如何在 Java 中使用 Pattern 拆分字符串

Question

I want to split a string using methods in Pattern,here is what i did

String s = "[[[0.093493,51.6037],[0.091015,51.5956]]]"
Pattern branchPattern = Pattern.compile("[...]");
String[] split = branchPattern.split(s);

I want to get the result that String[] contains [0.093493,51.6037] and [0.091015,51.5956]. But the result of this code is not the result i want. How can i split this string? Or is there any way like use matcher to split this string in the format i want?

Answer 1

Just do matching instead of splitting.

Pattern p = Pattern.compile("\\[[^\\[\\]]+\\]");
Matcher m = p.matcher(s);
while(m.find())
{
System.out.println(m.group());
}

DEMO

or

string.replaceAll("^\\[{2}|\\]{2}$", "").split(",(?=\\[)");

or

string.replaceAll("^\\[+|\\]+$", "").split("\\],\\[)");

Answer 2

I am trying to answer with another perspective. Have a look.Suppose you have placed parameters in between two special charaters like : #parameter# or parameter or even two differnt signs at a time like *paramter#. We can have list of all these parameters between those signs by this code :

import java.util.ArrayList;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import org.apache.commons.lang.StringUtils;

public class Splitter {

    public static void main(String[] args) {

        String pattern1 = "#";
        String pattern2 = "#";
        String text = "(#n1_1#/#n2_2#)*2/#n1_1#*34/#n4_4#";

        Pattern p = Pattern.compile(Pattern.quote(pattern1) + "(.*?)" + Pattern.quote(pattern2));
        Matcher m = p.matcher(text);
        while (m.find()) {
            ArrayList parameters = new ArrayList<>();
            parameters.add(m.group(1));
            System.out.println(parameters);
            ArrayList result = new ArrayList<>();
            result.add(parameters);
            // System.out.println(result.size());
        }

    }
}

Here list result will have parameters n1_1,n2_2,n4_4.