Matching whole words with special characters with a dynamically built pattern

Question

I need to match an exact substring in a string in Java. I've tried with

String pattern = "\\b"+subItem+"\\b";

But it doesn't work if my substring contains non alphanumerical characters. I want this to work exactly as the "Match whole word only" function in Notepad++. Could you help?

What can `subItem` look like when "it does not work"? What do you consider a word? — Wiktor Stribiżew, Jun 21 '21 at 12:02
Any combination of alphanumerical and non alphanumerical characters. Like: Data[3] or pst->numberOfTx. — Robert Ilin, Jun 21 '21 at 12:04
On this string: Data[3]=(uint8)Data[1]; It doesn't work for searching Data[3] — Robert Ilin, Jun 21 '21 at 12:08
Ok, if the words with special chars at the start/end are not to be glued to word chars, try `String pattern = "(? — Wiktor Stribiżew, Jun 21 '21 at 12:09
My guess is your `subItem` content isn't escaped. So when it is `Data[3]`, it looks for `Data3` instead of `Data[3]`, because `[` and `]` are special characters in Regex. — Nino DELCEY, Jun 21 '21 at 12:15
Does this answer your question? [How to escape text for regular expression in Java](https://stackoverflow.com/questions/60160/how-to-escape-text-for-regular-expression-in-java) — tevemadar, Jun 21 '21 at 12:16
It might be more complex than just replacing `subItem` with `Pattern.quote(subItem)` — Wiktor Stribiżew, Jun 21 '21 at 12:24
@WiktorStribiżew what's the difference between you last comment and the first one? Because the first one seems to work fine. I mean `String pattern = "(? — Robert Ilin, Jun 21 '21 at 12:57

Wiktor Stribiżew · Accepted Answer · 2021-06-23T08:35:29.920

I suggest either unambigous word boundaries (that match a string only if the search pattern is not enclosed with letters, digits or underscores):

String pattern = "(?<!\\w)"+Pattern.quote(subItem)+"(?!\\w)";

where (?<!\w) matches a location not preceded with a word char and (?!\w) fails if there is no word char immediately after the current position (see this regex demo), or, you can use a variation that takes into account leading/trailing special chars of the potential match:

String pattern = "(?:\\B(?!\\w)|\\b(?=\\w))" + Pattern.quote(subword) + "(?:(?<=\\w)\\b|(?<!\\w)\\B)";

See the regex demo.

Details:

(?:\B(?!\w)|\b(?=\w)) - either a non-word boundary if the next char is not a word char, or a word boundary if the next char is a word char
Data\[3\] - this is a quoted subItem
(?:(?<=\w)\b|(?<!\w)\B) - either a word boundary if the preceding char is a word char, or a non-word boundary if the preceding char is not a word char.

Matching whole words with special characters with a dynamically built pattern

1 Answers1