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 should be able to use non-greedy quantifiers, specifically *?. You're going to probably want the following:

Pattern MY_PATTERN = Pattern.compile("\\[(.*?)\\]");

This will give you a pattern that will match your string and put the text within the square brackets in the first group. Have a look at the Pattern API Documentation for more information.

To extract the string, you could use something like the following:

Matcher m = MY_PATTERN.matcher("FOO[BAR]");
while (m.find()) {
    String s = m.group(1);
    // s now contains "BAR"
}

You can achieve it using regex. With the regex (US\\d+) you will get the group that match "US" followed by an integer of arbitrary length (minimium of 1)

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

...

Matcher matcher = Pattern.compile("(US\\d+)").matcher(str1);
if (matcher.find()) {// if it matched the pattern
    String result = matcher.group(0);// the group captured by the regex
}

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);
}