I guess it is wrong in your sense. I suspect you’re looking for this. String pattern = “(\\D*)(\\d+)(.*)”; Maybe you want to check reluctant quantifiers too. http://docs.oracle.com/javase/tutorial/essential/regex/quant.html 13 solved Is this regex pattern wrong? [closed]
If you try to parse input date, find first non numeric character [0-9]+([^0-9]).* If you try to find element in “mask/template/…” then first character not in set [dMy]+(\.).* solved Use regex to find which date separator a user has inputted
Currency amount US & EU (cents optional) Can use US-style 123,456.78 notation and European-style 123.456,78 notation. Optional thousands separators; optional two-digit fraction Match; JGsoft: ^[+-]?[0-9]{1,3}(?:[0-9]*(?:[.,][0-9]{2})?|(?:,[0-9]{3})*(?:\.[0-9]{2})?|(?:\.[0-9]{3})*(?:,[0-9]{2})?)$ Reference: here 2 solved Java Regular expression for money
Check out regex 101. It is a wonderfull Tool to create regular expressions and understand them. From there i got this solution for you: %^\d{1,2}:\d{1,2}([ap])m\[(Mon|Tue|Wed|Thu|Fri|-){1,} Only\]$%mg The first bit (\d{1,2}) accepts a number between 0 and 99, then a : and another number between 0 and 99. Then it is either a or p followed … Read more
Use /\?[^\”]+/ like bellow: <?php $string = ‘<a href=”https://stackoverflow.com/title/tt3110958/?ref_=nm_flmg_act_2″ style=”color:#666″ >’; $result = preg_replace(“/\?[^\”]+/”, “”, $string); echo $result; ?> Output : <a href=”https://stackoverflow.com/title/tt3110958/” style=”color:#666″ > 0 solved How to remove query string from “href” attribute of anchor using php?
please find below working code import java.util.regex.Matcher; import java.util.regex.Pattern; public class Main { public static void main(String[] args) { String line = “This cat MAN-1243-23445 placed OK? This cat MAN-1243-23445 placed OK? This cat MAN-1243-23445 placed OK? This cat MAN-1243-23445 placed OK?This cat MAN-1243-23445 placed OK? This cat MAN-1243-23445 placed OK?”; String pattern = “(MAN-\\d{4}-\\d{5})”; … Read more
It captures 13-16 digits, each followed by zero or more spaces or dashes (the [ -]*?). In other words, the {13,16} applies to the entire group (?:\d[ -]*?). So, it could capture, for example, 1–2-3–4-5-6 7—–8-9-0-1-2-3-4-5-6-. In your sample case, it captures these 16 chunks: 4 5 6 4- 1 2 3 4- 4 3 … Read more
sically if there is a space after the capital A, I would to ignore the A and the space and capture everything else. But if there is no space after the A I want to include the A in the capture. You can try this regex in PCRE using match reset \K: \bA(?: \K)?\S+ \K … Read more
Something like that? re = new RegExp(/\d{2}\/\d{2}\/\d{4}/); “21/02/2017”.match(re) /* [“21/02/2018”, index: 0, input: “21/02/2018”] */ “21/2/2017”.match(re) /* null */ solved What is the regex ?
^[1]*(0[1]*0[1]*)*$ which can be interpreted in the following way: find any number of 1 followed by pairs of 00 with optional one or more 1s in between. solved How to determine if a number is binary and has an even amount of zeros with only one regex?
Unless your data size is small (< a few MB), you probably want something like Apache Lucene, which can be used to index and search large document sets, and can do fuzzy searches using Levenshtein distance and other algorithms for similarity matches. solved Is there a library in Java for SQL like string searches? [closed]
You can do that using the following regex: /^(?!.{36,})[a-z’-]{2,}\s[a-z’-]{2,}/gmi See this in action at Regex101 Explanation This matches strings that suit the following criteria: ^(?!.{36,}) It is not 36 characters or longer [a-z’-]{2,} Starts with a word containing only “a” – “z”, “‘” and “-“ \s followed by a whitespace [a-z’-]{2,} followed by a word … Read more
Edit: The data you provided is a JSON string. You can convert it to a dictionary using the json package: import json payload = u'{“encrypted_sender_transaction_id”:”514658451″,…}’ obj = json.loads(payload) print obj[‘donation_info’][‘amount’] # 1 obj is a nested dictionary in this case, amount is a key in the subdictionary under the key donation_info 2 solved How to … Read more
This has the function of introducing a backslash character \ before certain characters in that regular expression set […]. The characters are: . * + ? ^ = ! : $ { } ( ) | [ ] / \ Note that some of these are necessarily escaped with \ because they have special meaning … Read more
You can use the below regex to achieve your requirement: [ ](?=[^\)]*?(?:\(|$)) Explanation of the above regex: [ ] – Represents a space character. (?=[^\)]*?(?:\(|$)) – Represents a positive look-ahead asserting everything inside of (). (?:) – Represents a non-capturing group. | – Represents alternation. $ – Represents the end of the test String. You … Read more