regex to remove all text before a character

The regular expression:

^[^_]*_(.*)$

Then get the part between parenthesis. In perl:

my var = "3.04_somename.jpg";
$var =~ m/^[^_]*_(.*)$/;
my fileName = $1;

In Java:

String var = "3.04_somename.jpg";
String fileName = "";
Pattern pattern = Pattern.compile("^[^_]*_(.*)$");
Matcher matcher = pattern.matcher(var);
if (matcher.matches()) {
    fileName = matcher.group(1);
}

...

Variant of Tim's one, good only on some implementations of Regex: ^.*?_

var subjectString = "3.04_somename.jpg";
var resultString = Regex.Replace(subjectString,
    @"^   # Match start of string
    .*?   # Lazily match any character, trying to stop when the next condition becomes true
    _     # Match the underscore", "", RegexOptions.IgnorePatternWhitespace);

^[^_]*_

will match all text up to the first underscore. Replace that with the empty string.

For example, in C#:

resultString = Regex.Replace(subjectString, 
    @"^   # Match start of string
    [^_]* # Match 0 or more characters except underscore
    _     # Match the underscore", "", RegexOptions.IgnorePatternWhitespace);

For learning regexes, take a look at http://www.regular-expressions.info

no need to do a replacement. the regex will give you what u wanted directly:

"(?<=_)[^_]*\.jpg"

tested with grep:

 echo "3.04_somename.jpg"|grep -oP "(?<=_)[^_]*\.jpg"
somename.jpg