This will do it in 1 shot, but not the way you’d expect: r’\D{n}’ That is, it matches string of length n that does not contain any digits; thus, if the match fails, there must be a digit in the first n characters. 1 solved Python regex to find digit in first n characters [closed]
“(?s).*(ST\\*214|ST\\*210).*” In Java you need to enable DOTALL mode (to make . match with line terminators too). This can be done by including (?s) modifier. You had match only in this ST*214*900063730~ particular part of first string. 1 solved Looking for Correct Java REGEX for this kind of payload
[*] Provisos and assumptions: OP mentions needing to process multiple files; for this answer I’m going to focus on a single file; OP can start another question if issues arise with a multi-file solution OP mentions wanting to replace some strings but it’s not clear (to me) if the original file is to be overwritten … Read more
So after reading your question many times I think I get where you are going, and it definitely sounds like an XY problem. An escaped character is just a character; escaping is part of the syntax, not the string itself. If you are building regex dynamically, you will have to double escape special characters. In … Read more
Using a variant over the regex provided by the answer of karthik manchala, and noticing that you want the same output as given in your question here is a complete code example: import re inputText = “””The dodo was one of the sturdiest birds. An educated termite may learn how to operate a phonograph, but … Read more
In order to match only strings containing 5 words or less, i.e. not any five words in a row, you also need to add anchoring to the beginning and end of the string: ^((?:\w+ ?){1,5})$ Example https://regex101.com/r/pN9nQ0/1 1 solved Want to find exact 5 strings sentences
This for matching full string: /^A\.\d{8}.\d{3}$/ to match part of string remove ^ or $. solved Please tell me JavaScript RegEx for : A.12345678.123 that is start with ‘A.’ followed by 8 digits [0-9] followed by ‘.’ and again by 3 digits[0-9] [closed]
Using perl or sed, you could try something like: /\d{3,11}/ \d is for digit, and {3,11} means that a digit must appear between 3 and 11 times. solved How can I write a Regex expression like this [closed]
Make sure to use replaceAll(). Also you could put all of the terms and replacements in an array. Like so: String text = “ALTER DATABASE, ALTER TABLE, ALTER VIEW, CREATE DATABASE, CREATE PROCEDURE, CREATE SCHEMA, CREATE TABLE”; String[] terms = {ALTER DATABASE, ALTER TABLE, ALTER VIEW, CREATE DATABASE, CREATE PROCEDURE, CREATE SCHEMA, CREATE TABLE}; String[] … Read more
There’s probably a better way to do this using regexes, but here’s what I came up with: a.split(/[^(]*\(([^)]+)\)[^(]*/).reject(&:empty?) 0 solved Get array of matching strings
Whenever you need some reasonably common matching problem resolve in Perl, you should always first check the Regexp::Common family on CPAN. In this case: Regexp::Common::Email::Address. From POD Synopsys: use Regexp::Common qw[Email::Address]; use Email::Address; while (<>) { my (@found) = /($RE{Email}{Address})/g; my (@addrs) = map $_->address, Email::Address->parse(“@found”); print “X-Addresses: “, join(“, “, @addrs), “\n”; } 1 … Read more
You should try to fix the source of improperly escaped strings instead of mucking around with a regex. If you can’t do that and are desperate to get something done, one quick and dirty approach would be to remove quotes that don’t border on commas or start/end of string: resultString = Regex.Replace(subjectString, “(?<!,|^)\”(?!,|$)”, “”); This … Read more
$dir=”http://www.example.com/directory”; $data = new DOMDocument(); @$data->loadHTMLFile($dir); $links = array(); foreach($data->getElementsByTagName(‘a’) as $link) { $url = $link->getAttribute(‘href’); if ($url[0] !== ‘?’) // skip column links { $links[] = $url; } } print_r($links); 7 solved [JS]Get file names of Apache Directory Listing [closed]
You selected the correct regexp, you just have to quote it to use it in Python. You also need to call re.findall(), it’s not a string method. import re txt=”I refer to https://google.com and i never refer http://www.baidu.com” print(re.findall(r'(?<=https:\/\/)([A-Za-z0-9.]*)’, txt)) 1 solved Which of the following regular expressions can be used to get the domain … Read more
You could try a regex replacement: var text=”Hello. World… Lorem? Ipsum 123″; var output = text.replace(/([.?!])(?!.*[.?!]).*$/, “$1”); console.log(output); The replacement logic here says to: ([.?!]) match and capture a closing punctuation (?!.*[.?!]) then assert that this punctuation in fact be the final one .* match and consume all remaining text $ until the end of … Read more