How to extract a substring using regex

I have a string that has two single quotes in it, the ' character. In between the single quotes is the data I want.

How can I write a regex to extract "the data i want" from the following text?

mydata = "some string with 'the data i want' inside";

Assuming you want the part between single quotes, use this regular expression with a Matcher:

"'(.*?)'"

Example:

String mydata = "some string with 'the data i want' inside";
Pattern pattern = Pattern.compile("'(.*?)'");
Matcher matcher = pattern.matcher(mydata);
if (matcher.find())
{
    System.out.println(matcher.group(1));
}

Result:

the data i want

You don't need regex for this.

Add apache commons lang to your project (http://commons.apache.org/proper/commons-lang/), then use:

String dataYouWant = StringUtils.substringBetween(mydata, "'");

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

public class Test {
	public static void main(String[] args) {
		Pattern pattern = Pattern.compile(".*'([^']*)'.*");
		String mydata = "some string with 'the data i want' inside";

		Matcher matcher = pattern.matcher(mydata);
		if(matcher.matches()) {
			System.out.println(matcher.group(1));
		}

	}
}

There's a simple one-liner for this:

String target = myData.replaceAll("[^']*(?:'(.*?)')?.*", "$1");

By making the matching group optional, this also caters for quotes not being found by returning a blank in that case.

See live demo.

Because you also ticked Scala, a solution without regex which easily deals with multiple quoted strings:

val text = "some string with 'the data i want' inside 'and even more data'"
text.split("'").zipWithIndex.filter(_._2 % 2 != 0).map(_._1)

res: Array[java.lang.String] = Array(the data i want, and even more data)

Since Java 9

As of this version, you can use a new method Matcher::results with no args that is able to comfortably return Stream<MatchResult> where MatchResult represents the result of a match operation and offers to read matched groups and more (this class is known since Java 1.5).

String string = "Some string with 'the data I want' inside and 'another data I want'.";

Pattern pattern = Pattern.compile("'(.*?)'");
pattern.matcher(string)
       .results()                       // Stream<MatchResult>
       .map(mr -> mr.group(1))          // Stream<String> - the 1st group of each result
	   .forEach(System.out::println);   // print them out (or process in other way...)

The code snippet above results in:

> > the data I want > another data I want >

The biggest advantage is in the ease of usage when one or more results is available compared to the procedural if (matcher.find()) and while (matcher.find()) checks and processing.

String dataIWant = mydata.replaceFirst(".*'(.*?)'.*", "$1");

as in javascript:

mydata.match(/'([^']+)'/)[1]

the actual regexp is: /'([^']+)'/

if you use the non greedy modifier (as per another post) it's like this:

mydata.match(/'(.*?)'/)[1]

it is cleaner.

String dataIWant = mydata.split("'")[1];

See Live Demo

In Scala,

val ticks = "'([^']*)'".r

ticks findFirstIn mydata match {
    case Some(ticks(inside)) => println(inside)
    case _ => println("nothing")
}

for (ticks(inside) <- ticks findAllIn mydata) println(inside) // multiple matches

val Some(ticks(inside)) = ticks findFirstIn mydata // may throw exception

val ticks = ".*'([^']*)'.*".r    
val ticks(inside) = mydata // safe, shorter, only gets the first set of ticks

Apache Commons Lang provides a host of helper utilities for the java.lang API, most notably String manipulation methods. In your case, the start and end substrings are the same, so just call the following function.

> StringUtils.substringBetween(String str, String tag) > > Gets the String that is nested in between two instances of the same > String.

If the start and the end substrings are different then use the following overloaded method.

> StringUtils.substringBetween(String str, String open, String close) > > Gets the String that is nested in between two Strings.

If you want all instances of the matching substrings, then use,

> StringUtils.substringsBetween(String str, String open, String close) > > Searches a String for substrings delimited by a start and end tag, > returning all matching substrings in an array.

For the example in question to get all instances of the matching substring

String[] results = StringUtils.substringsBetween(mydata, "'", "'");

you can use this i use while loop to store all matches substring in the array if you use

if (matcher.find()) { System.out.println(matcher.group(1)); }

you will get on matches substring so you can use this to get all matches substring

Matcher m = Pattern.compile("[a-zA-Z0-9_.+-]+@[a-zA-Z0-9-]+\\.[a-zA-Z0-9-.]+").matcher(text);
   // Matcher  mat = pattern.matcher(text);
    ArrayList<String>matchesEmail = new ArrayList<>();
        while (m.find()){
            String s = m.group();
            if(!matchesEmail.contains(s))
                matchesEmail.add(s);
        }

    Log.d(TAG, "emails: "+matchesEmail);

add apache.commons dependency on your pom.xml

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-io</artifactId>
    <version>1.3.2</version>
</dependency>

And below code works.

StringUtils.substringBetween(String mydata, String "'", String "'")

Some how the group(1) didnt work for me. I used group(0) to find the url version.

Pattern urlVersionPattern = Pattern.compile("\\/v[0-9][a-z]{0,1}\\/");
Matcher m = urlVersionPattern.matcher(url);
if (m.find()) {	
	return StringUtils.substringBetween(m.group(0), "/", "/");
}
return "v0";