In Javascript ,why
/^(\d{1}){3}$/.exec(123)
returns ["123", "3"], but
/^(\d{1})$/.exec(123)
returns null rather than ["3"].
Also, why is the first expression returning 3, when 1 is the digit that follows ^ ?
1 Answer
First case
Noting that \d{1} is equivalent to just \d,
/^(\d{1}){3}$/
can be simplified to
/^(\d){3}$/
which means
- begin of the string
- match a three-digit string
- end of the string
The parenthesis around \d define a capture group. As explained here and here, when you repeat a capturing group, the usual implementation keeps only the last capture.
That's why the final result is
[
"123", // the whole matched string
"3", // the last captured group
]
Second case
/^(\d{1})$/
can again be simplified to
/^(\d)$/
which means
- begin of the string
- match a single digit
- end of the string
Being 123 a three-digit string, it's not matched by the regex, so the result is null.
answered Sep 28, 2013 at 5:43
Gabriele Petronella
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^and$anchors... Google it!{3}) and capture groups (why the first regex includes "3" in the results).