Question

I am new to Java. Can somebody help me?

Is there any method available in Java which escapes the special characters in the below regex automatically?

Before escaping ***(.*) and after escaping \\*\\*\\*(.*)

I don’t want to escape (.*) here.

Answer 1

On the face of it, Pattern.quote appears to do the job.

However, looking at the detail of your question, it appears that you want / expect to be able to escape some meta-characters and not others. Pattern.quote won’t do that if you apply it to a single string. Rather, it will quote each and every character. (For the record, it doesn’t use backslashes. It uses “\E” and “\Q”.\ which neatly avoids the cost of parsing the string to find characters that need escaping.)

But the real problem is that you haven’t said how the quoter should decide which meta-characters to escape and which ones to leave intact. For instance, how does it know to escape the first three ‘‘ characters, but not the “.“?

Without a clearer specification, your question is pretty much unanswerable. And even with a specification, there is little chance of finding an easy way to do this.

IMO, a better approach would be to do the escaping before you assemble the pattern from its component parts … assuming that’s what is going on here.