You could do something like this : let str = “_h Hello _h _h World _h” let range = NSRange(str.startIndex…, in: str) let regex = try! NSRegularExpression(pattern: “(?<=_h).+?(?=_h)”) let matches = regex.matches(in: str, range: range) let results: [String] = matches.compactMap { match in let subStr = String(str[Range(match.range, in: str)!]).trimmingCharacters(in: .whitespacesAndNewlines) return subStr.isEmpty ? nil : … Read more
You should add \ in front of + to recognize the symbol, not as an increment r’ID\s+([0-9]+)\s+linea\+b\s+([0-9]+)\s+’ solved how to extract data with regex in python? [duplicate]
There must be an error somewhere else in your program. This works as expected: $s1 = ‘M&M chocolates’; $s2 = ‘Tom & Jerry’; printf(“$s1\n”) if $s1 =~ m/\w\W\w\s\w+/; printf(“$s2\n”) if $s2 =~ m/\w+\s&\s\w+/; outputs M&M chocolates Tom & Jerry solved Perl regex for char ‘&’ [closed]
The solution is the data cutting between {}. Try this: $result = preg_replace(“#\{img:(.*)\}#U”,'<img src=\’$1\’ style=\’width:100%\’ />’,$content); 0 solved Match content between parenthesis and get the url and change it to img tag using RegExp PHP [duplicate]
Try this: import re lst = [‘a’, ‘4’, ‘add’, ‘e’, ‘a’, ‘c0a8d202’, ‘128’, ‘4’, ‘0’, ’32’] pattern = re.compile(r’c[0-9a-fA-F]?’) for i in lst: if re.search(pattern, i): print(lst.index(i)) Note: this is as per your desired output but i am agree with @Jean-François Fabre who said that what’s wrong with lst.index(‘c0a8d202’) ? what’s the point of regular … Read more
Your requirements are perhaps not totally complete, but if I assume that you only want six digit characters at the end, something like the regex /H\d{3}-\d{4}-\d{6}/ would work. You can see it working here or in the snippet below: const text=”nonsense nonsense lorem ipsum H005-2007-652764 and then more nonsense”; const regex = /H\d{3}-\d{4}-\d{6}/; console.log(text.match(regex)); solved … Read more
You can try this regex pattern: (?<=mailfrom=)[^;]+(?=;) (?<=mailfrom=) // assert ‘mailfrom=’ on the left [^;]+ // 1 or more characters not ‘;’ (?=;) // assert ‘;’ on the right Demo 0 solved Find email address between specific substring and semicolon
If DSGSQ is the part that is variable in your question (see my comment below question), a possible Regex is Find: (<img src =”https://stackoverflow.com/questions/54358159/wsg://i)([^”]*)(“>) Replace : <img src =”http://localhost/images/$2.jpg”> 1 solved Replacing text using Regex expression in Notepad ++
You can try this regex, although your question is not clear to me. ^students\/([\w\-\d]+)\/data\/sessions$ Check here https://regex101.com/r/xnxwCX/1 you can grab the data in between students/, /data/session. solved How to make regex for 3 slashes?
note that DETECTLANGUAGE does not work with array/range so only: =IF(DETECTLANGUAGE(A1)=”ja”, “Japanese”, ) but you could use a script: function NIPPON(input) { var output = []; for (i = 0 ; i < input.length; i++){ try { output[i] = LanguageApp.translate(input[i], ”, ‘ja’); } catch(err) { output[i] = err.message; } } return output; } =ARRAYFORMULA(FILTER(A1:A, IF(LEN(A1:A)=LEN(NIPPON(A1:A)), … Read more
Figured out a way to filter based on router eigrp or ospf with lookbehind and then loop through using ansible. ^interface (\S+)(?:\n\s{2}.)(?<=eigrp) solved Regex cisco interface parent and child
You can do this using groups and back-references. My examples are in Java’s regex dialogue, but you can adapt it to any dialect that supports back-references. For example some engines need \(…\) instead of (…). First you need a group, which you denote by surrounding in parentheses: (.) This matches any single character. If you … Read more
Introduction Regular expressions (RegEx) are a powerful tool for matching patterns in text. They can be used to check if a text contains only one type of letter, such as in the example “KKKK”. This can be done by creating a RegEx pattern that looks for any character that is repeated four times in a … Read more
In Java there are lots of ways you can code to get the contents between brackets (or any two specific characters for that matter) but you sort of want to dabble with a regular expression which can be rather slow for this sort of thing especially when dealing with multiple instances of bracket pairs within … Read more
You can try this $lines = array(); $lines = explode(“2013”,$string); foreach($lines as $key => $value) { $data = array() $data = explode(“;;”,$value); $lines[$key][‘data’] = $data } solved split string with preg_split [closed]