regex

[Solved] Use regular expressions to remove unwanted html from array [closed]

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PHP or JS? JS: http://jsfiddle.net/mplungjan/fpzEn/ var str = “description\”:\”SOME RELEVANT TEXT…<img width=”1″ height=”1″ src=”http:\/\/someurl.com.feedsportal.com\/c\/33818\/f\/608449\/s\/1c52b2b5\/mf.gif” border=”0″\/><div class=”mf-viral”><table border=”0″><tr><td valign=’middle’><a href=\\\”http:\/\/share.feedsportal.com\/viral\/sendEmail.cfm?lang=en&title=some_title&link=http%3A%2F%2Fwww.someurl.com%2Fworld-newssome_title%2FArticle1-805340.aspx\\\” target=\\\”_blank\\\”><img src=\\\”http:\/\/res3.feedsportal.com\/images\/emailthis2.gif\\\” border=\\\”0\\\” \/><\/a><\/td>” alert(str.split(/description”:”/)[1].split(/</)[0]) solved Use regular expressions to remove unwanted html from array [closed]

[Solved] How to search a document for IP addresses [closed]

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If your addresses are always on the end of a line, then anchor on that: ip_at_end = re.compile(r'(?:[0-9]{1,3}\.){3}[0-9]{1,3}$’, re.MULTILINE) This regular expression only matches dotted quads (4 sets of digits with dots in between) at the end of a line. Demo: >>> import re >>> ip_at_end = re.compile(r'(?:[0-9]{1,3}\.){3}[0-9]{1,3}$’, re.MULTILINE) >>> example=””‘\ … Only addresses on … Read more

[Solved] Need a preg_replace or best option for this conversion [closed]

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Your question is very unclear and I may be way off, but here is my attempt to your question. $link = array(‘S1, Ep1’, ‘S1, Ep2’, ‘S1, Ep3’, ‘S1, Ep10’); $results = preg_replace_callback(‘~^S(\d+)\D+(\d+)$~m’, function($m) { $which = strlen($m[2]) > 1 ? ’00’ : ‘000’; return $m[1] . $which . $m[2]; }, array_values($link)); print_r($results); Output Array ( … Read more

[Solved] Regex for replacing certain string in python [closed]

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You need regular expressions for this: import re s=”abc xyz.xyz(1, 2, 3) pqr” re.sub(r'[a-z]{3}\.{[a-z]{3}\([^)]*\)’, ‘NULL’, s) Explanation: [a-z]{3} stands for 3 small letters (lower case) \. escapes the dot, as it is special character [a-z]{3} again 3 small letters \( escapes the left parenthesis [^)]* any character but right parenthesis, 0 or more time \) … Read more

[Solved] Regular Expressions:JavaScript

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Once possible solution is the following: ” Hi @raju how do you do? “.replace(/\s+|[@\?]/g,’-‘) .replace(/–+/g,’-‘) .replace(/^[\s-]+|[\s-]+$/g,”) >> output: “Hi-raju-how-do-you-do” You can add any other ‘special’ chars to the [] in the first replace. 1 solved Regular Expressions:JavaScript

[Solved] date extraction through regex [closed]

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Without regex: s = “20160204094836A” year = s[:4] day = s[4:6] month = s[6:8] print(year, day, month) With Regex: import re s = “20160204094836A” result = re.search(r”^(\d{4})(\d{2})(\d{2})”, s) year = int(result.group(1)) day = int(result.group(2)) month = int(result.group(3)) print(year, day, month) solved date extraction through regex [closed]

[Solved] How can I validate a regular expression using jQuery Validation plugin?

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No need for jQuery. Use the RegExp constructor and catch the exception : try { new RegExp(someString); console.log(‘good’); } catch (e) { console.log(‘bad’); } Demonstration (type something in the input to know if it’s a well formed regular expression) 2 solved How can I validate a regular expression using jQuery Validation plugin?

[Solved] Program is saying all my valid inputs are invalid (REGEX code Issue maybe?) [closed]

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I think you need to add Pattern. import java.util.*; public class lab2q2 { public static void main(String[] args) { String RegularExp = “((Mr|Ms))?[A-Z][a-z]+([A-Z]([a-z]+\\.))?([A-Z](a-z)+)”; Pattern pattern = Pattern.compile(RegularExp); Scanner keyboard = new Scanner(System.in); for(int i = 0; i< 11; i++) { System.out.println(“Please enter a name: “); String inputString = keyboard.nextLine(); Matcher matcher = pattern.matcher(inputString ); if(!matcher.matches()) … Read more