Replacing a character sequence

Hi,
I am writing a program for a client which involves tokenizing strings based on a delimiter, usually "*" but it could be anything. An example string would be GH*ASD*FD**D2*f**123*45;

The problem I am facing is I would need to extract every element between the "*"s and in the case where there are two "*" as above (FD**D2 and f**123) I would need to be able to determine that there is an empty element.
I thought I could do it by replacing the ** with *!* or sthg to that effect. But using String.replace I am getting Pattern Error.

My code is :
String checkit = "GS*sf**a";
String elementDelimiter = ""+checkit.charAt(2);
checkit = checkit.replaceAll(elementDelimiter+elementDelimiter,elementDelimiter+"!"+elementDelimiter);
System.out.println(checkit);

I am getting the following error:
java.util.regex.PatternSyntaxException: Dangling meta character '*' near index 0
**
^
at java.util.regex.Pattern.error(Pattern.java:1542)
at java.util.regex.Pattern.sequence(Pattern.java:1659)
at java.util.regex.Pattern.expr(Pattern.java:1559)
at java.util.regex.Pattern.compile(Pattern.java:1293)
at java.util.regex.Pattern.<init>(Pattern.java:1049)
at java.util.regex.Pattern.compile(Pattern.java:793)
at java.lang.String.replaceAll(String.java:2038)
at com.h2h.core.ComplianceCheckerBean.main(ComplianceCheckerBean.java:23)

PLEASE HELP!!

Thanks
Karthik

The * in regular expressions doesn't mean *, it means any number of repetitions including 0.

The * character along with + ? - . and quite a lot of others is called a meta-character, so you have to escape it by passing \* instead. If you want \* to appear in a Java String literal you have to escape the \ too, so you would write "\\*".

Hope that helps.

Look in the Java Tutorial for more information.

There's no need to pre-process the string anyway. Instead of StringTokenizer, use split() to tokenize it. See the API doc for the split() method for an explanation of the -1 parameter.

Thanks a lot guys.

Karthik