问题描述

正则表达式截取字符串急急急急

@:2120100301@+@:2120100302@+@:2120100303@+@:2120100304@截取@:与@之间的数字串，再截取运算符号+

解决方案

String regEx = ""(@:)(d{120})(@)"";    String url = ""@:2120100301@+@:2120100302@+@:2120100303@+@:2120100304@"";    Pattern p = Pattern.compile(regEx);    Matcher m = p.matcher(url);    long sum = 0;    while (m.find()) {        String number=m.group(2);        sum+=Integer.parseInt(number);        System.out.println(number);    }    System.out.println(sum);    [快速掌握正则表达式](http://blog.csdn.net/eyishion/article/details/51075984 ""快速掌握正则表达式"")

解决方案二：
急急急
急急急
急急急 谢谢