I am wondering if there is a better to represent a fix amount of repeats in a regular expression. For example, if I just want to match exactly 14 letters/digits, I am using ^\w\w\w\w\w\w\w\w\w\w\w\w\w\w$
which will match a word like UNL075BE499135
and not match UNL075BE499135AAA
is there a handy way to do it? In am currently doing it in java but I guess this may apply to other language as well. Thanks in advance.
The finite repetition syntax uses {m,n}
in place of star/plus/question mark.
From java.util.regex.Pattern
:
X{n} X, exactly n times
X{n,} X, at least n times
X{n,m} X, at least n but not more than m times
All repetition metacharacter have the same precedence, so just like you may need grouping for *
, +
, and ?
, you may also for {n,m}
.
ha*
matches e.g. "haaaaaaaa"
ha{3}
matches only "haaa"
(ha)*
matches e.g. "hahahahaha"
(ha){3}
matches only "hahaha"
Also, just like *
, +
, and ?
, you can add the ?
and +
reluctant and possessive repetition modifiers respectively.
System.out.println(
"xxxxx".replaceAll("x{2,3}", "[x]")
); "[x][x]"
System.out.println(
"xxxxx".replaceAll("x{2,3}?", "[x]")
); "[x][x]x"
Essentially anywhere a *
is a repetition metacharacter for "zero-or-more", you can use {...}
repetition construct. Note that it's not true the other way around: you can use finite repetition in a lookbehind, but you can't use *
because Java doesn't officially support infinite-length lookbehind.
.*
and .*?
for regexregex{n,}?
== regex{n}
?a{1}b{0,1}
instead of ab?
In Java create the pattern with Pattern p = Pattern.compile("^\\w{14}$");
for further information see the javadoc
^\w{14}$
in Perl and any Perl-style regex.
If you want to learn more about regular expressions - or just need a handy reference - the Wikipedia Entry on Regular Expressions is actually pretty good.
Source: Stackoverflow.com