Your regex doesn’t say what you think it says. ^([a-z0-9_\.-])+@[yahoo]{5}\.([com]{3}\.)?[com]{3}$ Says any characters a-z, 0-9, ., – one or more times. That later part where you are trying match yahoo.com is incorrect. It says y, a, h, or o, any of those characters are allowed 5 times. Same with the com so aaaaa.ooo would be … Read more
That’s very simple. Use this pattern: /(\d{2})(\d{2})(\d{4})/ And replace it with $1/$2/$3 Online Demo 4 solved Regex String to Date
The given regex is a combination of multiple regex, keep in mind that you can break this complex regex into multiple smaller one, and each smaller one can be easily translated Refer to this wiki page for finding the meaning of each small regex parts http://en.wikipedia.org/wiki/Regular_expression Your regex can be broken into /^ Start of … Read more
If you can’t use developer tools and really need the regex to search through the source file (and it’s an actual text character x), then this (probably inelegant) regex should do it: (\Wx\W)|(^x\W)|(\Wx$) The \W means “not a ‘word’ character”. This may or may not quite work for you depending on what you have in … Read more
The question is not completely clear. If you mean that you can use between 3 and 10 characters, and these characters can be alphanumerical characters (digits and [A-Za-z]), you can use: /^(?=.*\d.*)[A-Za-z0-9]{3,10}$/ regex101 demo. The regex works as follows: ^[A-Za-z0-9]{3,10}$ says the regex consists out of 3 to 10 characters that can be digits and/or … Read more
yourPrefix([a-zA-Z0-9]+) Matches any alphabetic/numeric characters after ‘yourPrefix’ For your example data in javascript: ‘[BBSHORTCODE_LINK’.match(‘BBSHORTCODE_([a-zA-Z0-9]+)’).pop() //’LINK’ ‘[BBSHORTCODE_BUTTON’.match(‘BBSHORTCODE_([a-zA-Z0-9]+)’).pop() //’BUTTON’ solved Regex take part of string after another part [closed]
Is that what you want? Find what: (\w+)?.+?(VARCHAR.+) NULL Replace with: cast\(NULL as $2\) as $1 Don’t forget to escape the parenthesis in the replacement part! 1 solved Find and Replace for a string in Notepad++ using regular expression
Search for = (Session\(“.*?”\))\.ToString and replace with = Cstr($1) It captures the Session(…) portion, which becomes the $1 replacement. solved Regex in Visual Studio 2013 [closed]
Why not just split/explode on ; and get the relevant pieces? If you insist on a regex, just use this: /;(.*?);;(.*)/ The first subpattern will be the username, the second subpattern will be the password. solved Simple regular expression to determine login and password from a line
For case 1: Search using: ^([^_]*_)[^_]* And replace by: $1 RegEx Demo 1 For case 2: Search using: ^((?:[^_]*_){2})[^_]* And replace by: $1 RegEx Demo 2 For case 3: Search using: ^((?:[^_]*_){3})[^_]* And replace by: $1 RegEx Demo 3 0 solved Regex Split String on Delimiter
A regular expression to validate if a string starts with a single letter A-Z in any case and has 1 or more digits and nothing else is: ^[A-Za-z]\d+$ Explanation: ^ … beginning of string (or beginning of line in other context). [A-Za-z] … a character class definition for a single character (no multiplier appended) matching … Read more
Your string is in fact html – you should use html parser instead. I suggest the excellent lxml.html parser. To answer your question, regexes are greedy by default, that means your .+ part will grab as many chars as it can to satisfy the condition. So you will get the first /subtitles/ and the last … Read more
the second declaration of $line is erroneous, drop my: $line =~ m/textolibre=(.*?)&translated/g; 2 solved Perl — regex issue
i’m not sure of the code i write but you can find something like this: String data=”blah blah blah”; int index = data.indexOf(“http”); String subString = str.substring(index, data.length()-1); solved Remove a word starting with x in a sentence [closed]
[*] Use a negative lookahead assertion: r’\b[a-z][*]{2}1(?!\d)’ The (?!…) part is the lookahead; it states that it cannot match at any location where a digit (\d) follows. Regex101 demo; note how only the second line is matched (blue). Python demo: >>> import re >>> pattern = re.compile(r’\b[a-z][*]{2}1(?!\d)’) >>> pattern.search(‘x**1’) <_sre.SRE_Match object at 0x10869f1d0> >>> pattern.search(‘x**10’) … Read more