Using regexp in Java to verify that a string contains a pattern

Question

Simple question. If I need to check if a Regexp contains a simple pattern, in my case it would be ^0039, if I write something like this:

    if(Pattern.matches("^(0039|\\+39)", "00392121")) System.out.println("Yes");
    else System.out.println("No");

I have obviously No as answer because the pattern is missing the 2121. Do I need to transform the pattern in ^(0039|\\+39).* or is there a method more suitable for this?

I'm asking because I'm writing a method for our ETL engine and I'm not sure that everyone who's gonna use this is aware on how regexp works...so I'm foreseeing a lot of email of complaing with an obvious answer: "append .* to your regexp"*

Bohemian · Accepted Answer · 2012-03-06 19:35:21Z

2

I would advise you to use the Matcher.find method:

public static void main(String[] args) {
    Matcher m = Pattern.compile("^(0039|\\+39)").matcher("00392121");

    if (m.find())
        System.out.println("Found");
}

Note: This will only find the first occurrence of the pattern since it starts with ^.

Also note that Pattern.matches(regex, input) boils down to:

Pattern.compile(regex).matcher(input).matches()

Which calls Matcher.matches. And according to the API it matches against the whole input:

The matches method attempts to match the entire input sequence against the pattern.

edited Mar 6, 2012 at 19:35

Bohemian♦

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answered Mar 6, 2012 at 11:27

dacwe

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Bohemian · Accepted Answer · 2012-03-06 11:21:57Z

1

The String.matches() method returns true if the regex matches the whole String.

So yes, you have to add .* to the end.

answered Mar 6, 2012 at 11:21

Bohemian♦

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Comments

Oleg Mikheev · Accepted Answer · 2012-03-06 11:34:57Z

0

For this particular case I would not use regexp but rather use String's startsWith(String prefix, int toffset) instead.

Especially if you're not sure that everyone who's gonna use this will be aware of how regexp works.

answered Mar 6, 2012 at 11:34

Oleg Mikheev

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