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x = "abcdefg"
x = x.match(/ab(?:cd)ef/)
shouldn't x be abef? it is not, it is actually abcdef
Why is it that my ?: not having any effect? (of course my understanding could very well be wrong)
842
capturing in regexps means indicating that you're interested not only in matching (which is finding strings of characters that match your regular expression), but you're also interested in using specific parts of the matched string later on.
The Difference Between \s and \s+ For example, expression X+ matches one or more X characters. Therefore, the regular expression \s matches a single whitespace character, while \s+ will match one or more whitespace characters.
Groups group multiple patterns as a whole, and capturing groups provide extra submatch information when using a regular expression pattern to match against a string. Backreferences refer to a previously captured group in the same regular expression.
(?:...) still matches, it just doesn't create a new group for purposes of \1/$1/.groups(1)/etc.
135
Your understanding is wrong. The group will still be part of the main capture, but it won't count as a sub-expression capture. The following would return an array of two matches:
x = "abcdefg"
x = x.match(/ab(cd)ef/)
Array index 0 would be "abcdef" (the complete match) and array index 1 would be "cd", the sub-expression capture. Adding the ?: tells the regex not to care about capturing the sub-expression, the full match is still fully captured.
From your other comments, there are a number of ways you could do what you're trying to do. For instance:
x.replace(/(ab)cd(ef)/, "$1$2");
x.slice(0, x.indexOf("cd")) + x.slice(x.indexOf("cd") + 2);
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