问题
I am trying to check whether a string contains a word as a whole, using Java. Below are some examples:
Text : "A quick brown fox"
Words:
"qui" - false
"quick" - true
"quick brown" - true
"ox" - false
"A" - true
Below is my code:
String pattern = "\\b(<word>)\\b";
String s = "ox";
String text = "A quick brown fox".toLowerCase();
System.out.println(Pattern.compile(pattern.replaceAll("<word>", s.toLowerCase())).matcher(text).find());
It works fine with strings like the one I mentioned in the above example. However, I get incorrect results if the input string has characters like %, ( etc, e.g.:
Text : "c14, 50%; something (in) bracket"
Words:
"c14, 50%;" : false
"(in) bracket" : false
It has something to do with my regex pattern (or maybe I am doing the entire pattern matching wrongly). Could anyone suggest me a better approach.
回答1:
It appears you only want to match "words" enclosed with whitespace (or at the start/end of strings).
Use
String pattern = "(?<!\\S)" + Pattern.quote(word) + "(?!\\S)";
The (?<!\S) negative lookbehind will fail all matches that are immediately preceded with a char other than a whitespace and (?!\s) is a negative lookahead that will fail all matches that are immediately followed with a char other than whitespace. Pattern.quote() is necessary to escape special chars that need to be treated as literal chars in the regex pattern.
回答2:
Try escape the special characters with the backslash. They can have other meanings in a pattern.
small correction: Probably you even need two backslash, since the backslash itself is a special character in a String.
来源：https://stackoverflow.com/questions/42904361/java-regex-match-whole-word-with-word-boundary