问题描述：

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

网友观点：

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\\\")

正则表达式从字符串中提取字符串怎么写?

C#