is where (strangely) .NET thinks it should match.

Categories : Dotnet

Scala Regex Pattern Match not able to get it to work
With the repeated group, you only get the last match. To get them all, use findFirstMatchIn or similar. There are certainly duplicate questions. scala> val r0 = "([a-z]+)".r.unanchored r0: scala.util.matching.UnanchoredRegex = ([a-z]+) scala> val m0 = r0 findFirstMatchIn x m0: Option[scala.util.matching.Regex.Match] = Some(abc) scala> val r1 = "(\.?[a-z]+)".r.unanchored r1: scala.util.matching.UnanchoredRegex = (.?[a-z]+) scala> val m1 = r1 findFirstMatchIn m0.get.after m1: Option[scala.util.matching.Regex.Match] = Some(.def) scala> r1 findFirstMatchIn m1.get.after res2: Option[scala.util.matching.Regex.Match] = Some(.hij)

Categories : Regex

Java Regex: Match any word from pattern
Split the user-specified phrase by s+ into, say, arr. Build the following pattern: "\b(?:" + Pattern.quote(arr[0]) + "|" + Pattern.quote(arr[1]) + "|" + Pattern.quote(arr[2]) + ... + "\b" Compile without the Pattern.LITERAL option. In other words, if you want your patterns to match words in a user-specified phrase, you have to use alternation (the pipes) so that any one of those words can be considered a match. However, using the Pattern.LITERAL option makes the alternation operators literal—therefore you have to "literalize" just the words themselves, using the Pattern.quote(...) method. The \b are word boundaries so that you do not match, say, a word in the user's phrase like "bar" when encountering text like "barrage". Edit. In response to your edit. If you want to match th

Categories : Java

Regex to match words in sentence after a pattern
Maybe the regex should be: How much is (.*)? Or if you want to match all the words but one word in each capture: How much is (?:(w+)s*)+? Regex regexWords = new Regex(@"How much is (?:(w+)s*)+?"); foreach(Capture word in regexWords.Match(input).Groups[1].Captures) { // word.Value contains one word. } Good luck with your quest.

Categories : C#

regex pattern to match words of a string
You can use array_filter in combination with a regular expression to achieve this: $array = array("hello word", "lovely child", "i am lost in paradise" ); $term = "lo"; // necessary in case $term contains characters with special meaning in a regex $term = preg_quote($term, '/'); $results = array_filter( $array, function($el) use($term) {return preg_match('/'.$term.'/', $el);} ); The regular expression uses the word boundary anchor to ensure that the search term appears at the beginning of a word.

Categories : PHP

RegEx pattern to match between nth and mth occurence of a word
K is so under-appreciated grep -oP '^(.*?one){2}K.*?(?=one)' <<< "There is one wherever you one and here is the one , nowhere is the one" and here is the

Categories : Regex

Why can't Regex match Zip Code pattern in Richtextbox1
The problem is you are "anchoring" the regex by using the beginning of input (^) and end of input ($) metacharacters. What you are saying, in essense, is match nothing BUT a zip code (with no surrounding text). Just remove the anchoring characters, and your solution will work: d{5}(?:[-s]d{4})?

Categories : C#

Get original Regex pattern for which match is found
You should use Regex.Matches as this will return all matches in a collection. foreach (Match m in Regex.Matches(value, pattern)) Console.WriteLine(m.Value);

Categories : C#

Regex Pattern for filter out anything that doesn't Match
Can't you just do: string matchedText = null; var match = Regex.Match(myString, @"MV:[0-9]+"); if (match.Success) { matchedText = Value; } Console.WriteLine((matchedText == null) ? "Not found" : matchedText); That should give you exactly what you need.

Categories : C#

Java using regex to match a pattern for quizzes
I personally wouldn't use a regex, I would just use a StringTokenizer on the , and just check if the first character is a numeric (since no other lines seem to start with a number). But to more specifically answer your question. You need to specify the MULTILINE flag on your pattern for ^ and $ to match the start and end of lines. Pattern pattern = Pattern.compile("^[0-9]{1}.+\?", Pattern.MULTILINE); This should allow your pattern to match lines within the text. Otherwise ^ and $ just match the start and end of the string as a whole.

Categories : Java

iOS regex pattern not working to match numbers
Try this regex: ^(?<!.)d*(.d+)?$ I added a negative look-behind assertion that means that no dot is allowed before that numbers. That should fix your problem.

Categories : IOS

HTML5 Pattern Regex Password Match
Here is what I would go with: (?=.*d)(?=.*[a-z])(?=.*[A-Z])(?!.*[!#$%&?])^D.{7} Note that the .* after each look-ahead term was superfluous. (?!...) is a negative look-ahead, to make sure there are no special characters. ^D requires that the first character be a non-digit. Then I simply require 7 characters after that, because the end is not enforced. But why exclude special characters from passwords? Usually just the opposite is encouraged.

Categories : Javascript

Regex pattern query to match text within string
As others have written, Regex is NOT meant for this case, and if you're looking for a robust solution, an XML parser is the way to go. For quick and dirty though, this will work: snames*=s*""(.*?)"" This matches the following: <level id="100" name="blaha blah blah" currency="USD" abbr="blh"> <level id="100" name ="blaha blah blah" currency="USD" abbr="blh"> <level id="100" name= "blaha blah blah" currency="USD" abbr="blh"> <level id="100" name = "blaha blah blah" currency="USD" abbr="blh">

Categories : Regex

Is it possible to create a regex that will find strings that DO NOT match a pattern?
Sure thing, boss /(?!_PROCESSED)/ This is a negative lookahead and it's supported in almost all regexp flavours I've adapted an answer from this question to further help you. public static File[] listFilesMatching(File root, String regex) { if(!root.isDirectory()) { throw new IllegalArgumentException(root+" is not directory."); } final Pattern p = Pattern.compile(regex); // careful: could also throw an exception! return root.listFiles(new FileFilter(){ @Override public boolean accept(File file) { return p.matcher(file.getName()).matches(); } }); } listFilesMatching(new File("/some/path"), "(?!_PROCESSED)") Here's the docs for FileFilter

Categories : Java

C# RegEx preventing greedy match with same pattern repeated
Using a "?" as a qualifier makes a "*" non-greedy. In fact, it may be better to use "+": [.+?] As @It'sNotALie said (paraphrased), you would benefit from a bit of explanation: A sample of a nice tutorial

Categories : C#

use regex to match a URL to a pattern, returning an array of keys
PHP on Rails, eh? ;-) Important note about the behavior of strpos: you should check using the strict === operator, because it may return false (source: http://php.net/manual/en/function.strpos.php ). After a cursory read/test this is all I see wrong with the script... <?php // routes-test.php echo "should be [ id => 123 ]: "; var_dump( match_path( 'user/123', 'user/:id' ) ); function match_path($path, $pattern) { ... } ?> // cmd line $ php routes-test.php # your implementation should be [ id => 123 ]: array(2) { ["ser"]=> string(4) "user" ["id"]=> string(3) "123" } $ php routes-test.php # using === should be [ id => 123 ]: array(1) { ["id"]=> string(3) "123" } You should take a YAGNI approach to using regex. If all you'd be doing is matching stuff

Categories : PHP

How can I match a regex with a pattern and arbitary amount of times?
Use re.findall if you want all the matches, re.search stops at the first match.: >>> strs = "52345:54325432:555:443:3:33" >>> re.findall(r"(d+):(d+)",strs) [('52345', '54325432'), ('555', '443'), ('3', '33')] If you want the exact same result as str.split then you could do: >>> re.split(r":",strs) ['52345', '54325432', '555', '443', '3', '33'] >>> re.findall(r"[^:]+",strs) ['52345', '54325432', '555', '443', '3', '33']

Categories : Python

Print line of pattern match in Perl regex
You can grep out the lines into an array, which will then also serve as your conditional: my @match = grep /line/mi, split / /, $input; if (@match) { # ... processing }

Categories : Regex

notepad++ ( perl ) regex match multiple line pattern
Regex <div[^>]+>(.*?)</div> Don't forget to check the option . matches newline like in the image below : Alternatively, you can use this regex also: <div[^>]+>([sS]*?)</div> with or without the checkbox checked. Discussion Since * metacharacter is greedy, you need to tell him to take as few as possible characters (use of ?). Check that the divs you want to remove DO NOT contain nested div. In that case, the regex at the start of my answer won't help you. If you face this case, I'd suggest you using an html parser.

Categories : Regex

Find all lines that match regex pattern and grab part of string
You can do something like this(no need of regex): Use str.startswith to check if a line starts with '\': >>> strs = "\BTLCMOODY01 MRA Server " >>> strs.startswith('\') True Then use a combination of str.split and str.lstrip to get the first word: >>> strs.split(None, 1) ['\BTLCMOODY01', 'MRA Server '] #apply str.lstrip on the first item >>> strs.split(None, 1)[0].lstrip('\') 'BTLCMOODY01' Code: >>> with open('abc1') as f: ... for line in f: ... if line.startswith('\'): #check if the line startswith `` ... print line.split(None,1)[0].lstrip('\') ... TESTHOSTDEV01 TESTHOSTDEVDB01 TESTHOSTDEVDBQA TESTHOSTDEVQA02 BTLCMOODY01 BTLCSTG05 BTLCWEB02 BTLCWSUS01 HIMSAPP01 SLVAPP01 TORAAPP01 HNSVA

Categories : Python

Showing which files that match a specific pattern have changed between git branches
You can pass the file names to git diff after a --, so: git diff --name-only branch1..branch2 -- '*.twig' Should do what you want. I put *.twig in single quotes so the shell won't expand it.

Categories : GIT

modify XML tags with specific pattern by Regex tools
Adding solution from JS: awk -F'>' '/table name/{$NF="slonyid="q x++ q FS}1' q='"' inputFile Try this: awk -F'>' '/table name/{print $(NF-1)" slonyid""=""""NR-1"""">"}' inputFile Adding test: $ cat temp.txt table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table name="dbname.tablename" lots of text here> table

Categories : Xml

Python regex to match a specific word
You should use re.search here not re.match. From the docs on re.match: If you want to locate a match anywhere in string, use search() instead. If you're looking for the exact word 'Not Ok' then use  word boundaries, otherwise if you're only looking for a substring 'Not Ok' then use simple : if 'Not Ok' in string. >>> strs = 'Test result 1: Not Ok -31.08' >>> re.search(r'Not Ok',strs).group(0) 'Not Ok' >>> match = re.search(r'Not Ok',strs) >>> if match: ... print "Found" ... else: ... print "Not Found" ... Found

Categories : Python

RegEx to match a string not containing a specific HTML tag
Description This expression will: find substrings which start with <br /> and end with the next </li> validate the substring does not contain a <li> capture the text between the start an end tags defined above <brs*/>(@(?:(?!<li>).)*?)</li> Example Sample Text Live Example: http://www.rubular.com/r/CIledJX54O Note the first line has the bad condition <br />@ Don't <li>find me.</li> <br />@ This is the content.</li> <br />@ more desired content.</li> Capture Groups [0] => Array ( [0] => <br />@ This is the content.</li> [1] => <br />@ more desired content.</li> ) [1] => Array ( [0] => @ This is the content. [1] =>

Categories : Regex

Regex to Match Specific URL with Query String
Don't use a Regex, use an URL parser. You could use purl Then, you'll do: url = "http://some.test.domain.com/home" // Or any other purl(url).attr('path') // is equal to "home" here. You'll just need to check .attr('path') against your accepted paths (seemingly "", "/", and "home"). Here's some sample output: purl("http://some.test.domain.com/?qs=1").attr('path') "/" purl("http://some.test.domain.com/other").attr("path") "/other" purl("http://some.test.domain.com/home").attr("path") "/home"

Categories : Javascript

How to remove words of a line upto specific character pattern...Regex
A pure bash one-liner: while read x; do [[ $x =~ test.* ]] && echo ${BASH_REMATCH[0]}; done <infile Input: infile This is a test page. My test work of test is complete. Output: test page. test work of test is complete. It reads all lines from file infile, checks if the line contains the string test and then prints the rest of the line (including test). The same in sed: sed 's/.(test.)/1/' infile (Oops! This is wrong! .* is greedy, so it cuts too much from the 2nd example line). This works well: sed -e 's/(test.*)/x03&/' -e 's/.*x03//' infile I did some speed testing (for the original (wrong) sed version). The result is that for small files the bash solution performs better. For larger files sed is better. I also tried this awk version, which is even better for

Categories : Regex

extract file content from match pattern to another match pattern
With awk: awk '/> myoccupation/,/> mygrosssalary/' file With sed: sed -n '/> myoccupation/,/> mygrosssalary/p' file And you can use output redirection to create another file, with comand ... > newfile

Categories : Linux

How do I force user to enter specific pattern in input field using regex and javascript
Try this: P[0-9]{8} That's a P, and then any number eight times. If a P followed by any number of digits is valid, try P[0-9]+ That's a P, and then any number at least once. To make sure that this is all they enter, add ^ and $ (as mentioned in the comments): ^P[0-9]+$ A ^ is 'start of input', and the $ is 'end of input'. So, your final code might be something like: <input type="text" /> <div class="fail" style="display: none;">Fail!</div> and $("input").keyup(function() { var patt=/^P[0-9]+$/g; var result=patt.test($(this).val()); if (result) { $(".fail").hide(); } else { $(".fail").show(); } }); http://jsfiddle.net/fuMg4/1/ Finally, make sure you check this on the server side as well. It's easy to by

Categories : Javascript

Regex to match spaces except for spaces inside url pattern
Description This regex will find an replace all multiple spaces with a single space, and will bypass the url section. In a sequence of X number of spaces, the first space is placed into group 1 which is fed to the output as 1 and the additional spaces are ignored. The URL section is bypassed because if it is encountered as part of the | or statement, then it is captured into group 2 which is then injected back into the output by the 2 replacement. Regex: (s)s*|("url":"[^"]*"), Replace with: 12 Source String "startofjunk junkjunkjunkjunk","url":"https://x.com/a/C25/XPS - Connection - May 2013.docx","contentsource":"AX","returpath":null,"detailpath":"https://ax.sample.com/Rep>ositories/form.aspx?path=C25/96/99&mode=Read","detailspath2":"samplepath" PHP example This php

Categories : Regex

Looking for difference between re.match(pattern, ...) and re.search(r'A' + pattern, ...)
There might be something I am not seeing here, but I think the difference is clear. re.match() returns a successful match only if the pattern you are looking for is at the start of the string, and from the look of the examples in the documentation it seems that re.match() uses A to anchor the match to the start of the string and not the start-of-line in multi-line mode. re.search() returns a successful match no matter where the pattern is inside the target string as long as there is a match, of course as long as you don't anchor the pattern intentionally. Now answering your main question, about what is the difference between re.match(pattern, …) and re.search(r'A' + pattern, …) ? Well there is no difference whatsoever, it is just a convenience method just so you don't have to type

Categories : Python

Regex - only matching wildcard of a certain length or less
To match a known length use .{2,5} where the 2 is the minimum number of characters and 5 is the max. both values are optional but you do need one or the other More can be read on this topic here

Categories : C#



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