Match text lines using Regex

Question

I have lines like this :

36600.10: [Host #255] utilization is 0.00%
36600.10: [Host #256] utilization is 21.64%
36600.10: [Host #257] utilization is 3.29%
36600.10: [Host #258] utilization is 0.94%
36600.10: [Host #260] utilization is 3.76%
36600.10: [Host #260] utilization is 1.21%
36600.10: [Host #260] utilization is 86.09%
36600.10: [Host #260] utilization is 7.32%

I need to get all numbers after utilization is. What I want is an array like this :

myArray[0] => 0.00,
myArray[1] => 21.64,
myArray[2] => 3.29,
myArray[3] => 0.94,
myArray[4] => 3.76,
myArray[5] => 1.21,
myArray[6] => 7.32

What I tried so far (it just works for first line) :

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class main {

    public static void main(String[] args) {
          String lines = "36600.10: [Host #256] utilization is 21.65% \n 36600.10: [Host #256] utilization is 91.78% \n 36600.10: [Host #256] utilization is 3.29%";
          String pattern = "(utilization is\\s)(\\d+\\.\\d*)(.*)";

          Pattern r = Pattern.compile(pattern);

          Matcher m = r.matcher(lines);
          if (m.find( )) {
             System.out.println(m.group(2));
          } else {
             System.out.println("NO MATCH");
          }

    }

}

Sorry I'm new in java and tried some patterns but not helped. Any helps would be appreciated.

The regex is correct. All you need to do is capture the group for numbers. Although I prefer using `\\s*` instead of hard coding spaces.. — Identity1, Jan 01 '16 at 11:21
Could you post the code using this pattern? The regex is fine. — Tunaki, Jan 01 '16 at 11:23
This is what you need http://stackoverflow.com/questions/6020384/create-array-of-regex-matches — Identity1, Jan 01 '16 at 11:27
you need only answer only by regular expression or any approach will work?? — Narendra Jaggi, Jan 01 '16 at 11:33

score 2 · Accepted Answer · answered Jan 01 '16 at 11:33

Use While loop. If statement will limit the match with the first occurence (Which is happening)

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class main {

    public static void main(String[] args) {
          String lines = "36600.10: [Host #256] utilization is 21.65% \n 36600.10: [Host #256] utilization is 91.78% \n 36600.10: [Host #256] utilization is 3.29%";
          String pattern = "(utilization is\\s)(\\d+\\.\\d*)(.*)";

          Pattern r = Pattern.compile(pattern);

          Matcher m = r.matcher(lines);
         while (m.find()) {
             System.out.println(m.group(2));
          } 
    }

}

Good answer. Thank you – Hamed Kamrava Jan 01 '16 at 11:36 — Hamed Kamrava, Jan 01 '16 at 11:36

Paul Vargas · Answer 2 · 2016-01-01T12:12:40.487

You can use more than one regular expression.* And... there are several solutions. One of them is to replace the redundant text by spaces. Then use these spaces to split the string. That is to say:

• Code:

String[] array = lines.replaceAll("(?m).*is\\s+|%\\s*", " ").trim().split("\\s+");
System.out.println(Arrays.toString(array));

• Output:

[0.00, 21.64, 3.29, 0.94, 3.76, 1.21, 86.09, 7.32]

_{────────────

* By the way, Regular-Expressions.info - Regex Tutorial, Examples and Reference - Regexp Patterns is a good place to start to learn regular expressions. Mastering Regular Expressions, 3rd Edition is a highly recommended book.}

score 0 · Answer 3 · answered Jan 01 '16 at 11:39

You can also do this using split function if you want to -

String lines = "36600.10: [Host #256] utilization is 21.65% \n 36600.10: [Host #256] utilization is 91.78% \n 36600.10: [Host #256] utilization is 3.29%";
        String lineArray[] = lines.split("\n");
        for(String temp:lineArray){
            String percentWithSign = temp.split("utilization is")[1];
            String percentStr  = percentWithSign.replace("%", "");
            Float percentFloat = Float.parseFloat(percentStr);
            System.out.println(percentFloat);
        }

score 0 · Answer 4 · answered Jan 01 '16 at 11:44

It can be solved without regular expression also at least in this case after seeing the pattern of your input.

List<String> list = new ArrayList<String>();
        list.add("36600.10: [Host #255] utilization is 0.00%");
        list.add("36600.10: [Host #256] utilization is 21.64%");
        list.add("36600.10: [Host #255] utilization is 3.29%");
        for(int i=0 ; i<list.size() ; i++){
            String str = list.get(i);
            int start = str.lastIndexOf(" ");
            String temp = str.substring(start+1, str.length()-1);
            System.out.println(temp);
        }

Match text lines using Regex

4 Answers4