Java 正则表达式匹配两个单词
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Regex match two words
提问by user3004412
I'm using Java regex and I need to match two words, but between these words can appear a space or another word, for e.g. "power energy", "power of energy", "power for energy". I have to obtain all variations of "power energy". Is it possible?
我正在使用 Java 正则表达式，我需要匹配两个词，但是在这些词之间可以出现一个空格或另一个词，例如“power energy”、“power of energy”、“power for energy”。我必须获得“动力能量”的所有变化。是否可以？
Further, I need to have something like "pow* ener*" to catch all the double words that start con pow and ener. For e.g "powers for energies".
此外，我需要像“pow* ener*”这样的东西来捕捉所有以 con pow 和 ener 开头的双字。例如“能量的力量”。
回答by Jorge Campos
Try this:
尝试这个：
/(?=power)(?=energy)/i
回答by megawac
回答by Danny
The following
下列
Pattern pattern = Pattern.compile("(\bpow\w+)(.*?)(\bener\w+)");
Matcher matcher = pattern.matcher("Test powers test energies test.");
if(matcher.find()) {
System.out.println(matcher.group());
}
Prints powers test energies. It should work with other variations that you posted. Obviously it will give you false positives if words start with pow, but that is what you seemed to want in the question.
打印powers test energies。它应该适用于您发布的其他变体。显然，如果单词以 开头，它会给你误报pow，但这正是你在问题中想要的。
回答by Casimir et Hippolyte
You can try this pattern:
你可以试试这个模式：
\bpowers?\s+(?:\w+\s+){0,2}energ(?:y|ies)\b
note: if you want you can allows more than two words between "power" and "energy" increasing (or decreasing) the quantifier value inside curly brackets.
注意：如果您愿意，您可以在“power”和“energy”之间允许两个以上的单词增加（或减少）大括号内的量词值。
\\bis a word boundary to avoid having other letters (or digit or underscore) after or before a word.
\\b是一个单词边界，以避免在单词之后或之前有其他字母（或数字或下划线）。
回答by Paolo Falabella
if you need to capture the words "power", "energy" and variations in groups, you can use:
如果您需要在组中捕获单词“power”、“energy”和变体，您可以使用：
(pow\S+\b).*(ener\S+\b)
In java:
在Java中：
Pattern pattern = Pattern.compile("(pow\S+\b).*(ener\S+\b)");
Matcher matcher = pattern.matcher("powers of energies");
if(matcher.find()) {
int count = matcher.groupCount();
for(int i=1;i<=count;i++){
System.out.println(matcher.group(i));
}
