How to represent a fix number of repeats in regular expression?

Question

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.

Answer 1

For Java:

Quantifiers documentation

X, exactly n times: X{n}
X, at least n times: X{n,}
X, at least n but not more than m times: X{n,m}

Answer 2

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.

References

• regular-expressions.info/Repetition

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