提取Java中的子字符串

Question

I have the following string:

"hello this.is.a.test(MainActivity.java:47)"

and I want to be able to extract the MainActivity.java:47 (everything that is inside '(' and ')' and only the first occurance).

I tried with regex but it seems that I am doing something wrong.

Thanks

Answer 1

You can do it yourself:

int pos1 = str.indexOf('(') + 1;
int pos2 = str.indexOf(')', pos1);

String result = str.substring(pos1, pos2)

Or you can use commons-lang which contains a very nice StringUtils class that has substringBetween()

Answer 2

I think Regex is a liitle bit an overkill. I would use something like this:

String input = "hello this.is.a.test(MainActivity.java:47)";
String output = input.subString(input.lastIndexOf("(") + 1, input.lastIndexOf(")"));

Answer 3

This should work:

^[^\\(]*\\(([^\\)]+)\\)

The result is in the first group.

Answer 4

Another answer for your question :

String str = "hello this.is.a.test(MainActivity.java:47) another.test(MyClass.java:12)";
Pattern p = Pattern.compile("[a-z][\\w]+\\.java:\\d+", Pattern.CASE_INSENSITIVE);
Matcher m=p.matcher(str);

if(m.find()) {
    System.out.println(m.group());
}

The RegExp explained :

[az][\\w]+\\.java:\\d+

[az] > Check that we start with a letter ...
[\\w]+ > ... followed by a letter, a digit or an underscore...
\\.java: > ... followed exactly by the string ".java:"...
\\d+ > ... ending by one or more digit(s)

Answer 5

Pseudo-code:

int p1 = location of '('
int p2 = location of ')', starting the search from p1
String s = extract string from p1 to p2

String.indexOf() and String.substring() are your friends.

Answer 6

Try this:

String input = "hello this.is.a.test(MainActivity.java:47) (and some more text)";
Pattern p = Pattern.compile("[^\\)]*\\(([^\\)]*)\\).*");
Matcher m = p.matcher( input );
if(m.matches()) {
  System.out.println(m.group( 1 )); //output: MainActivity.java:47
}

This also finds the first occurence of text between ( and ) if there are more of them.

Note that in Java you normally have the expressions wrapped with ^ and $ implicitly (or at least the same effect), ie the regex must match the entire input string. Thus [^\\\\)]* at the beginning and .* at the end are necessary.