Use negative lookahead assertion. foo(?:(?!foo).)*?boo DEMO (?:(?!foo).)*? – Non-greedy match of any character but not of foo zero or more times. That is, before matching each character, it would check that the character is not the letter f followed by two o‘s. If yes, then only the corresponding character will be matched. Why the regex … Read more
You’re looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?. By default grep doesn’t support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
This is an excellent question, and it took me a while to see the point of the lazy ?? quantifier myself. ? – Optional (greedy) quantifier The usefulness of ? is easy enough to understand. If you wanted to find both http and https, you could use a pattern like this: https? This pattern will … Read more
For a regular expression like .* or .+, append a question mark (.*? or .+?) to match as few characters as possible. To optionally match a section (?:blah)? but without matching unless absolutely necessary, use something like (?:blah){0,1}?. For a repeating match (either using {n,} or {n,m} syntax) append a question mark to try to … Read more
Instead of .* use .\{-}. %s/style=”.\{-}”//g Also, see :help non-greedy
The non-greedy regex modifiers are like their greedy counter-parts but with a ? immediately following them: * – zero or more *? – zero or more (non-greedy) + – one or more +? – one or more (non-greedy) ? – zero or one ?? – zero or one (non-greedy)
Common credit card vendor regular expressions: Amex Card: ^3[47][0-9]{13}$ BCGlobal: ^(6541|6556)[0-9]{12}$ Carte Blanche Card: ^389[0-9]{11}$ Diners Club Card: ^3(?:0[0-5]|[68][0-9])[0-9]{11}$ Discover Card: ^65[4-9][0-9]{13}|64[4-9][0-9]{13}|6011[0-9]{12}|(622(?:12[6-9]|1[3-9][0-9]|[2-8][0-9][0-9]|9[01][0-9]|92[0-5])[0-9]{10})$ Insta Payment Card: ^63[7-9][0-9]{13}$ JCB Card: ^(?:2131|1800|35\d{3})\d{11}$ KoreanLocalCard: ^9[0-9]{15}$ Laser Card: ^(6304|6706|6709|6771)[0-9]{12,15}$ Maestro Card: ^(5018|5020|5038|6304|6759|6761|6763)[0-9]{8,15}$ Mastercard: ^(5[1-5][0-9]{14}|2(22[1-9][0-9]{12}|2[3-9][0-9]{13}|[3-6][0-9]{14}|7[0-1][0-9]{13}|720[0-9]{12}))$ Solo Card: ^(6334|6767)[0-9]{12}|(6334|6767)[0-9]{14}|(6334|6767)[0-9]{15}$ Switch Card: ^(4903|4905|4911|4936|6333|6759)[0-9]{12}|(4903|4905|4911|4936|6333|6759)[0-9]{14}|(4903|4905|4911|4936|6333|6759)[0-9]{15}|564182[0-9]{10}|564182[0-9]{12}|564182[0-9]{13}|633110[0-9]{10}|633110[0-9]{12}|633110[0-9]{13}$ Union Pay Card: ^(62[0-9]{14,17})$ Visa Card: ^4[0-9]{12}(?:[0-9]{3})?$ Visa Master Card: … Read more
The non-greedy ? works perfectly fine. It’s just that you need to select dot matches all option in the regex engines (regexpal, the engine you used, also has this option) you are testing with. This is because, regex engines generally don’t match line breaks when you use .. You need to tell them explicitly that … Read more
You seek the all-powerful *? From the docs, Greedy versus Non-Greedy the non-greedy qualifiers *?, +?, ??, or {m,n}? […] match as little text as possible.
In this specific case, you can get the job done without using a non-greedy regex. Try this non-greedy regex [^/]* instead of .*?: sed ‘s|\(http://[^/]*/\).*|\1|g’