capturing-group | 易学教程

Regex .NET attached named group

阅读更多关于 Regex .NET attached named group

Javascript RegExp non-capturing groups

阅读更多关于 Javascript RegExp non-capturing groups

问题 I am writing a set of RegExps to translate a CSS selector into arrays of ids and classes. For example, I would like '#foo#bar' to return ['foo', 'bar']. I have been trying to achieve this with "#foo#bar".match(/((?:#)[a-zA-Z0-9\-_]*)/g) but it returns ['#foo', '#bar'], when the non-capturing prefix ?: should ignore the # character. Is there a better solution than slicing each one of the returned strings? 回答1: You could use .replace() or .exec() in a loop to build an Array. With .replace() :

Vim / sed regex backreference in search pattern

阅读更多关于 Vim / sed regex backreference in search pattern

问题 Vim help says that: \1 Matches the same string that was matched by */\1* *E65* the first sub-expression in $ and $. {not in Vi} Example: "$[a-z]$.\1" matches "ata", "ehe", "tot", etc. It looks like the backreference can be used in search pattern. I started playing with it and I noticed behavior that I can't explain. This is my file: <paper-input label="Input label"> Some text </paper-input> <paper-input label="Input label"> Some text </paper-inputa> <aza> Some text </az> <az> Some text <

Regex - Saving Repeating Captured Group

阅读更多关于 Regex - Saving Repeating Captured Group

问题 This is what I'm doing a = "%span.rockets#diamonds.ribbons.forever" a = a.match(/(^\%\w+)([\.|\#]\w+)+/) puts a.inspect This is what I get #<MatchData "%span.rockets#diamonds.ribbons.forever" 1:"%span" 2:".forever"> This is what I want #<MatchData "%span.rockets#diamonds.ribbons.forever" 1:"%span" 2:".rockets" 3:".#diamonds" 4:".ribbons" 5:".forever"> help? I tried and failed :( 回答1: Generally, you can't get an arbitrary number of capturing groups, but if you use scan you can get a match for

Vim / sed regex backreference in search pattern

阅读更多关于 Vim / sed regex backreference in search pattern

Vim help says that: \1 Matches the same string that was matched by */\1* *E65* the first sub-expression in $ and $. {not in Vi} Example: "$[a-z]$.\1" matches "ata", "ehe", "tot", etc. It looks like the backreference can be used in search pattern. I started playing with it and I noticed behavior that I can't explain. This is my file: <paper-input label="Input label"> Some text </paper-input> <paper-input label="Input label"> Some text </paper-inputa> <aza> Some text </az> <az> Some text </az> <az> Some text </aza> I wanted to match the lines where the opening and closing tags are matching i

Regex - Saving Repeating Captured Group

阅读更多关于 Regex - Saving Repeating Captured Group

This is what I'm doing a = "%span.rockets#diamonds.ribbons.forever" a = a.match(/(^\%\w+)([\.|\#]\w+)+/) puts a.inspect This is what I get #<MatchData "%span.rockets#diamonds.ribbons.forever" 1:"%span" 2:".forever"> This is what I want #<MatchData "%span.rockets#diamonds.ribbons.forever" 1:"%span" 2:".rockets" 3:".#diamonds" 4:".ribbons" 5:".forever"> help? I tried and failed :( Generally, you can't get an arbitrary number of capturing groups, but if you use scan you can get a match for every token you want to capture: a = "%span.rockets#diamonds.ribbons.forever" a = a.scan(/^%\w+|\G[.|#]\w+/)

PHP How to set the preg-groups to “non-capture” (?:…)

阅读更多关于 PHP How to set the preg-groups to “non-capture” (?:…)

In HTML page, I remove HTML comments like this $contentHTML = preg_replace("#(?=#", "", $contentHTML); But on a huge page for preg_replace , I got " PHP Fatal error: Allowed memory size ... " Perhaps, one solution, would use the non-matching group to avoid capturing text? Could someone explain how use on-matching group ?: Or how can I suppress HTML comments in huge page without preg_replace ? Just unroll the regex as $contentHTML = preg_replace("##", "", $contentHTML); See the regex demo . Comapre with yours taking about 3 times as more steps as

what does this django regex mean? `?P`

阅读更多关于 what does this django regex mean? `?P`

问题 I have the following regex in my urls.py and I'd like to know what it means. Specifically the (?P<category_slug> portion of the regex. r'^category/(?P<category_slug>[-\w]+)/$ 回答1: (?P<name>regex) - Round brackets group the regex between them. They capture the text matched by the regex inside them that can be referenced by the name between the sharp brackets. The name may consist of letters and digits. Copy paste from: http://www.regular-expressions.info/refext.html 回答2: In django, named

Regex fails to capture all groups

阅读更多关于 Regex fails to capture all groups

问题 Using java.util.regex (jdk 1.6), the regular expression 201210(\d{5,5})Test applied to the subject string 20121000002Test only captures group(0) and does not capture group(1) (the pattern 00002 ) as it should, given the code below: Pattern p1 = Pattern.compile("201210(\\d{5,5})Test"); Matcher m1 = p1.matcher("20121000002Test"); if(m1.find()){ for(int i = 1; i<m1.groupCount(); i++){ System.out.println("number = "+m1.group(i)); } } Curiously, another similar regular expression like 201210(\d{5

Regex - capture all repeated iteration

阅读更多关于 Regex - capture all repeated iteration

I have a variable like this var = "!123abcabc123!" i'm trying to capture all the '123' and 'abc' in this var. this regex (abc|123) retrieve what i want but... My question is: when i try this regex !(abc|123)*! it retrieve only the last iteration. what will i do to get this output MATCH 1 1. [1-4] `123` MATCH 2 1. [4-7] `abc` MATCH 3 1. [7-10] `abc` MATCH 4 1. [10-13] `123` https://regex101.com/r/mD4vM8/3 Thank you!! If your language supports \G then you may free to use this. (?:!|\G(?!^))\K(abc|123)(?=(?:abc|123)*!) DEMO 来源： https://stackoverflow.com/questions/31992558/regex-capture-all