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I was taking a look at this @Sergio response here at stackoverlow:
There was an excerpt of code like this:
/^(\d)\1+$/.test(111)
What exactly does the \1 in that regular expression?
2 answers
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If you have one or more capture groups in the regex \1 fetch the value captured in the first group. As a variable that takes the value of the group. If there is more than one you can use \2, etc..
For example:
/^(\w)\1(\d)\2$/
accepts strings with two equal letters followed by two numbers equal to two.
For example aa11 or bb33 (example).
/^(\w)(\d)\1\2$/
is similar to the example above but accepts interspersed patterns.
For example a1a1 or b3b3 (example).
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It is the result of the first group that is representing the ( ), then number bar (\1) get its value. Other languages use dollar sign, e.g.: $1.
Regular expressions have utilities on the day, say you have a text file full of Inserts where dates are in format dd/mm/yyyy with a suitable editor it is possible to convert to yyyy-mm-dd replace.
23/10/2015
Utilize ([0-9]{2})\/([0-9]{2})/([0-9]{4}) to cover the date and \3\2\1 this will change the position of the year by the day, then just change the tab of \ for - and date will stay 2015-10-23.
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The interesting thing is that usually the
$1is used "on the other side" ofregexp. I mean, after capture. Equalpreg_replace('/(\w+)/, 'Peguei isso $1', 'Aqui')
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are called
retrovisores, because they use something that has already been captured.– Guilherme Lautert
Just to finish solving the question: This only exists in javascript?
– Wallace Maxters
@Wallacemaxters this exists in almost all programming languages. Apparently RE2 does not support, but most support. Take a look here, in the column "Backreference".
– Sergio