仅当字符串位于两个分隔符之间时，如何将字符串与正则表达式匹配？

Question

My goal is to delete all matches from an input using a regular expression with Java 7:

input.replaceAll([regex], "");

Given this example input with a target string abc- :

<TAG>test-test-abc-abc-test-abc-test-</TAG>test-abc-test-abc-<TAG>test-abc-test-abc-abc-</TAG>

What regex could I use in the code above to match abc- only when it is between the <TAG> and </TAG> delimiters? Here is the desired matching behaviour, with <--> for a match:

               <--><-->     <-->                                       <-->     <--><-->
<TAG>test-test-abc-abc-test-abc-test-</TAG>test-abc-test-abc-<TAG>test-abc-test-abc-abc-</TAG>

Expected result:

<TAG>test-test-test-test-</TAG>test-abc-test-abc-<TAG>test-test-</TAG>

The left and right delimiters are always different. I am not particularly looking for a recursive solution (nested delimiters).

I think this might be doable with lookaheads and/or lookbehinds but I didn't get anywhere with them.

Answer 1

You can use a regex like

(?s)(\G(?!^)|<TAG>(?=.*?</TAG>))((?:(?!<TAG>|</TAG>).)*?)abc-

See the regex demo . Replace with $1$2 . Details :

• (?s) - a Pattern.DOTALL embedded flag option
• (\G(??^)|<TAG>(.=?*?</TAG>)) - Group 1 ( $1 ): either of the two:
  • \G(?!^) - end of the previous successful match
  • | - or
  • <TAG>(?=.*?</TAG>) - <TAG> that is immediately followed with any zero or more chars, as few as possible, followed with </TAG> (thus, we make sure there is actually the closing, right-hand delimiter further in the string)
• ((?:(?.<TAG>|</TAG>)?)*?) - Group 2 ( $2 ): any one char ( . ), zero or more repetitions, but as few as possible ( *? ) that does not start a <TAG> or </TAG> char sequences (aka tempered greedy token )
• abc- - the pattern to be removed, abc- .

In Java:

String pattern = "(?s)(\\G(?!^)|<TAG>(?=.*?</TAG>))((?:(?!<TAG>|</TAG>).)*?)abc-";
String result = text.replaceAll(pattern, "$1$2");