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How to replace character *, using .replaceAll

eileen keeney
Ranch Hand

Joined: May 04, 2009
Posts: 51
ok, I am trying to replace the character "*", with " [A-Za-z0-9]*"

String oldStr = "XYZ*";
String newStr = oldStr.replaceAll("\*", "[A-Za-z0-9]*");
System.out.println("newStr=" + newStr + "\n");

This one gives me illegal escape character

String oldStr = "XYZ*";
String newStr = oldStr.replaceAll("*", "[A-Za-z0-9]*");
System.out.println("newStr=" + newStr + "\n");

This one compiles, but then gives me a dangling meta character error when I run it.

Trying to replace another character, such as Y, works fine doing this:
String oldStr = "XYZ*";
String newStr = oldStr.replaceAll("Y", "[A-Za-z0-9]*");
System.out.println("newStr=" + newStr + "\n");

That compiles and runs and works. So my basic syntax (for replacing non meta characters) is correct here.

So how do I take a string, and replace every occurrence of * with A-Za-z0-9]*
If I do not escape it, it is read as a meta character.
If I do escape it, it tells me it is an illegal use of an escape character.


eileen keeney
Ranch Hand

Joined: May 04, 2009
Posts: 51
I found the answer in another forum,
So in case anyone else stumbles on this doing a search, the answer is to use double escape characters.

This works:
String oldStr = "XYZ*";
String newStr = oldStr.replaceAll("\\*", "[A-Za-z0-9]*");
System.out.println("newStr=" + newStr + "\n");

Campbell Ritchie

Joined: Oct 13, 2005
Posts: 38045
Welcome to JavaRanch

You need to use \\ instead of \.

The reason is that the String literal turns the \\ into a single \ before it analyses the regular expression so you get a proper \*. You will occasionally have to use \\\\ in regular expressions, or even \\\\\\\\
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