正则表达式截取字符串急急急急

Question

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

Answer 1

String regEx = "(@:)(\\d{1,20})(@)";
    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")