[英]Regular Expression :match string containing only non repeating words
我有這種情況(Java代碼):1)字符串如:“狂野冒險”應匹配。 2)帶有相鄰重復單詞的字符串:“狂野野外冒險”不應該匹配。
使用此正則表達式:。* \\ b(\\ w +)\\ b \\ s * \\ 1 \\ b。*我可以匹配包含相鄰重復單詞的字符串。
如何扭轉這種情況,即如何匹配不包含相鄰重復單詞的字符串
使用負前瞻斷言, (?!pattern)
?! (?!pattern)
。
String[] tests = {
"A wild adventure", // true
"A wild wild adventure" // false
};
for (String test : tests) {
System.out.println(test.matches("(?!.*\\b(\\w+)\\s\\1\\b).*"));
}
REGEX: (?!.*\b(\w+)\s\1\b).*
NODE EXPLANATION
--------------------------------------------------------------------------------
(?! look ahead to see if there is not:
--------------------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
\b the boundary between a word char (\w)
and something that is not a word char
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
\w+ word characters (a-z, A-Z, 0-9, _) (1
or more times (matching the most
amount possible))
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
\s whitespace (\n, \r, \t, \f, and " ")
--------------------------------------------------------------------------------
\1 what was matched by capture \1
--------------------------------------------------------------------------------
\b the boundary between a word char (\w)
and something that is not a word char
--------------------------------------------------------------------------------
) end of look-ahead
--------------------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
只有當你想要積極匹配的其他模式時,否定斷言才有意義(參見上面的例子)。 否則,你可以使用布爾補碼運算符!
用你之前使用的任何模式否定matches
。
String[] tests = {
"A wild adventure", // true
"A wild wild adventure" // false
};
for (String test : tests) {
System.out.println(!test.matches(".*\\b(\\w+)\\s\\1\\b.*"));
}
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