How to escape all regex special chars but not all at once (by Pattern.quote()), just one-by one

Question

Here's the problem: user is presented with a text field, unto which may he or she type the filter. A filter, to filter unfiltered data. User, experiencing an Oracle Forms brainwashing, excpets no special characters other than %, which I guess more or less stands for ".*" regex in Java.

If the User Person is well behaved, given person will type stuff like "CTHULH%", in which case I may construct a pattern:

Pattern.compile(inputText.replaceAll("%", ".*"));

But if the User Person hails from Innsmouth, insurmountably will he type ".+\[a-#$%^&*(" destroying my scheme with a few simple keystrokes. This will not work:

Pattern.compile(Pattern.quote(inputText).replaceAll("%", ".*"));

as it will put \Q at the beginning and \E at the end of the String, rendereing my % -> .* switch moot.

The question is: do I have to look up every special character in the Pattern code and escape it myself by adding "\\" in front, or can this be done automagically? Or am I so deep into the problem, I'm omitting some obvious way of resolution?

Answer 1

I think this algorithm should work for you:

• Split on %
• Quote each part separately using Pattern.quote
• Join the string using .*