java regex match a string

Question

Here is a string

* '''A happy man is too satisfied with the present to dwell too much on the future.'''
** &quot;My Future Plans&quot; an essay written at age 17 for school exam (18 September 1896) ''The Collected Papers of Albert Einstein'' Vol. 1 (1987) Doc.22

I need to get the text which starts with only one * till the end of that line

Pattern pattern = Pattern.compile("\\*(.+?)\\n");
    Matcher matcher = pattern.matcher(text);
    List<String> s = new ArrayList<String>();
    while (matcher.find()) {
        s.add(matcher.group(1));
    }

I tried the above code, but it matches even 2 * and also the text from * to new line is not coming up properly, what am I missing here?

Casimir et Hippolyte · Accepted Answer · 2013-11-16 00:15:21Z

3

You can try this:

Pattern pattern = Pattern.compile("(?m)^\\*([^*].*)");

(?m) set the pattern in multiline mode where ^ means "begining of the line"

Adding \\n at the end is useless since the dot can't match a newline. (it's the reason why a lazy quantifier is useless too)

edited Nov 16, 2013 at 0:15

answered Nov 16, 2013 at 0:10

Casimir et Hippolyte

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2 Comments

Casimir et Hippolyte Over a year ago

@Naren: The quantifiers * and + are greedy by default and take all possible characters. Since . doesn't match a newline (or the end of the string) .* will stop automaticaly before the new line.

Casimir et Hippolyte Over a year ago

@Naren: take a look at the site: www.regular-expressions.info to have an explanation about lazy/greedy quantifiers

Naveen · Accepted Answer · 2013-11-16 00:20:40Z

1

which will make some sense i think,

Pattern pattern = Pattern.compile("^(\\*[^*].+)");

answered Nov 16, 2013 at 0:20

Naveen

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