Tag: regex
For powershell v3 i need to find items in the range of Test100 to Test200.
For example it should work as this;
Test001 - not valid
Test099 - not valid
Test100 - valid
Test150 - valid
Test200 - valid
Test201 - not valid
I tried -match (Test[100,200]) but this does not work. How can I do this?
Test(?:1\d\d|200)\b
You can simply use this.See demo.
Using sqlite3 from bash on OS X seems fairly straightforward (I'm no expert at this, by the way). You will need to find out which table you need. You can do this with an interactive session. I'll show you with the database you suggested: /Users/fredbloggs> sqlite3 ~/Library/Application\ Support/Dock/desktoppicture.db SQLite version...
Find what: ^(.{30}) Replace with: \1\n ...
You turn off case sensitivity using an i inline flag in regex: (?i)https?:\\/.* See Foundation Framework Reference for more information on available regex features. (?ismwx-ismwx) Flag settings. Change the flag settings. Changes apply to the portion of the pattern following the setting. For example, (?i) changes to a case insensitive...
Try using the unicode character code, \u2022, instead: message.replace(/\u2022/, "<br />\u2022"); ...
Just get the dot outside of the captruing group and then make it as optional. @"[+-]?\d+\.?\d*" Use anchors if necessary. @"^[+-]?\d+\.?\d*$" ...
This is about as simple as I can get it: \b\w+\.?\w* See demo...
The quotes are an issue but not the issue you are running into when you escape them. Your delimiter is terminating your regex just before the closing a which is giving you the unknown modifier error. It appears you don't have error reporting on though so you aren't seeing that....
You can make use of the multiline flag, and ^ and $ anchors that will match at the string start and string end repsectively: (?m)^.*lonfksa\.newsvine\.com.*$ Mind that you need to escape a dot in regex to make it match a literal dot. Your regex (?s)lonfksa.newsvine.com(?s) contains unescaped dots that match...
python,regex,algorithm,python-2.7,datetime
What about fuzzyparsers: Sample inputs: jan 12, 2003 jan 5 2004-3-5 +34 -- 34 days in the future (relative to todays date) -4 -- 4 days in the past (relative to todays date) Example usage: >>> from fuzzyparsers import parse_date >>> parse_date('jun 17 2010') # my youngest son's birthday datetime.date(2010,...
This matches all given examples as well: ^\$?\d+(?:[.,:]\d+)?%?$ See it in action: RegEx101 Please comment, if adjustment / further detail is required....
Let's say the domain is as following String domain[] = { a, b, .., z, A, B, .. Z, 0, 1, 2, .. 9 }; Let's say the password size is 8 ArrayList allCombinations = getAllPossibleStrings2(domain,8); This is going to generate SIZE(domain) * LENGTH number of combinations, which is in...
What you're looking for is called a character class. A character class is a group of characters that you're saying can be matched at that position in the string. To make a character class you enclose your list of matchable characters in square brackets, like this: [.,:;!?] This will match...
Could not figure out a regex solution, but here's a non-regex solution. It involves parsing numbers (not in curly braces) before each comma (unless its the last number in the string) and parsing strings (in curly braces) until the closing curly brace of the group is found. If regex solution...
$pattern = '! ^ # start of string \d{5} # five digits [[:alpha:]]{2} # followed by two letters - # followed by a dash \d{2} # followed by two digits $ # end of string !x'; $matches = preg_match($pattern, $input); ...
jquery,regex,string,substring,substr
You could convert this to a slightly more maintainable format, without getting into regular expressions. This is one way to use an array to accomplish your goal: // Super-quick one-liner: var str = '2042038423408'; var matchCount = $.grep(['12', '23', '34', '45', '56', '67', '78', '89', '90', '01'], function(num, i) {...
Some notes about your original regex: a lookahead only makes sense at the end of the string; you were probably looking for a non-capturing group, e.g. T(?:P-) instead of T(?=P-), but you don't even need those if they appear exactly once (i.e. if there's no need to put a *,...
Use {} instead of () because {} are not used in XPath expressions and therefore you will not have confusions.
There's a bug in your code. As a string is immutable in JavaScript, replace doesn't change it but returns a new one. You do the replacement but you doesn't take the returned value Change val.replace(/[\-\[\]{}()*+?.,\\\^$|#\s]/g, "\\$&"); to val = val.replace(/[\-\[\]{}()*+?.,\\\^$|#\s]/g, "\\$&"); Demonstration...
Your first regular expression has a black slash followed by the letter b because of that @. The second one has the character that represents backspace. Just put an @ in front string bound = @"\b"; ...
You can use: Public\s+Const\s+g(?<Name>[a-zA-Z][a-zA-Z0-9]*)\s+=\s+(?<Value>False|True) demo ...
python,regex,string,loops,twitter
Store slangNames and riskNames as sets, split the strings and check if any of the words appear in both sets slangNames = set(["Vikes", "Demmies", "D", "MS", "Contin"]) riskNames = set(["enough", "pop", "final", "stress", "trade"]) d = {1: "Vikes is not enough for me", 2:"Demmies is okay", 3:"pop a D"} for...
Use the alternation with $: import re mystr = 'HelloWorldToYou' pat = re.compile(r'([A-Z][a-z]*)') # or your version with `.*?`: pat = re.compile(r'([A-Z].*?)(?=[A-Z]+|$)') print pat.findall(mystr) See IDEONE demo Output: ['Hello', 'World', 'To', 'You'] Regex explanation: ([A-Z][a-z]*) - A capturing group that matches [A-Z] a capital English letter followed by [a-z]* -...
You could use a negative lookahead which will exclude those having _FX following the initial alpha string ^ABD_DEF_GHIJ(?!_FX)(?:_\d{8})?$ see example here...
It’s quite trivial: RegEx string.match(/\$((?:\d|\,)*\.?\d+)/g) || [] That || [] is for no matches: it gives an empty array rather than null. Matches $99 $.99 $9.99 $9,999 $9,999.99 Explanation / # Start RegEx \$ # $ (dollar sign) ( # Capturing group (this is what you’re looking for) (?: #...
You can use this simplified regex: /^[\s0]*11\s*$/ ...
Put dot and / inside a character class so that it would match .png or /png strings. var imageReg = /[\/.](gif|jpg|jpeg|tiff|png)$/i; Your regex would return true if there is a dot exists before png but here there exists a forward slash , so it fails....
python,regex,string,list,python-2.7
With such a small range you could just iterate the move_order and check if each element exists in the allowed moves def start(): move_order=[c for c in raw_input("Enter your moves: ")] moves = ['A','D','S','C','H'] for c in move_order: if c not in moves: print "That's not a proper move!" return...
Try this regex: (?<=[a-zA-Z])(\n) I used parentheses to capture the newline character. https://regex101.com/r/zS9pB4/3...
I would try with: Pattern regex = Pattern.compile("(?<= )\\d(\\d*[,\\.]?\\d+)*(?=[ .]?sq)"); where: (?<= ) - there is space before \d - starts with digit (\d*[,\.]?\d+)* - next is digit or digits, there could be comma or point with more digits - and it can repeats like in 100,000,000 (?=[ .]?sq) -...
How about using JFormattedTextField with MaskFormatter. JFormattedTextField formattedTextField = new JFormattedTextField("00:00:00;00"); try { MaskFormatter maskFormatter = new MaskFormatter("##:##:##;##"); maskFormatter.install(formattedTextField); } catch (ParseException e) { e.printStackTrace(); } More info at http://docs.oracle.com/javase/tutorial/uiswing/components/formattedtextfield.html Demo code: JFrame frame = new JFrame(""); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); JPanel panel = new JPanel(); JFormattedTextField...
I don't understand why it would give me two hellos back? Because the first entry in the array is the overall match for the expression, which is then followed by the content of any capture groups the expression defines. Since the expression defines one capture group, you get back...
Use \d+ to match one or more digits. \b(?:http:\/\/)?(?:www\.)?example\.com\/g\/(\d+)\/\w put http:// and www. inside a capturing or non-caturing group and then make it as optional by adding ? quantifier next to that group. For both http and https, it would be (?:https?:\/\/)? DEMO...
You can use this regex to test. It will ensure that after the @ there is .xx. but may also match the string @.xx.* .*@[^.]*[.]xx[.] Or this one to ensure that there is at least one character before and after the @. [email protected][^.]+[.]xx[.] ...
php,regex,apache,.htaccess,mod-rewrite
You can add a new rule for +/- conversion: Options -MultiViews RewriteEngine On RewriteBase /indianrealitybytes/ RewriteCond %{THE_REQUEST} /search_advance\.php\?keywords=([^&]+)&f=([^\s&]+) [NC] RewriteRule ^ search/%1/%2? [R=301,L] RewriteRule ^([^+]*)\+(.*)$ $1-$2 [R=302,NE,L] RewriteRule ^search/([^/]+)/([^/]+)/?$ search_advance.php?keywords=$1&f=$2 [QSA,L,NC] ...
A variant of n-dru pattern since you don't need to describe all the string: SELECT '#hellowomanclothing' REGEXP '(^#.|[^o]|[^w]o)man'; Note: if a tag contains 'man' and 'woman' this pattern will return 1. If you don't want that Gordon Linoff solution is what you are looking for....
python,regex,nsregularexpression
You're not capturing the whole filename in the group. You can also use noncapturing groups with (?:...). .*\.(rom|[0-9]{3})+ # from this (.*\.(?:rom|[0-9]{3})) # to this ...
You are trying to write a python code using ruby syntax. This is not a best approach to GTD. Slashes are handled right-to-left, yielding not what you expected. As soon as one finds herself putting three or more backslashes inside the string, she should admit, she’s doing it wrong. At...
The pattern attribute has to match the entire string. Assertions check for a match, but do not count towards the total match length. Changing the second assertion to \w+ will make the pattern match the entire string. You can also skip the implied ^, leaving you with just: <input pattern="(?!34)\w+"...
You can use the regex in a negative look-ahead and then add a \w shorthand class to match alphanumeric symbols, or [a-zA-Z] with \b word boundaries: (?![0-9-+*/()x]|abs|pow|ln|pi|e|a?(?:sin|cos|tan)h?)\b[a-zA-Z]+\b See regex demo Since we are only allowing letters with [a-zA-Z], we can reduce this further to (?!x|abs|pow|ln|pi|e|a?(?:sin|cos|tan)h?)\b[a-zA-Z]+\b See another demo...
A work-around for the lack of variable-length lookbehind is available in situations when your strings have a relatively small fixed upper limit on their length. For example, if you know that strings are at most 100 characters long, you could use {0,100} in place of * or {1,100} in place...
You can make use of a Unicode category \p{Zs}: Zs Space separator $string = preg_replace('~\p{Zs}~u', ' ', $string); The \p{Zs} Unicode category class will match these space-like symbols: Character Name U+0020 SPACE U+00A0 NO-BREAK SPACE U+1680 OGHAM SPACE MARK U+2000 EN QUAD U+2001 EM QUAD U+2002 EN SPACE U+2003 EM SPACE...
This is one way to do it, using preg_match: $string ="SomeStringExample"; preg_match('/^[b-df-hj-np-tv-z]*/i', $string, $matches); $count = strlen($matches[0]); The regular expression matches zero or more (*) case-insensitive (/i) consonants [b-df-hj-np-tv-z] at the beginning (^) of the string and stores the matched content in the $matches array. Then it's just a matter...
sql,regex,oracle,oracle11g,regexp-like
Try this one: SELECT * FROM employee WHERE REGEXP_LIKE (fname, '^pr(*)'); Fiddle This one also seems to work as far as I can tell: SELECT * FROM employee WHERE REGEXP_LIKE (fname, '^pr.'); Or another one that works: SELECT * FROM employee WHERE regexp_like(fname,'^pr'); ...
python,regex,parsing,beautifulsoup,python-requests
In order to match the string with a literal backlash, you need to double-escape it in a raw string, e.g.: re.search(r'@CAD_DTA\\">(.+?)@[email protected]@CAD_LBL',result.text) ^ ^ In order to get the index of the found match, you can use start([group]) of re.MatchObject IDEONE demo: import re obj = re.search(r'@CAD_DTA\\">(.+?)@[email protected]@CAD_LBL', 'Some text [email protected]_DTA\\">I WANT...
Updated: This will check for the existence of a sentence followed by special characters. It returns false if there are no special characters, and your original sentence is in capture group 1. Updated Regex101 Example r"(.*[\w])([^\w]+)" Alternatively (without a second capture group): Regex101 Example - no second capture group r"(.*[\w])(?:[^\w]+)"...
To only allow digits, comma and spaces, you need to remove (, ) and -. Here is a way to do it with Matcher.find(): Pattern pattern = Pattern.compile("^[0-9, ]+$"); ... if (!m.find()) { evt.consume(); } And to allow an empty string, replace + with *: Pattern pattern = Pattern.compile("^[0-9, ]*$");...
Use a different set of delimiters for the regex. For example, you can write preg_match_all('~[^/\s]+/\S+\.(jpg|png|gif)~', $string, $results ...
python,html,regex,wordpress,beautifulsoup
At least, you can rely on the tag names and text, navigating the DOM tree horizontally - going sideways. These are all strong, p and span (with id attribute set) tags you are showing. For example, you can get the strong text and get the following sibling: >>> from bs4...
You can use python's built-in csv module to do this. j = next(csv.reader([string])); Now j is each item delimited by a , and will include commas if the value is wrapped in ". See j[2]....