zoukankan      html  css  js  c++  java
  • 正则--零宽断言



    public static void main(String[] args) {
    String pattern = "(?<=">)(.*?)(?=\</base:Paragraphs)";
    Pattern r = Pattern.compile(pattern);
    Matcher m =r.matcher("<base:Paragraphs id="p0057" num="0057"><base:TableDetails num="0001"><base:Table></base:Table></base:TableDetails></base:Paragraphs>");
    if (m.find()) {
    System.out.println("Found value: " + m.group(0) );
    }
    }

    举例来说(?=exp)代表匹配以exp结尾的字符串,但匹配出来的结果并不带exp,?<=exp)匹配以exp开头的字符串,但结果不带exp

    中间的表达式".*?"

    ".*"(任意字符匹配0次或多次)

    "?"(前面的内容匹配0次或1次)

    参考地址:https://blog.csdn.net/lindiwo/article/details/81012819

  • 相关阅读:
    Java集合的Stack、Queue、Map的遍历
    LinkedHashMap的实现原理
    HashSet的实现原理
    HashMap的实现原理
    leetcode526
    leetcode406
    leetcode413
    leetcode513
    leetcode338
    leetcode419
  • 原文地址:https://www.cnblogs.com/xgyweb/p/15494545.html
Copyright © 2011-2022 走看看