Java正则表达式，点匹配

Question

I have a string in the form of:

{something here}{something here}{something here}

etc In summary there are brackets with text or anything inside (any character). I want to split it into an array (using Java's String.split(regex) function). The regex I used is \\\\{.*\\\\} but it's not working.

Any ideas?

Answer 1

public static void main(String[] args) {
    String input="{something here}{something here}{something here}";
    String[] parts=input.substring(1,input.length()-1).split("\\}\\{");
    for(String s:parts)
        System.out.println(s);
}

output:

something here
something here
something here

Answer 2

In regex, the * is greedy, meaning that it will consume as many characters as possible. This means that the regex: \\{.*\\} will match the entire string, as the .* will match "something here}{something here}{something here". putting a ? after a * will make it behave in an un-greedy fashion, meaning that it will only consume characters up until the next expression can match. therefore, try \\{.*?\\} as your regex instead

Answer 3

The argument in the split() method specifies the separator between the parts, not the parts themselves.

As suggested by Sotirios Delimanolis, you can achieve your goal by repeatedly matching a pattern. The example code below gets the the text inside the braces.

    String val = "{alpha}{beta}{delta\nepsilon}";
    Pattern pattern = Pattern.compile( "\\{(.*?)\\}", Pattern.DOTALL);
    Matcher matcher = pattern.matcher( val );
    while ( matcher.find() ) {
        String part = matcher.group(1);
        System.out.print( String.format("%s,", part));
    }

The .*? expression provides a reluctant match, which prefers to match as few characters from the string as possible. If you just use .* , that's a greedy match - the first match will be the entire string.

You mentioned in a comment on a deleted answer that your "something here" strings can contain new lines. For the '.' to match newlines, you need to use the Pattern.DOTALL flag, as shown above.

The call to matcher.group(1) gives you the text matching the capturing group (.*?) . If you wanted the braces included, you could omit the parantheses and simply call matcher.group(0) to get the entire match.

Answer 4

You know the quote about regexps ... Well, it doesn't help.

What does help, however, are (like usual) tests. And regular expressions can be tested online using websites such as RegExp Planet

So, when using as test string {a}{long}{text with spaces} I can find as a "good" regexp the following \\{([^\\}]*)\\} . And, to quote the source site

Regular Expression {([^}] )}
as a Java string "\\{([^\\}] )\\}"

Cause one shouldn't forget to add the double backslashes