Java Regex /在groovy上运行：如何提取所有捕获的块？

Question

我正在尝试使用Java正则表达式从文本文件中提取两个代码块。 但是，我只能提取最后一个块。 有人可以指出mycode有什么问题吗？

谢谢。

这里是

import java.util.regex.*;

INPUT_START_OR_BLANK_LINE = /(?:\A|\n\n)/
FOUR_SPACES_OR_TAB = /(?:[ ]{4}|\t)/
CODE = /.*\n+/
CODE_LINES = /(?:$FOUR_SPACES_OR_TAB$CODE)/
LOOKAHEAD_FOR_NON_CODE_LINE = /(?:(?=^[ ]{0,4}\S)|\Z)/

// this regular expression will find all of the consecutive code lines in a markdown file
// in a markdown file, if the line starts with a tab or at least 4 spaces, it's a code line
// slightly modified from one in markdownj
// see: http://github.com/myabc/markdownj/tree/master/src/java/com/petebevin/markdown/MarkdownProcessor.java
MARKDOWN_CODE_BLOCK = "(?m)" + 
                      "$INPUT_START_OR_BLANK_LINE" +
                      "($CODE_LINES+)" +
                      "$LOOKAHEAD_FOR_NON_CODE_LINE"

def text="""
  Normal paragraph ....


    first Code block begin
       all codes
    first code block end

    how about this line?

   that is not good but what we have are very important
for the purpose of text. yes, we are good.


    second Code block begin
       all codes
    second code block end
    how about this line?

Normal returns
"""

 Pattern p = Pattern.compile(MARKDOWN_CODE_BLOCK);
 Matcher m = p.matcher(text); 

while (m.find() == true){
 m.group().eachLine {println it}
}

该代码来自http://naleid.com/blog/2009/01/01/using-groovy-regular-expressions-to-parse-code-from-a-markdown-file/

Answer 1

Pattern regex = Pattern.compile("((    |\t).*(\r\n|\r|\n))*");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
    System.out.println(regexMatcher.group());
}