Regex pattern java with commas

Question

I have a below string which comes from an excel column

  "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\""

I would like to set regex pattern to retrieve the entire string,so that my result would be exactly like

"USE CODE ""Gef, sdf"" FROM 1/7/07"

Below is what I tried

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

public class RegexMatches
{
    public static void main( String args[] ){

      // String to be scanned to find the pattern.
      String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
      String line2 = "Test asda ds asd, tesat2 . test3";

      String dpattern = "(\"[^\"]*\")(?:,(\"[^\"]*\"))*,|([^,]+),";
      // Create a Pattern object
      Pattern d = Pattern.compile(dpattern);
      Matcher md = d.matcher(line2);

      Pattern r = Pattern.compile(dpattern);

      // Now create matcher object.
      Matcher m = r.matcher(line);
      if (m.find( )) {
         System.out.println("Found value: 0 " + m.group(0) );
       //  System.out.println("Found value: 1 " + m.group(1) );
         //System.out.println("Found value: 2 " + m.group(2) );
      } else {
         System.out.println("NO MATCH");
      }
   }
}

and the result out of it breaks after ,(comma) and hence the output is

Found value: 0 "USE CODE ""Gef,

It should be

Found value: 0 "USE CODE ""Gef sdf"" FROM 1/7/07",

and for the second line Matcher m = r.matcher(line2); the output should be

Found value: 0 "Test asda ds asd",

Note: In the real scenario,I cannot manipulate on the given string like removing the quotes and then apply regex. All I can do is to get it via Regex pattern .So this java is to test the regex pattern . — eshaa
– eshaa, Commented Jun 15, 2016 at 8:22
im not sure what your trying to do with the non capturing group ?: and the OR operator |, but it looks like you could just use (\"[^\"]*\"){3} and it would match 3 quoted groups. — ug_
– ug_, Commented Jun 15, 2016 at 8:22
Do you mean you need to get the double quoted substring only? Like "[^"]*(?:""[^"]*)*"? What is the expected output for the line1 and line 2? Maybe (?:"[^"]*(?:""[^"]*)*"|[^,])+ might do? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jun 15, 2016 at 8:25
Why not select simply everything between two \"? like \".*\". You can do a pre computation check of validity string assuring that \" is a pair number in order to avoid false positives. — morels
– morels, Commented Jun 15, 2016 at 8:25
See ideone.com/1AOvhZ - I just modified it to return the first match. Does it work for you? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jun 15, 2016 at 8:32

Wiktor Stribiżew · Accepted Answer · 2016-06-15 08:42:30Z

You may use

(?:"[^"]*(?:""[^"]*)*"|[^,])+

See the regex demo

Explanation:

" - leading quote
[^"]* - 0+ chars other than a double quote
(?:""[^"]*)* - 0+ sequences of a "" text followed with 0+ chars other than a double quote
" - trailing quote

OR:

[^,] - any char but a comma

And the whole pattern is matched 1 or more times as it is enclosed with (?:...)+ and + matches 1 or more occurrences.

IDEONE demo:

String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
String line2 = "Test asda ds asd, tesat2 . test3";
Pattern pattern = Pattern.compile("(?:\"[^\"]*(?:\"\"[^\"]*)*\"|[^,])+");
Matcher matcher = pattern.matcher(line);
if (matcher.find()){                        // if is used to get the 1st match only
    System.out.println(matcher.group(0)); 
}
Matcher matcher2 = pattern.matcher(line2); 
if (matcher2.find()){
    System.out.println(matcher2.group(0)); 
}

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