You’ll have much more succes using an XML parser, for example, XML::Parser. Parsing XML using regular expressions is very difficult (impossible?) and unless your use case is trivial, a proper XML parser is the reliable solution. 1 solved How can I match XML tags and attributes with a regular expression in Perl?
How about something like this? It would only handle one specific format but the regex can be adjusted for other formats. Regex rg = new Regex(@”[01]?\d[/-][0123]?\d[/-]\d{2}”); Match m = rg.Match(string); m.ToString(); Here is a question with a bunch of date regexes that may help. Regular Expression to match valid dates solved Extract a date from … Read more
\B is the negated version of \b. \B matches at every position where \b does not. Effectively, \B matches at any position between two word characters as well as at any position between two non-word characters. \b allows you to perform a “whole words only” search using a regular expression in the form of \bword\b. … Read more
You can use like ^[a-zA-Z]+[ ][a-zA-Z0-9]+$ function CheckValidation() { var str = document.getElementById(“txtInput”).value; var pattern = new RegExp(“^[a-zA-Z]+[ ][a-zA-Z0-9]+$”); var res = pattern.test(str); document.getElementById(“para”).innerHTML = res; } <input type=”text” id=”txtInput” /> <button onclick=”CheckValidation()”>Check</button> <p id=”para”></p> solved Regex pattern for following example test test123 [closed]
You can try this also 2.1.2 :007 > s = “The Fashion Project – The Project of Fashion-Technology and Science-institute” => “The Fashion Project – The Project of Fashion-Technology and Science-institute” 2.1.2 :008 > s.split(/\s[-\b]\s/) => [“The Fashion Project”, “The Project of Fashion-Technology and Science-institute”] 2.1.2 :009 > 1 solved extract string using ruby regex
I found this book “Mastering Regular Expressions, 3rd Edition by Jeffrey E.F. Friedl ” really helpful in understanding regular expression. You can get your copy here In initial chapters, book details the syntax to write regular expression. Last four chapters are really helpful in using regular expression in different languages like a> Perl b> Java … Read more
Here’s a solution using regex if that’s what you really need. I search to see if any of your 3 substrings are present inside any given string from the list. Using https://docs.python.org/3/library/re.html as the Python regex library. import re for word in wordList: m = re.search(‘.*(ra|dec|lat).*’, word) if m: <youve matched here> solved regex multiple … Read more
quote | space || ——-+——-++—– 0 | 0 || ^[a-z]*$ 0 | 1 || ^[a-z]* [a-z]*$ 1 | 0 || ^[a-z]*'[a-z]*$ 1 | 1 || ^[a-z]* [a-z]*'[a-z]*$ or ^[a-z]*'[a-z]* [a-z]*$ Eg. final Pattern regex = Pattern .compile(“^([a-z]*|[a-z]* [a-z]*|[a-z]*'[a-z]*|[a-z]* [a-z]*'[a-z]*|[a-z]*'[a-z]* [a-z]*)$”, CASE_INSENSITIVE); for(String x: asList(“”, “‘”, ” “, “‘ ‘”, ” a “, “p%q”, “aB cd’Ef”, … Read more
The regex /\d\d:\d\d(:\d\d)/g selects 01:01:00 and would capture :00, this applies to all situations accept 01-01-01 01:01, ignore me please which is ignored. solved Regex to select seconds from time string (all times not in the same format) without lookbehind
Try with this: \d+\s*((?:Apple|Banana|Orange|Pineapple)s?\b[\s\S]*?)(?=$|\d+\s*(?:Apple|Banana|Orange|Pineapple)s?\b) See: Regex demo The code: import re regex = r”\d+\s*((?:Apple|Banana|Orange|Pineapple)s?\b[\s\S]*?)(?=$|\d+\s*(?:Apple|Banana|Orange|Pineapple)s?\b)” test_str = “I have 2 apples in my bag and apples are great food toeat. you shud eat apples daily. it is very good for health. 3 bananas are also good. it reduces fat.” matches = re.findall(regex, test_str, re.MULTILINE | re.IGNORECASE) … Read more
^[^ ]([\w- \.]+)[^ ]$ [^ ] = cannot start or end with space ([\w- .]+) = all characters allowed (or in this case, \w, hyphen, space and dot) matches: ABC BCD 1 solved How to write a regular expression for javascript valiadation [closed]
What this big regex commands mean? Pattern #1 Breakdown: / #start of pattern delimiter (@.*@) #Capture Group #1: match an @ sign, then zero or more (as many as possible) of any non-newline character, then another @ sign | #or (\.\.) #Capture Group #2: match a literal dot, then another literal dot | #or (@\.) … Read more
You might use a positive lookahead and test if the string contains /*? If so, match any character one or more times .+ from the beginning of the string ^ until the end of the string $ ^(?=.*\/\*).+$ Explanation ^ Begin of the string (?= Positive lookahead that asserts what is on the right .*\/\* … Read more
Finally I got answer from here How to combine multiple regex into single one in python? Working fine this import re re1 = r’\d+\.\d*[L][-]\d*\s[A-Z]*[/]\d*’ re2 = ‘\d*[/]\d*[A-Z]*\d*\s[A-Z]*\d*[A-Z]*’ re3 = ‘[A-Z]*\d+[/]\d+[A-Z]\d+’ re4 = ‘\d+[/]\d+[A-Z]*\d+\s\d+[A-z]\s[A-Z]*’ sentences = [string1, string2, string3, string4] generic_re = re.compile(“(%s|%s|%s|%s)” % (re1, re2, re3, re4)).findall(sentence) solved How to check with more RegEx for … Read more
You’d probably want to create a regex as follows: ^\d+\.\d+\.\d+$ The ^ means “start of phrase”, the $ means “end of phrase”, \d+ says “digit one or more times in a row”, and \. means “.” but must be escaped with the leading \ due to . having a special meaning in regex. 0 solved … Read more