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

Try

Pattern p = Pattern.compile("#([^,]*)");
Matcher m = p.matcher(myString);
if (m.find()) {
  doSomethingWith(m.group(1));  // The matched substring
}

public static void main(String[] args) {
    String abcd1 = "/abcd/";
    String abcd2 = "/abcd";
    String abcd3 = "abcd/";
    String abcd4 = "abcd";
    System.out.println(abcd1.replaceAll("(^/)?(/$)?", ""));
    System.out.println(abcd2.replaceAll("(^/)?(/$)?", ""));
    System.out.println(abcd3.replaceAll("(^/)?(/$)?", ""));
    System.out.println(abcd4.replaceAll("(^/)?(/$)?", ""));
}

Will work.

Matches the first (^/)? means match 0 or 1 '/' at the beginning of the string, and (/$)? means match 0 or 1 '/' at the end of the string.

Make the regex "(^/*)?(/*$)?" to support matching multiple '/':

String abcd5 = "//abcd///";
System.out.println(abcd1.replaceAll("(^/*)?(/*$)?", ""));

Start by reading the javadocs:

Pattern

Matcher

You can use negative look behind to get file name

((?:.(?<!/))+)\"

and below regex to get full path

/(.*)\"

Sample code

public static void main(String[] args) {
    Pattern pattern = Pattern.compile("/(.*)\"");
    Pattern pattern1 = Pattern.compile("((?:.(?<!/))+)\"");
    String matchString = "/slm/attachment/63338424306/Note.jpg\"xxxxxxxx";
    Matcher matcher = pattern.matcher(matchString);
    String fullString = "";
    while (matcher.find()) {
        fullString = matcher.group(1);
    }
    matcher = pattern1.matcher(matchString);
    String fileName = "";
    while (matcher.find()) {
        fileName = matcher.group(1);
    }

    System.out.println(fullString + " " + fileName);
}

Declare a temp variable before the if condition and then append the matched string to that temp variable.

String temp = "";
Pattern pattern = Pattern.compile("Gold(\\s{9})(\\d{4})\\.(\\d{2})");
Matcher matcher = pattern.matcher(str);

if (matcher.find())
{
    temp = temp + matcher.group();
}
else {
    System.out.println("No string found");
}

You can try using the simple regex:

image/\w+

String str = "data:image/png;base64,iVBO....";
Matcher m = Pattern.compile("image/\\w+").matcher(str);

while (m.find())
    System.out.println(m.group());

image/png

If you want the contents of \w+ (which I imagine might be the case), you can place it in a capturing group:

image/(\w+)

and call m.group(1).

Full example:

private static final Pattern p = Pattern.compile("^([a-zA-Z]+)([0-9]+)(.*)");
public static void main(String[] args) {
    // create matcher for pattern p and given string
    Matcher m = p.matcher("Testing123Testing");

    // if an occurrence if a pattern was found in a given string...
    if (m.find()) {
        // ...then you can use group() methods.
        System.out.println(m.group(0)); // whole matched expression
        System.out.println(m.group(1)); // first expression from round brackets (Testing)
        System.out.println(m.group(2)); // second one (123)
        System.out.println(m.group(3)); // third one (Testing)
    }
}

Since you're looking for the first number, you can use such regexp:

^\D+(\d+).*

and m.group(1) will return you the first number. Note that signed numbers can contain a minus sign:

^\D+(-?\d+).*