Java Regex replace all

Question

My text will look like this

| birth_date = {{birth date|1925|09|2|df=y}}
| birth_place = [[Bristol]], [[England]], UK
| death_date = {{death date and age|2000|11|16|1925|09|02|df=y}}
| death_place = [[Eastbourne]], [[Sussex]], England, UK
| origin = 
| instrument = [[Piano]]
| genre = 
| occupation = [[Musician]]

I would like to get everything that is inside of [[ ]]. I tried to use replace all to replace everything that is not inside the [[ ]] and then use split by new line to get a list of text with [[ ]].

input = input.replaceAll("^[\\[\\[(.+)\\]\\]]", "");

Required output:

[[Bristol]]
[[England]]
[[Eastbourne]]
[[Sussex]]
[[Piano]]
[[Musician]]

But this is not giving the desired output. What am I missing here?. There are thousands of documents and is this the fastest way to get it? If no, do tell me the optimum way to get the desired output.

• ajb: In addition to other problems, please note that (.+) is a "greedy" quantifier that will grab as many characters as it can between [[ and ]], meaning that for birth_place you'll get "Bristol]], [[England" as one of the matches. Adding ? after .+, as in falsetru's answer, prevents this.

Answer 1

You need to match it not replace

Matcher m=Pattern.compile("\\[\\[\\w+\\]\\]").matcher(input);
while(m.find())
{
 m.group();//result
}

Answer 2

Use Matcher.find. For example:

import java.util.regex.*;
...
String text =
 "| birth_date = {{birth date|1925|09|2|df=y}}\n" +
 "| birth_place = [[Bristol]], [[England]], UK\n" +
 "| death_date = {{death date and age|2000|11|16|1925|09|02|df=y}}\n" +
 "| death_place = [[Eastbourne]], [[Sussex]], England, UK\n" +
 "| origin = \n" +
 "| instrument = [[Piano]]\n" +
 "| genre = \n" +
 "| occupation = [[Musician]]\n";
Pattern pattern = Pattern.compile("\\[\\[.+?\\]\\]");
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
 System.out.println(matcher.group());
}

Answer 3

Just for fun, using replaceAll:

 String output = input.replaceAll("(?s)(\\]\\]|^).*?(\\[\\[|$)", "1ドル\n2ドル");