regex-lookarounds

问题 Below is the Content: Subject: Security ID: S-1-5-21-3368353891-1012177287-890106238-22451 Account Name: ChamaraKer Account Domain: JIC Logon ID: 0x1fffb Object: Object Server: Security Object Type: File Object Name: D:\ApacheTomcat\apache-tomcat-6.0.36\logs\localhost.2013-07-01.log Handle ID: 0x11dc I need to capture the words after the Object Name: word in that line. Which is D:\ApacheTomcat\apache-tomcat-6.0.36\logs\localhost.2013-07-01.log . I hope somebody can help me with this. ^.*

问题 I\'m working on a grammar extension in VS Code, and I\'m having difficulty with a look behind regex pattern. Given the following string, I want to only return cmp when it\'s preceded by the @fmt( @fmt(cmp,foo) The matching string I used in another editor was this: (?<=[@|©](fmt)\\()(\\w+) However, this is not working in VS Code, when I do a regex search it comes back with the error that it\'s not a valid expression. Playing around with it, the problem is the <= characters, which indicate the

问题 This question already has answers here : Javascript: negative lookbehind equivalent? (13 answers) Closed 3 months ago . Consider: var re = /(?<=foo)bar/gi; It is an invalid regular expression in Plunker. Why? 回答1: JavaScript lacks support for lookbehinds like (?<=…) (positive) and (?<!…) (negative), but that doesn't mean you can't still implement this sort of logic in JavaScript. Matching (not global) Positive lookbehind match: // from /(?<=foo)bar/i var matcher = mystring.match( /foo(bar)/i

问题 For regex what is the syntax for search until but not including? Kinda like: Haystack: The quick red fox jumped over the lazy brown dog Expression: .*?quick -> and then everything until it hits the letter \"z\" but do not include z 回答1: The explicit way of saying "search until X but not including X " is: (?:(?!X).)* where X can be any regular expression. In your case, though, this might be overkill - here the easiest way would be [^z]* This will match anything except z and therefore stop

问题 I would like to write a regex for searching for the existence of some words, but their order of appearance doesn\'t matter. For example, search for \"Tim\" and \"stupid\". My regex is Tim.*stupid|stupid.*Tim . But is it possible to write a simpler regex (e.g. so that the two words appear just once in the regex itself)? 回答1: See this regex: /^(?=.*Tim)(?=.*stupid).+/ Regex explanation: ^ Asserts position at start of string. (?=.*Tim) Asserts that "Tim" is present in the string. (?=.*stupid)

问题 I want to match all strings ending in \".htm\" unless it ends in \"foo.htm\". I\'m generally decent with regular expressions, but negative lookaheads have me stumped. Why doesn\'t this work? /(?!foo)\\.htm$/i.test(\"/foo.htm\"); // returns true. I want false. What should I be using instead? I think I need a \"negative look behind \" expression (if JavaScript supported such a thing, which I know it doesn\'t). 回答1: The problem is pretty simple really. This will do it: /^(?!.*foo\.htm$).*\.htm$