This is a neat well documented regular expression, easy to understand, maintain and modify.
text = text.replace(/
( // Wrap whole match in $1
(
^[ \t]*>[ \t]? // '>' at the start of a line
.+\n // rest of the first line
(.+\n)* // subsequent consecutive lines
\n* // blanks
)+
)
/gm,
But how do you go about working with these?
text = text.replace(/((^[ \t]*>[ \t]?.+\n(.+\n)*\n*)+)/gm,
Is there a beautifier of some sort that makes sense of it and describes its functionality?
It's worth the effort to become adept at reading regexs in the one line form. Most of the time there are written this way
RegexBuddy will "translate" any regex for you. When fed your example regex, it outputs:
((^[ \t]*>[ \t]?.+\n(.+\n)*\n*)+)
Options: ^ and $ match at line breaks
Match the regular expression below and capture its match into backreference number 1 «((^[ \t]*>[ \t]?.+\n(.+\n)*\n*)+)»
Match the regular expression below and capture its match into backreference number 2 «(^[ \t]*>[ \t]?.+\n(.+\n)*\n*)+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Note: You repeated the capturing group itself. The group will capture only the last iteration.
Put a capturing group around the repeated group to capture all iterations. «+»
Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match a single character present in the list below «[ \t]*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
The character “ ” « »
A tab character «\t»
Match the character “>” literally «>»
Match a single character present in the list below «[ \t]?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
The character “ ” « »
A tab character «\t»
Match any single character that is not a line break character «.+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match a line feed character «\n»
Match the regular expression below and capture its match into backreference number 3 «(.+\n)*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Note: You repeated the capturing group itself. The group will capture only the last iteration.
Put a capturing group around the repeated group to capture all iterations. «*»
Match any single character that is not a line break character «.+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match a line feed character «\n»
Match a line feed character «\n*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
This does look rather intimidating in text form, but it's much more readable in HTML form (which can't be reproduced here) or in RegexBuddy itself. It also points out common gotchas (such as repeating capturing groups which is probably not wanted here).
I like expresso
After a while, I've gotten used to reading the things. There is not much to most regexes, and I recommend the site http://www.regular-expressions.info/ if you want to use them more often.
Regular expressions are just a way to express masks, etc. At the end it's just a "language" with its own syntax.
Comment every bit of your regular expression would be the same thing as comment every line of your project.
Of course it would help people who doesn't understand your code, but it's just useless if you (the developer) do understand the meaning of the regex.
For me, reading regular expressions is the same thing as reading code. If the expression is really complex an explanation below could be useful but most of the time it isn't necessary.
Related
I'm building a regular expression which have to extract strings from brackets. This is an example string:
((?X is parent ?Y)(?X is child ?Z))
I need to get strings: '?X is parent ?Y' and also '?X is child ?Z'. This is what I've created yet:
^(\((.*?)\))+$
The problem is that it matches only the string in the second bracket. Could anybody help me to improve the expression so that it matches both strings in brackets?
Note: brackets can contain any content, like ((AAA)(BBB)). In this case 'AAA' and 'BBB' should be matched.
Thanks forward.
Based on your comments, it seems that you just want to match anything inside the brackets, for that you can use:
String Sample1 = "((something)(world)(example))";
Pattern regex = Pattern.compile("\\(?\\((.*?)\\)\\)?");
Matcher regexMatcher = regex.matcher(Sample1);
while (regexMatcher.find()) {
System.out.print(regexMatcher.group(1));
// something world example
}
Demo
Regex Explanation
Match the character “(” literally «\(?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match the character “(” literally «\(»
Match the regular expression below and capture its match into backreference number 1 «(.*?)»
Match any single character that is not a line break character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “)” literally «\)»
Match the character “)” literally «\)?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
This seems to work:
Pattern.compile("[\\(]{0,1}(\\((.*?)\\))")
Thanks all for replies and comments.
I'm looking for a regex that will match a word only if all its characters are unique, meaning, every character in the word appears only once.
Example:
abcdefg -> will return MATCH
abcdefgbh -> will return NO MATCH (because the letter b repeats more than once)
Try this, it might work,
^(?:([A-Za-z])(?!.*\1))*$
Explanation
Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match the regular expression below «(?:([A-Z])(?!.*\1))*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match the regular expression below and capture its match into backreference number 1 «([A-Z])»
Match a single character in the range between “A” and “Z” «[A-Z]»
Assert that it is impossible to match the regex below starting at this position (negative lookahead) «(?!.*\1)»
Match any single character that is not a line break character «.*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match the same text as most recently matched by capturing group number 1 «\1»
Assert position at the end of a line (at the end of the string or before a line break character) «$»
You can check whether there are 2 instances of the character in the string:
^.*(.).*\1.*$
(I just simply capture one of the character and check whether it has a copy elsewhere with back reference. The rest of .* are don't-cares).
If the regex above match, then the string has repeating character. If the regex above doesn't match, then all the characters are unique.
The good thing about the regex above is when the regex engine doesn't support look around.
Apparently John Woo's solution is a beautiful way to check for the uniqueness directly. It assert at every character that the string ahead will not contain the current character.
This one would also provide a full match to any length word with non-repeating letters:
^(?!.*(.).*\1)[a-z]+$
I slightly revised the answer provided by #Bohemian to another question a while ago to get this.
It has also been a while since the question above has been asked but I thought it would be nice to also have this regex pattern here.
Given these urls:
1: http://site/page-name-one-123/
2: http://site/page-name-set2/
3: http://site/set20
I wrote this expression that will be applied to last url segment:
(?(?<=set[\d])([\d]+)|([^/]+))
What I'd want to do is to catch every digits followed by 'set' only if the url segment starts with 'set' and a digit immediately after; otherwise i want to use the whole segment (excluding slashes).
As I wrote this regex, it matches any character that is not a '/'. I think I'm doing something wrong in test statement.
Could anyone point me right?
Thanks
UPDATE
Thanks to Josh input I played around for a bit and found that this one fits better my needs:
set-(?P<number>[0-9]+)|(?P<segment>[^/]+)
I hope this pattern can help you out, I put it together based on your requirements. You may want to play around with setting some of the groups to not capture so that you only get the segments that you need. However, it does seperate capture your set URL's without set at the start.
((?<=/{1})(((?<!set)[\w|-]*?)(\d+(?=/?))|((?:set)\d+)))
I suggest using RegExr to pick it apart if you need to.
Try this:
((?<=/)set\d+|(?<=/)[^/]+?set\d+)
Explanation
<!--
Options: ^ and $ match at line breaks
Match the regular expression below and capture its match into backreference number 1 «((?<=/)set\d+|(?<=/)[^/]+?set\d+)»
Match either the regular expression below (attempting the next alternative only if this one fails) «(?<=/)set\d+»
Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=/)»
Match the character “/” literally «/»
Match the characters “set” literally «set»
Match a single digit 0..9 «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «(?<=/)[^/]+?set\d+»
Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=/)»
Match the character “/” literally «/»
Match any character that is NOT a “/” «[^/]+?»
Between one and unlimited times, as few times as possible, expanding as needed (lazy) «+?»
Match the characters “set” literally «set»
Match a single digit 0..9 «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
-->
A minor inconvenience my users have found is that if they use a smilie such as >_> at the end of parentheses (kind of like this: >_>) then during processing it is run through htmlspecialchars(), making it >_>) - you can see the problem, I think. The ;) at the end is then replaced by the "Wink" smilie.
Can anyone give me a regex that will replace ;) with the smilie, but only if the ; is not the end of an HTML entity? (I'm sure it would involve a lookbehind but I can't seem to understand how to use them >_>)
Thank you!
Handling smileys like ;) is always a bit tricky - the way I would do it is transform it to the "canonical" :wink: before encoding HTML entities, and then changing only canonical-form :{smileyname}: smileys afterwards.
Like this: (?<!&[a-zA-Z0-9]+);\)
The (?>!...) is a zero-width assertion that will only allow the following construct to match text that isn't preceded by the ....
You should probably handle it along these lines, which sidesteps the issue of replacing replacements entirely:
Break the string apart wherever a smilie occurs, convert the smilies into tokens
HTML escape all the text nodes
Convert all the smilie tokens into their HTML tag equivalents
Glue everything back together
That's a bit non-trivial though. :)
Find: (&#?[a-z0-9]+;)\)
Replace: $0)
We're looking for:
Match the regular expression below and capture its match into backreference number 1 «(&#?[a-z0-9]+;)»
Match the character “&” literally «&»
Match the character “#” literally «#?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match a single character present in the list below «[a-z0-9]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
A character in the range between “a” and “z” «a-z»
A character in the range between “0” and “9” «0-9»
Match the character “;” literally «;»
Match the character “)” literally «\)»
Created with RegexBuddy
well if your intrested in a regex solution try this maybe
(?!t)([A-Za-z0-9]| );)
If it's in php (preg_replace you said ?), you can use preg_replace_callback :
preg_replace_callback('#(&[a-z0-9]+)?;\)#i', 'myFunction', 'myText');
in the "myFunction" function, you just have to check if you got some html entity in the capturing bracket.
function myFunction($matches) {
if(!empty($matches[1]) {
return $matches[0];
}
return '[Smilie]';
}
I use a library to parse an iCalendar file, but I don't understand the regex to split property.
iCalendar property has 3 different style:
BEGIN:VEVENT
DTSTART;VALUE=DATE:20080402
RRULE:FREQ=YEARLY;WKST=MO
The library uses this regex that I would like to understand:
var matches:Array = data.match(/(.+?)(;(.*?)=(.*?)((,(.*?)=(.*?))*?))?:(.*)$/);
p.name = matches[1];
p.value = matches[9];
p.paramString = matches[2];
Thanks.
That's a terrible regular expression! .* and .*? mean to match as many (greedy) or as few (lazy) of anything. These should only be used as a last resort. Improper use will result in catastrophic backtracking when the regex cannot match the input text. All you need to understand about this regular expression that you don't want to write regexes like this.
Let me show how I would approach the problem. Apparently the iCalendar File Format is line-based. Each line has a property and a value separated by a colon. The property can have parameters that are separated from it by a semicolon. This implies that a property cannot contain line breaks, semicolons or colons, that the optional parameters cannot contain line breaks or colons, and that the value cannot contain line breaks. This knowledge allows us to write an efficient regular expression that uses negated character classes:
([^\r\n;:]+)(;[^\r\n:]+)?:(.+)
Or in ActionScript:
var matches:Array = data.match(/([^\r\n;:]+)(;[^\r\n:]+)?:(.+)/);
p.name = matches[1];
p.value = matches[3];
p.paramString = matches[2];
As explained by RegexBuddy:
Match the regular expression below and capture its match into backreference number 1 «([^\r\n;:]+)»
Match a single character NOT present in the list below «[^\r\n;:]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
A carriage return character «\r»
A line feed character «\n»
One of the characters “;:” «;:»
Match the regular expression below and capture its match into backreference number 2 «(;[^\r\n:]+)?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match the character “;” literally «;»
Match a single character NOT present in the list below «[^\r\n:]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
A carriage return character «\r»
A line feed character «\n»
The character “:” «:»
Match the character “:” literally «:»
Match the regular expression below and capture its match into backreference number 3 «(.+)»
Match any single character that is not a line break character «.+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»