This is the simple expression I could come up from your question (\d+):(\d+):(\d+) BattlEye Server: \(\w+\) \w+: !rpt \w+ \w+ \w+ Try regexpal.com for validating your regex 1 solved Matching Regex to string [closed]
Match a word beginning with a @, reset matching output, match more than one whitespace afterwards then do a replacement: preg_replace(‘~@\w+\K\s{2,}~’, ‘ ‘, $input_string); Regex explanation: @\w+ # Match a `@` fowlling a word \K # Reset matching process output \s{2,} # Match double or more whitespaces PHP live demo solved PHP regex to replace … Read more
Try out this: ^(\d)\1*$ or try this ^([0-9])\1*$ please modify the above regex according to your problem.pattern matching if the user enters same digit. \1 matches the first capture group, so the pattern matches whether the digits are repeated in the string. 0 solved Regular Expression to count number of pairs in a string [closed]
Maybe, this expression would also function here, even though your original expression is OK. ^\/src\/main\/java(\/[^\/]+)?\/\*\.java$ Demo 1 My guess is that here we wish to pass: /src/main/java/{any_package}/*.java /src/main/java/*.java If the second one is undesired, then we would simply remove the optional group: ^\/src\/main\/java(\/[^\/]+)\/\*\.java$ Demo 2 and it might still work. Test import java.util.regex.Matcher; import java.util.regex.Pattern; … Read more
This is using glob rather than listdir, but this might be a possible method. No regular expressions involved here either though. import glob folder_path = “C:\Temp” file_pattern = “\*.txt” search_string = “hello” match_list = [] folder_contents = glob.glob(folder_path + file_pattern) for file in folder_contents: print(“Checking”, file) read_file = open(file, ‘rt’).read() if search_string in read_file: match_list.append(file) … Read more
Disclaimer: Note that the rules which define a “valid domain” constitute a moving target. The answer below deals only with the “old school” DNS rules (using exclusively ASCII characters) and does not attempt to deal with international domains (as laid out in RFC3490). Note also that there will soon be lots of new top level … Read more
This solution uses 2 Replace calls which might be OK if the code is not executed too frequently: string input = @”<div class=””tr-summaryinfo””>’ <p class=””tr-summaryitem””>test </> </div>”; string result = Regex.Replace(input, “<div class=\”tr-summaryinfo\”>(.*?)</div>”, “<ul class=\”tr-summaryinfo\”>$1</ul>”, RegexOptions.Singleline); result = Regex.Replace(result, “<p class=\”tr-summaryitem\”>(.*?)</p>”, “<li class=\”tr-summaryitem\”>$1</li>”, RegexOptions.Singleline); Note that you need ? in the patterns to avoid the … Read more
My warning in the comments section being said, you could use a combination of preg_replace_callback() and str_replace(): $str=”<input data-content=”This is a text string with a <br /> inside of it” />”; $regex = ‘/data-content=”([^”]*)/i’; $str = preg_replace_callback($regex, function($matches) { return str_replace(array(‘<br/>’, ‘<br />’), ”, $matches[0]); }, $str); echo $str; // output: <input data-content=”This is a … Read more
Use the “current state of art” regexp package. It has support for \w and \s and also handles non capturing groups (?:re). 2 solved Poor regular expressions in Go [closed]
This is something to get you started, but you’ll need to tweak it and adapt it to what you need: [a-zA-Z]\w{5,14} ^ ^ | match an alphanumeric or underscore character 5 to 14 times | match a single alphabetic character 6 solved Regular Expression for nickname [closed]
You can use String‘s methods replaceAll and split together here. String one = “show ip interface brief | include 1234**\n Gi1/23.1234 xxx.225.xxx.106 YES manual up up”; String[] tokens = one.replaceAll(“[^\\n]+\\n\\s*”, “”).split(“\\s+”); System.out.println(Arrays.toString(tokens)); Output [Gi1/23.1234, xxx.225.xxx.106, YES, manual, up, up] solved How to split a string if any chracter is found in java
To retrieve the value of foo3, you can just use JSON.Parse with a callback function that will only return the value of foo3 if it is present. Here is a working snippet: var text=”{“foo1″:”1123″,”foo2″:”332″,”foo3″:”23-SEP-15″}”; JSON.parse(text, function(k,v){ if (k === ‘foo3’) { document.write(v); } return v; }); Here, in function(k,v), k stands for the key and … Read more
Try this. I am assuming path is a real word. This also includes the \ after loading so it would not change paths where loading would be loadingPlatform or somewordloading which may produce undesirable results. Note that the back slashes need to be escape once for the String and once for the regex engine. String … Read more
May be you get some idea from this: Pattern : ^\d{1,2}\.\d{1,2}$|^\d{1,2}$ Match 1 Full match 0-4 0.01 Match 2 Full match 5-10 99.99 Match 3 Full match 11-16 01.01 https://regex101.com/r/p03CtT/4 5 solved Regular expression Range with decimal 0.10 – 99.99
Try this: import java.util.regex.Matcher; import java.util.regex.Pattern; … String text = ” Hi Guys \”Good Morning\””; Pattern pattern = Pattern.compile(“\”(.+?)\””); Matcher matcher = pattern.matcher(text); matcher.find(); System.out.println(matcher.group(1)); 1 solved How to split the text using selenium webdriver? [closed]