ORO demo
这里主要说明一下关于正则表达式的分组操作，如果要对一个pattern进行分组可以使用分组操作符号"()"来完成，比如：\d\w，因为没有显式的进行分组则\d\w是一组对应的group的index为0，但是如果我们处理其为(\d)(\w)则就执行了一次分组为0,1,2其中group(0)的效果等同于\d\w，group(1)匹配\d,group(2)匹配\w的内容。下面使用一个例子来说明分组的情况：public static void checkMatchGroup(String patternStr, String inputStr) {try {pattern = compiler.compile(patternStr,Perl5Compiler.CASE_INSENSITIVE_MASK);PatternMatcher matcher = new Perl5Matcher();boolean isExist = matcher.contains(inputStr, pattern);if (isExist) {MatchResult res = matcher.getMatch();int length = res.length();for (int i = 0; i < length; i++) {String g = res.group(i);if(g!=null)System.out.println(i + "=:" + g);}}} catch (MalformedPatternException e) {System.out.println("Error occur when checkMatchGroup:"+ e.getMessage());}System.out.println("---------------------");}
测试的代码如下：
/* * for this if we us the charset "?:" the group of (?:\\w+) will be ignored * so the pattern (\\w)(\\d)(?:\\w+) has 3 groups,but we have on right to access * the third part(\\w+), * the result of "j2se333" as fallows: * 0=:j2se333 * 1=:j * 2=:2 * if we didn't add ?: as the prefix to \\w+ the result will be : * 0=:j2se333 * 1=:j * 2=:2 * 3=:se333 */checkMatchGroup("(\\w)(\\d)(?:\\w+)","j2se333");/* Reluctant Qualifiers * for using ? the result of "\d+?" for string "1234" will be get 1 * if we remove the '?' the result will be 1234 */checkMatchGroup("\\d+?","1234");/* **/checkMatchGroup("(?=^255).*","255.0.0.1");checkMatchGroup("(?=^255).*","223.0.0.1");
需要注意的一点是?:符号主要是告诉oro，对于匹配的内容不记忆其状态，也就是说无论是存在与否只要使用了?:我们无法通过group(i)来得到相应的内容
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