u011785114 于 2016.04.14 10:13 提问
- 正则表达式截取字符串急急急急
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@:2120100301@+@:2120100302@+@:2120100303@+@:2120100304@截取@:与@之间的数字串，再截取运算符号+
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- eyishion 2016.04.14 10:41
- 已采纳
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")
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- CSDNXIAON 2016.04.14 10:45
准确详细的回答，更有利于被提问者采纳，从而获得C币。复制、灌水、广告等回答会被删除，是时候展现真正的技术了！