regex-lookarounds

问题 I'm building a chatbot subset of RiveScript and trying to build the pattern matching parser with regular expression. Which three regexes match the following three examples? ex1: I am * years old valid match: - "I am 24 years old" invalid match: - "I am years old" ex2: what color is [my|your|his|her] (bright red|blue|green|lemon chiffon) * valid matches: - "what color is lemon chiffon car" - "what color is my some random text till the end of string" ex3: [*] told me to say * valid matches: -

问题 I need to modify this regex href=\"(.*)\" which matches this... href="./pothole_locator_map.aspx?lang=en-gb&lat=53.153977&lng=-3.533306" To NOT match this... href="./pothole_locator_map.aspx?lang=en-gb&lat=53.153977&lng=-3.533306&returnurl=AbandonedVehicles.aspx" Tried this, but with no luck href=\"(.*)\"(?!&returnurl=AbandonedVehicles.aspx) Any help would be much appreciated. Thanks, Al. 回答1: Lookaheads should be placed before the string is consumed by matching, i.e. href=\"(?!.*&returnurl

问题 I'm writing a DSL using Scala's parser combinators. I have recently changed my base class from StandardTokenParsers to JavaTokenParsers to take advantage of the regex features I think I need for one last piece of the puzzle. (see Parsing a delimited multiline string using scala StandardTokenParser) What I am trying to do is to extract a block of text delimited by some characters ( {{ and }} in this example). This block of text can span multiple lines. What I have so far is: def docBlockRE =

问题 I'm writing a DSL using Scala's parser combinators. I have recently changed my base class from StandardTokenParsers to JavaTokenParsers to take advantage of the regex features I think I need for one last piece of the puzzle. (see Parsing a delimited multiline string using scala StandardTokenParser) What I am trying to do is to extract a block of text delimited by some characters ( {{ and }} in this example). This block of text can span multiple lines. What I have so far is: def docBlockRE =

问题 Using SQL Anywhere 16 regex (Perl 5 equivalent) I would like to match all non-digit characters in the following string: This 100 is 2 a 333 test. such that the match results in This is a test. I have a feeling this requires lookarounds but I'm not sure of the exact syntax. Some of my failed attempts: \D* matches This \D*(?=\d) matches This (\D*(?=\d*)|(?<=\d)\D*)* matches This 来源： https://stackoverflow.com/questions/36020273/regex-to-match-all-non-digit-characters-which-are-interspersed-with

问题 I'm trying to invert this expression: ([0-9]{14} ) , so all 14 digit numbers followed by a space. I looked everywhere, and it seems that the best way should be using negative lookahead . But when I try apply q(?!u) to my case >> (?!([0-9]{14} )) , it doesn't work. What am I doing wrong? I will appreaciate any advice, thank you. The point is to remove everything that is not a 14-digit chunk of text while preserving those 14-digit chunks. 回答1: If you want to delete text other than 14 digits

问题 I have some trouble understanding how the following regular expression is working. ,(?=([^\"]*\"[^\"]*\")*[^\"]*$) The expression basically matches all the commas that are NOT enclosed in quotes. For example: apple, banana, pineapple, "tropical fruits like mango, guava, key lime", peaches Will be split into: apple banana pineapple "tropical fruits like mango, guava, key lime" peaches Can someone provide me with a good breakdown of the expression? I don't understand how positive look-ahead is

问题 My current regex battle is replacing all commas before a number in a string. The regex must then ignore all following commas. I've been screwing around on rubular for about an hour and can't quite seem to get something working. Test String... 'this is, a , sentence33 Here, is another.' Desired Output... 'this is comma a comma sentence33 Here, is another.' So something along the lines of... testString.gsub(/\,*\d\d/,"comma") To give you some background, I'm doing a little scraping sideproject.

问题 I've been trying to create Regex for WhatsApp chat log. So far I've been able to achieve this Click Here for the test link By creating the following Regex: (?P<datetime>\d{2}\/\d{2}\/\d{4},\s\d(?:\d)?:\d{2} [pa].m.)\s-\s(?P<name>[^:]*):(?P<message>.*) The problem with this regex is, it is not able to match big messages which span multiple lines with line breaks. You can see the issue in the link provided above. Help would be appreciated. Thank you. 回答1: There you go: ^ (?P<datetime>\d{2}/\d{2

问题 I'm looking for a way to make the finditer function of the python re module or the newer regex module to match all possible variations of a particular pattern, overlapping or otherwise. I am aware of using lookaheads to get matches without consuming the search string, but I still only get one regex per index, where I could get more than one. The regex I am using is something like this: (?=A{2}[BA]{1,6}A{2}) so in the string: AABAABBAA it should be able to match: AABAA AABAABBAA AABBAA but