posix-ere

问题 I thought that in regular expressions, the "greediness" applies to quantifiers rather than matches as a whole. However, I observe that grep -E --color=auto 'a+(ab)?' <(printf "aab") returns aab rather than aa b. The same applies to sed. On the other hand, in pcregrep and other tools, it is really the quantifier that is greedy. Is this a specific behaviour of grep? N.B. I checked both grep (BSD grep) 2.5.1-FreeBSD and grep (GNU grep) 3.1 回答1: In the description of term matched, POSIX states

问题 How come my regex pattern isn't lazy? It should be capturing the first number, not the second. Here is a working bash script.. #!/bin/bash text='here is some example text I want to match word1 and this number 3.01 GiB here is some extra text and another number 1.89 GiB' regex='(word1|word2).*?number[[:blank:]]([0-9.]+) GiB' if [[ "$text" =~ $regex ]]; then echo 'FULL MATCH: '"${BASH_REMATCH[0]}" echo 'NUMBER CAPTURE: '"${BASH_REMATCH[2]}" fi Here is the output... FULL MATCH: word1 and this

问题 How come my regex pattern isn't lazy? It should be capturing the first number, not the second. Here is a working bash script.. #!/bin/bash text='here is some example text I want to match word1 and this number 3.01 GiB here is some extra text and another number 1.89 GiB' regex='(word1|word2).*?number[[:blank:]]([0-9.]+) GiB' if [[ "$text" =~ $regex ]]; then echo 'FULL MATCH: '"${BASH_REMATCH[0]}" echo 'NUMBER CAPTURE: '"${BASH_REMATCH[2]}" fi Here is the output... FULL MATCH: word1 and this

问题 Consider the following commands: text <- "abcdEEEEfg" sub("c.+?E", "###", text) # [1] "ab###EEEfg" <<< OKAY sub("c(.+?)E", "###", text) # [1] "ab###EEfg" <<< WEIRD sub("c(.+?)E", "###", text, perl=T) # [1] "ab###EEEfg" <<< OKAY The first does exactly what I expect, basically matching just the first E. The second one should essentially be identical to the first, since all I'm doing is adding a capturing group (though I'm not using it), yet for some reason it captures an extra E. That said, it

问题 I have the following regex to match the last pair of braces in a string, .+(?={)(.+)(?=}) The example string is, abc{abc=bcd}{gef=hij} I want the contents within the last braces (gef=hij) inside the captured group. This works in a regex tester available in the web http://regexpal.com/ When I use regcomp to compile the same regex, it doesnt. Any ideas? int reti = regcomp(&regex, ".+(?={)(.+)(?=})", REG_EXTENDED); if (reti) { fprintf(stderr, "Could not compile regex\n"); exit(1); } 回答1: Anyway,

问题 I have the following regex to match the last pair of braces in a string, .+(?={)(.+)(?=}) The example string is, abc{abc=bcd}{gef=hij} I want the contents within the last braces (gef=hij) inside the captured group. This works in a regex tester available in the web http://regexpal.com/ When I use regcomp to compile the same regex, it doesnt. Any ideas? int reti = regcomp(&regex, ".+(?={)(.+)(?=})", REG_EXTENDED); if (reti) { fprintf(stderr, "Could not compile regex\n"); exit(1); } 回答1: Anyway,

问题 I would like to use a regular expression like this in Java : [[=a=][=e=][=i=]] . But Java doesn't support the POSIX classes [=a=], [=e=] etc . How can I do this? More precisely, is there a way to not use US-ASCII? 回答1: Java does support posix character classes. The syntax is just different, for instance: \p{Lower} \p{Upper} \p{ASCII} \p{Alpha} \p{Digit} \p{Alnum} \p{Punct} \p{Graph} \p{Print} \p{Blank} \p{Cntrl} \p{XDigit} \p{Space} 回答2: Quoting from http://download.oracle.com/javase/1.6.0

问题 I would like to use a regular expression like this in Java : [[=a=][=e=][=i=]] . But Java doesn't support the POSIX classes [=a=], [=e=] etc . How can I do this? More precisely, is there a way to not use US-ASCII? 回答1: Java does support posix character classes. The syntax is just different, for instance: \p{Lower} \p{Upper} \p{ASCII} \p{Alpha} \p{Digit} \p{Alnum} \p{Punct} \p{Graph} \p{Print} \p{Blank} \p{Cntrl} \p{XDigit} \p{Space} 回答2: Quoting from http://download.oracle.com/javase/1.6.0

问题 I would like to use a regular expression like this in Java : [[=a=][=e=][=i=]] . But Java doesn't support the POSIX classes [=a=], [=e=] etc . How can I do this? More precisely, is there a way to not use US-ASCII? 回答1: Java does support posix character classes. The syntax is just different, for instance: \p{Lower} \p{Upper} \p{ASCII} \p{Alpha} \p{Digit} \p{Alnum} \p{Punct} \p{Graph} \p{Print} \p{Blank} \p{Cntrl} \p{XDigit} \p{Space} 回答2: Quoting from http://download.oracle.com/javase/1.6.0

问题 I am working on a MySQL database and noticed that it doesn't natively support PCRE (requires a plugin). I wish to use these three for some data validation (these are actually the values given to the pattern attribute): ^[A-z\. ]{3,36} ^[a-z\d\.]{3,24}$ ^(?=^.{4,}$)(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?!.*\s).*$ How do I do this? I looked on the web but couldn't find any concrete examples or answers. Also there seem to exist no utilities that could do this automatically. I am aware that some times,