Grep perl regex7/21/2023 ![]() ![]() Or used fixed Lookbehind combined with Lookahead. echo 'foosomethingbar' | grep -Po 'foo\K.*?(?=bar)' To interpret the pattern as an extended regular expression, use the -E ( or -extended-regexp) option. In its simplest form, when no regular expression type is given, grep interpret search patterns as basic regular expressions. To get a substring only between two patterns, you can add Positive Lookahead into the mix. GNU grep supports three regular expression syntaxes, Basic, Extended, and Perl-compatible. So you could do: echo 'foosomething' | grep -Po '\w \Ksomething' grep understands three different versions of regular expression syntax: basic (BRE), extended (ERE), and Perl-compatible (PCRE). Number between them doesn't have a fixed length, while it can't be done with \K since we need "foo"Īny further suggestions are still appreciated :)Īfter some testing I found out, that the pattern inside the look-behind assertion needs to be fixed length (something like (? grep: lookbehind assertion is not fixed length Regular expressions are constructed analogously to arithmetic expressions, by using various operators to combine smaller expressions. There is no way to match all above examples while output "foobar", since the If "foo" and "bar" are place-holders for words that don't always have the same length, The best way so far is either using perl or combine a combination of look-behind assertions and the \K but it still seems to have some limitations. Will not return the letter "a", only the digit following it, but it needs a fixed length, for example it cannot be used as: (?<=\w )\d The problem is nearly solved using "look-behind assertions" such as: (?<=a)\d I need to able to get any part of the match, for instance: ~]# ~]# echo $var |grep -oP "(a)\d"īut the wanted output in this case would be "a" Is not really the solution, since I don't necessarily need the last part of the match. Probably it's also possible without the -P (some not-perl regex format), I'm thankful for every answer :) When user learns regular expression then there might be a need for quick look of those concepts which he didn’t use often. In this, set of characters together form the search pattern. It is used for searching the specified text pattern. I also tried a non-capturing group to ignore the "a" in the output, but no effect in perl regex: '(?:a)\d'įor reference, this is the full command in my terminal: ~]# ~]# echo $var |grep -oP "a(\d)" Regex or Regular Expressions are an important part of Perl Programming. Now what I want to do is match only "a" follow by any digit, but I only want it to return the number after "a"Īnd since I only need the digit, I tried with 'a(\d)'Īnd maybe it did capture it somewhere, but I don't know where, the output here is still "a1" ![]() So I've done some research on the subject and I didn't quite find the perfect solution.įor example I have a string inside a variable. ![]()
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